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concept 1
The of numbers.
In this chapter we will explore the early approaches to counting, arithmetic and the understanding of numbers. This study will lead us from the concrete to the abstract almost from the very beginning. We will also see how simple problems about numbers bring us very rapidly to analyzing really big numbers. In section 7 we will look at a modern application of large numbers to cryptography (public key codes). In this chapter we will only be dealing with whole numbers and fractions. In the next chapter we will study geometry and this will lead us to a search for more general types of numbers.
1.1
Representing numbers and basic arithmetic.
Primitive methods of counting involve using a symbol such as | and counting by hooking together as many copies of the symbol as there are objects to be counted. Thus two objects would correspond to ||, three to |||, four to ||||, etc. In prehistory, this was achieved by scratches on a bone (a wolf bone approximately 30,000 years old with 55 deep scratches was excavated in Czechoslovakia in 1937) or possibly piles of stones. Thus if we wish to record how many dogs we have we would, say, mark a bone with lines, one for each dog. That is 5 dogs would correspond to |||||. Notice, that we are counting by assigning to each dog an abstract symbol for one dog. Obviously, the same method could have been used for cats or cows, etc. Thus the mark | has no unit attached. One can say |||||| dogs (dogs being the unit). Notice that you need exactly the same number of symbols as there are objects that you are counting. Although this system seems very simple, it contains the abstraction of unitless symbols for concrete objects. It uses the basic method of set theory to tell if two sets have the same number of elements. That is, if A and B are sets (collections of objects called elements) then we say that they have the same number of elements (or the same cardinality) if there is a way of assigning to each element of the set A a unique element of the set B and every element of the set B is covered by this assignment. Primitive counting is done by using sets whose elements are copies of | to be numbers. Although each of the symbols | is indistinguishable from any other they must be considered dierent. This primitive method of counting and attaching symbols to numbers basically involves identifying sets with the same cardinality with one special set with that cardinality. In modern mathematics, one adds one level of abstraction and says that the set of all sets with the same cardinality constitutes one cardinal number. There is no limit to the size of a set in this formalism. We will come back to this point later. Early methods of representing numbers more concisely than what we have called the primitive system are similar to Roman numerals which are still used today for decorative purposes. In this system, one, two, three are represented by I, II, III. For ve there is a new symbol V (no doubt representing one hand) and four is IV (to be considered one before V and probably representing a hand with the thumb covering the palm). Six, seven and eight are given as VI, VII, 1
Figure 1: VIII. Then there is a separate symbol for ten, X (two hands) and nine is IX. This pattern continues, so XII is twelve, XV is f ifteen, XIV is fourteen, XIX is nineteen. Twenty and thirty are XX, XXX. Fifty is L. Forty is XL. One hundred is C, f ive hundred is D and a thousand is M. Thus 1998 is MCMXCVIII. This system is adequate for counting (although cumbersome). It is, however, terrible for arithmetic. Here we note that one has a dramatic improvement in the number of symbols necessary to describe the number of elements in a set. Thus one symbol M corresponds to the cardinal with 1000 of the symbols | in it in the most primitive system. The ancient Egyptians (beginning about 3500 BC) used a similar system except that they had no intermediate symbols for f ive, f ifty or f ive hundred. But they had symbols for large numbers such as ten thousand, one hundred thousand, one million and ten million. The below is taken from the Rhind Papyrus (about 1600 BC). Our number system derives from the Arabic positional system which had its precursor in the Babylonian system (beginning about 3000 BC). Before we describe the Babylonian system it is useful to recall our method of writing numbers. We use symbols 1,2,3,4,5,6,7,8,9 for one element, two elements,...,nine elements. we then write 10 for ten elements, 11 for eleven, ..., 19 for nineteen. This means that we count by ones, then by tens, then by hundreds, then by thousands, etc. This way we can write an arbitrarily large number using ten symbols (we also need 0 which will be discussed later). Our system has base ten. That, is we count to nine then the next is ten which is one ten, 10, then we count by ones from 11 to 19 and the next number is two tens, 20. When we get to 9 tens and 9 ones (99) the next number is 10 tens which we write as 100 (hundred). 10 hundreds is then 1000 etc. Thus by hooking together 10 symbols
2
we can describe all numbers. One could do the same thing using a base of any positive integer. For example, if we worked with base 2 then we would count 1, then 10 for two, then 11 for three (one two and one one), then 100 (2 twos), 101, 110, 111, 1000 (two (two twos)). Thus we would only need 2 symbols in juxtaposition to describe all numbers. For example, 1024 would need 1024 of the units, | ,in the most primitive system, it is 4 symbols long in ours, and base 2 it is 10000000000. Still a savings of 1013 symbols. The Roman method would be MXXIV so in this case slightly worse than ours. However, if we try 3333 in Roman notation we have MMMCCCXXXIII. How long is the expression for 3333 in base 2? The Babylonians used base 60 which is called sexagesimal. We should note that for some measurements we still use this system: 60 seconds is a minute, 60 minutes is an hour. Their system is preserved in clay tablets in various excavations. Their method of writing (cuneiform) involved making indentations in soft clay tablets by a wedge shaped stylus.
They used two basic symbols, one equivalent with | for one. and one for 10 which we will represent as . Thus six is ||||||. Normally written in the form: ||| ||| and thirty seven is ||| ||| . | But 61 is | |. 3661 is | | |. Thus, except that they used only symbols for 1 and 10 and had to juxtapose them to get to 59, they used a system very similar to ours. They did not have a symbol for 0. We will see that this is a concept that would have to wait more than 3000 years. So when they saw |, they would have to deduce from the context whether it represented 1, 60, 3600, etc. For example if I said that a car cost ||| then you would be pretty sure (in 2003) that I meant 10,800, not 180 or 3. They later (200 BC) had a symbol that they could use for a place marker in all but the last digit (but still no 0). // Thus they could write | // | and mean 3601. There is still an ambiguity in the symbol | which can still mean 1, 61, 3601, etc.
3
Exercises. 1. Write out the number 1335 in Egyptian notation, binary, sexagesimal and in Roman numerals. 2. For computers one kilobit (1K) is actually 1024. Why is that? 3. The early computer programmers used base 16 they therefore needed 16 symbols which they wrote as 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F. For example,AF = 10 16 + 15 = 175. What number is F F F F ? Write it in binary. Why was it important to 16 bit computers? F F F F F + 1 is called a megabit. Why is that? 4. In writing numbers in the Egyptian system what is the maximum repetition necessary for each symbol?
1.2
1.2.1
Arithmetic.
Addition.
We return to the most primitive method of counting. If you have ||| sheep and you have purchased |||| sheep, then you have ||||||| sheep. That is, to add ||| and |||| we need only hook |||| onto |||. For cardinal numbers we have thus described a method of addition: If A corresponds (i.e. is an element of) to the cardinal a and if B corresponds to the cardinal b, and if no element of A is an element of B then a + b is the cardinal number that contains A B (with A B the set that consists of the elements of A combined with those of B).This can be made rigorous (independent of the choice of A and B) we will look into this point later in the book. Thus the abstraction of primitive addition is set theoretic union of disjoint (no element in common) sets. In the Roman system there is one more degree of abstraction since for example |||| is represented as IV and ||||| is represented as V so IV + V = ||||||||| = IX. Obviously, one must remember much more if one uses the more abstract method of the Romans than the direct primitive method. In our system for the same addition we are looking at 4 + 5 and we must remember that this is 9. Thus the situation is analogous to that of the Romans. However, if we wish to add XXXV to XVI, then in Roman numerals we have LI. In our system we have 35 + 16. We add 5 + 6 and get 11 (memorization). We now know that the number has a 1 in the ones position we carry the other 1 to see that for the tens position. We have 1 + 3 + 1 = 5. The sum is therefore 51. Thus we need only remember how to add pairs of numbers up to 9 in our system and all other additions are done following a prescribed method. The Roman system clearly involves much more memorization. We next look at the Babylonian system. For this we will use a method of expressing numbers to base 60 that is due to O. Neugebauer(a leader in the history of mathematics). We write 23,14,20 for 20 plus 14 sixties plus 23 3600. Thus in the Babylonian base 60 system we must memorize all additions of numbers up to 59. If we wish to add 21,13 and 39,48 then we add 48 + 13 and get 1,1 (this is memorized or in an addition table) 21+39 and get 1 (remembering the context). Thus the full sum is 1,1,1. Here we must remember a very large addition table. However, we have grown up thinking in terms of base 10 and we
4
do the additions of pairs of numbers below 59 in our method and then transcribe them to our version of the Babylonian notation. Exercises. 1. Do the addition 1, 2 + 32, 21, 3 + 43, 38, 1 in Neugebauers notation. 2. How do you think that an Egyptian would add together 3076 and 9854? 1.2.2 Multiplication.
Multiplication is a more sophisticated operation than addition. There isnt any way to know when and how the notion arose. However, the Egyptians and the Babylonians knew how to multiply (however as we shall see the Egyptian method is not exactly what one would guess). We understand multiplication as repeated addition. That is, if we wish to multiply a times b, a b, then we add b to itself a times. That is 3 5 is 5 + 5 + 5 = 15. If we attempt to multiply a times b in the primitive system we must actually go through the full juxtaposition of b with itself a times (or vice-versa). In a system such as the Roman system we must memorize a great deal. For example XVLI = DCCLXV. For us the multiplication is done using a system: 51 15 255 . 510 765 We usually leave out the 0 in the 510 and just shift 51 into the position it would have if there were a 0. We see that we must memorize multiplication of pairs of numbers up to 9. The Babylonian system is essentially the same. However, one must memorize multiplication of pairs of numbers up to 59. This is clearly a great deal to remember and there are tablets that have been excavated giving this multiplication table. The Egyptian system is different. They used the method of duplication. For example if we wish to multiply 51 by 15 then one would proceed as follows: 51 51 + 51 = 102 102 + 102 = 204 204 + 204 = 408 1 2 4 8
Now 1 + 2 + 4 + 8 = 15 so the product is 51 + 102 + 204 + 408 = 765. Notice that they are actually expanding 15 in base 2 as 1111. If the problem had been multiply 51 by 11 then the answer would be 51 + 102 + 408 = 561 (in base 2, 11 is 1011). So their multiplication system is a combination of doubling and addition. We note that this method is used in most computers. Since, in base 2, multiplication by 2 is just putting a 0 at the end of the number. In base 2, 5
51 = 110011. Thus the same operations are 110011 1100110 11001100 110011000 1 10 100 1000
The basic dierence is that we must remember many carries in binary. Thus it is better to proceed as follows. 110011 1100110 10011001 11001100 110011000 1001100100 10011001 1001100100 1011111101 Addition is actually an operation that involves adding pairs of numbers. In our system we rarely have to carry numbers to more than one column to the left (that is when a column adds up to more than 99). In binary it is easy if we add 4 numbers with 1 in the same digit we will have a double carry. Exercise. 1. Multiply 235 by 45 using the Egyptian method. Also do it in binary. 1.2.3 Subtraction.
If we wish to subtract ||| from ||||| then the obvious thing to do is to remove the lines one by one from each of these primitive numbers. ||| ||||| || |||| | ||| ||. With this notion we have ||||| ||| = ||. If we do this procedure to subtract a from b and run out of tokens in b then we will say that the subtraction is not possible. This is because we have no notion of negative numbers. We will see that the concept of 0 and negative numbers came relatively late in the history of mathematics. In any event, we will write a < b if subtraction of a from b is possible. If a < b then we say that a is strictly less than b. In our notation subtraction is an inverse process to addition. This is because our number notation has a higher degree of abstraction than the primitive one. Thus we memorize such subtractions as 3 2 = 1. If we are calculating 23 12 then we subtract 2 from 3 and 1 from 2 to get 11. For 23 15 we do in initial borrow and calculate 135 and 11. So the answer is 8. Obviously, we can only subtract a smaller number from a larger one if we expect to get a number in the sense we have been studying. Both the Egyptians and the Babylonians used a similar system. For the Egyptians it would be somewhat more complicated, since every new power of 10 entailed a new symbol. 6
1.2.4
Division and fractions.
For us division is the inverse operation to multiplication in much the same way as subtraction is the inverse operation to addition. Thus a is the number such b that if we multiply it by b we have a. Notice that if b is 0 this is meaningless and that even if a and b are positive integers then a is not always an integer. b Integer division can be implemented as repeated subtraction thus in the primitive notation ||||||||| - ||| = ||||||, |||||| - ||| = ||| thus |||||||||/||| = |||. However, the Egyptians and Babylonians understood how to handle divisions that do not yield integers. 1 The ancient Egyptians created symbols for the fractions n (i.e. reciprocals). 2 They also had a symbol for 3 . However, if they wished to express, say, 7 5 1 then they would write it as a sum of reciprocals say 1 + 1 + 15 . Also they 3 limited their expressions to distinct reciprocals (or 2 ). Thus 1 + 1 + 1 was not 3 5 5 a valid expression. Note that such an expression is not unique. For example, 7 1 1 1 5 = 1 + 4 + 10 + 20 . Notice that one allows any number of reciprocals in the expression. With a method such as this for handling fractions, there was a necessity for tables of fractions. One also had to be quite ingenious to handle fractions. An ancient Egyptian problem asks: If we have seven loaves of bread to distribute among 10 soldiers, how would we do it?
7 We would instantly say that each soldier should get 10 of a loaf. However, this makes no sense to the ancient Egyptians. Their answer was (answers were 1 supplied with the problems) 2 + 30 . 3
The mathematician Leonardo of Pisa (Fibonacci 1175-1250 A.D.) devised an ingenious method of expressing fractions in the Egyptian form. In order to see that the method works in general several basis properties of numbers will be used here. They will be considered in more detail later He starts by observing b that we need only consider a with 1 < a < b. We rst observe that a > 1 so b there exists a positive whole number n such that n1< So b < n. a
1 Thus a = n + anb . We observe that anb > 0 and a(anb) = ba(n1) = b nb b a( a n 1) > 0. Thus a > (an b) > 0. Set a1 = an b, b1 = bn. Then 0 < a1 < a and 1 a1 < b1 . If a1 = 1 then we are done. Otherwise, we repeat 1 the process with a1 . Call n, n1 . If a1 = 1 then we see that b1 = n1 b > n1 so b a 1 1 b = n1 + n1 b is a desired expression. Assume that a1 > 1. Do the same for a1 b1 . This is Fibonaccis method. A full proof that this always works didnt get published until the nineteenth century and is attributed to J.J.Sylvester. What
a 1 1 < < n b n1
7
has not been shown is that if 1 a1 1 < < n b1 n1 then n > n1 . We do this by showing that that an1 b < a < b. Thus
a1 b1
<
1 n1 .
To see this we observe
a1 an1 b a b 1 = . < < < b1 n1 b n1 b n1 b n1 In this argument we used an assertion about not necessarily whole numbers that says that if we have a number then it lies between two consecutive integers.
7 7 Consider, for example, 10 then n1 = 2 and 10 1 = 1 . Thus we get as an 2 5 1 1 answer to the Egyptian problem 2 + 5 which seems preferable to the answer given in the original Papyrus. Fibonacci was one of the leading European mathematicians of the Middle Ages. He was instrumental in introducing the Arabic number system (the one we use) to the West. However, he preferred the Egyptian method of fractions to our decimal notation (below). Clearly one must be much cleverer to deal with Egyptian fractions than with decimals. Also, as we will see, strange and impractical problems have propelled mathematics to major new theories (some of which are even practical).
The ancient Babylonians used a method that was analogous to our decimal notation. In our decimal method we would express a fraction such as 1 as 8 follows: We rst try to divide 8 into 1 this fails so we multiply by 10. We can divide 8 into 10 once with remainder 2. We must multiply by 10 and divide 8 ani 20 getting 2 with remainder 4. We now multiply 4 by 10 and get 40. Divide by 8 and get 5. The numbers for the three divisions are 1, 2, 5. We write 1 = .125. 8 We can express this as follows:
We will use Neugebauer notation in our description of their method. The fraction 7 = 1+ 24 . Our version of their notation would be 1; 24. In our decimal 5 60 notation this is 1.4. We could use exactly the same process (though it is harder 8
for us to do the intermediate steps in our heads). We must divide 5 into 2. So we multiply by 60 and do the division. That is divide 120 by 5. We get 24 and no remainder. If we were to write 1 we could work as follows:1 60 divided 8 by 8 is 7 with remainder 4. 4 60 = 240 which divided by 8 is 30 with no remainder. Thus we have ; 7, 30. There were also bad fractions. In our decimal notation 1 = 0.33333... 3 That is we must write the symbol 3 forever. In the Babylonian form f irst bad fraction and it is given by ; 8, 34, 17, 8, 34, 17, ... repeating 8,34,17 forever. Suddenly the Egyptian way doesnt seem to be so silly! We also think of fractions as expressions p with p and q positive integers. q p = r means ps = rq. Addition is given by p + r = ps+qr . Multiplication is q s q s qs given by p r = pr . We note that 1 = 2 = 3 = ... That is we identify all of q s qs 2 4 6 n the symbols 2n with 1 . This is similar to our def inition of cardinal number. 2 Usually, to rid ourselves of this ambiguity, we insist that p and q are in lowest terms. That is they have no common factor other than 1 1.2.5 Exercises.
2 3 1 7
is the
1. Why do you think that the Egyptians preferred
+
1 30
to
1 2
+
1 5
for
7 10 ?
4 2. Use the Fibonacci method to write 17 as an Egyptian fraction. n 3.Make a table in Egyptian fractions of 10 for n equal to 1, 2, 3, 4, 5, 6, 7, 8, 9 1 4. Among { 1 , 1 , 1 , ..., 19 } which are the bad fractions in base 10? What do 2 3 4 1 they have in common. Can you guess a property of n that guarantees that n is a good fraction to base 10? How about base 60?
5. The modern fame of Fibonacci is the outgrowth of a problem that he proposed: Suppose that it takes a rabbit exactly one month from birth before it is sexually mature and that a sexually mature pair (male and female) of rabbits will give birth to two rabbits (male and a female) every month. If we start with newly born male and female rabbits how many rabbits will there be at the end of one year? What is the answer to his question? 6. If b is a positive integer then we can represent any integer to base b in the form a0 + a1 b + a2 b2 + ... + ak bk with 0 ai < b.. This if b = 10 then 231 means 1 + 3 10 + 2 102 . If b = 60 then 231 means 1 + 3 60 + 2 602 . Show that if n < b then the square of 111...1 (n ones) is 123...n...321 that is the digits increase to n then decrease to 1. What happens if n > b?
9
Three examples of early Algebra. At this point we have looked at counting methods and developed the basic operations of arithmetic. We have studied one simple Egyptian exercise in arithmetic and given a method of Fibonacci to express a fraction as an Egyptian fraction. The start with a practical problem or applied mathematics. Whereas, by the time Fibonacci devised his method, there was no reason to use Egyptian fractions. It is what we now call pure mathematics. The method is clever and has an underlying simplicity that is much more pleasant than using trial and error. Obviously, ancient peoples had many uses for their arithmetic involving counting, commerce, taxation, measurement, construction, etc. But even in the early cultures there were mathematical puzzles and techniques developed that seem to have no practical use. An Egyptian style problem: A quantity added to two thirds of it is 10. What is the quantity? We would say set the quantity equal to x (we will see that this small but critical step would not be discovered for thousands of years). Then we have 2 x + x = 10. 3 Hence 5 x = 10. 3 So x = 6. Since the Egyptians had no notion of how to deal with unknown quantities, they would do something like. If the quantity were 3 then the sum of the quantity plus two thirds of the quantity is 5. Since the sum we desire is 10, the answer is 2 times 3 or 6. In other words, they would use a convenient value for the quantity and see what the rule gave for that value. Then re-scale to get the answer. We will now discuss a Babylonian style problem (this involves basic geometry which we will assume now and discuss in context later). Before we write it out we should point out that multiplication as repeated addition was probably not an important motivation for doing multiplication. More likely they multiplied two numbers because the outcome is the area of the rectangle whose sides were the indicated number of whatever units they were using. I add the area of a square to two thirds of its side and I have ;35. What is the side of the square? Solution: Take 1 multiply by 2 take half of this and we have ; 20. You multiply this 3 by ; 20 and the result is ; 6, 40. You add to this ;35 to have ; 41, 40. This is the area of the square of side ; 50. You subtract ; 20 from ; 50 and you have ; 30 the side of the square.
10
In modern notation if we set the side equal to x then we are solving 2 7 = 0. x2 + x 3 12 The quadratic formula tells us that if we are solving x2 + ax b = 0
In our notation what we have done is taken 2 . Next divided by 2 to get 1 . 3 3 7 The square of 1 is 1 now add 35 = 12 to get 25 . The square root of this is 5 . 3 9 60 36 6 7 7 Subtract 12 and we have 1 . Thus if a = 2 , b = 12 then the answer is 2 3 r a a 2 +b . 2 2
a2 + 4b x= . 2 If a > 0,b > 0 then the positive solution is exactly the Babylonian answer. Their method of solving such problems put a premium on the ability to calculate ex pressions of the form a2 + b. They had an approximate method of doing such calculations which corresponds to what we will see is the second iteration of a method of Newton method applied to this simple case. They use the approxib b mation a + 2a . Notice that if 2a is small then this is a good approximation. Thus the Babylonians were aware of general methods to solve quadratic equations. They, however, could only express their method in words. What they wrote out is except for the order (and the absorption of the 1 ) exactly 2 what we would write. It is hard to imagine how either of these methods could be used in practical applications. However, one of the most interesting exercises in pure mathematics can be found in a tablet in the Yale collection (Plimpton 322). This tablet is a tabulation of 15 triples of numbers a, b, c with the property that a a2 + b2 = c2 . The simplest example that we know of this is 32 + 42 = 52 . This triple appears on the tablet as number 11 and in the form 602 + 452 = 752 . The tablet is thus using some strange rule for generating these numbers (usually called Pythagorean triples). We will discuss the Pythagorean theorem later. Here we will study the tablet as a collection of relationships between numbers. The table is arranged as follows: there are 15 existent rows and 4 readable columns. The f irst column contains a fraction and the fractions are decreasing. 11
then
The second and third contain integers and the last is just the numbers 1,...,15 in order. If we label an element of the second column a and the element of the third column in the same row c then c2 a2 = b2 with b a positive integer and 2 the element of the f irst column in the same row is c2 . Also the f irst column b contains only regular sexagesimal rational numbers. It seems clear that the Babylonians were aware of a method of generating Pythagorean triples.
y
5
4
3
2
-5
-2.5
1
0
2.5 x
5
In our modern notation we know how to generate all Pythagorean triples a, b, c (a2 + b2 = c2 ) with a,b,c having no common factor. Indeed, consider y = c , x = a then y 2 x2 = 1. We are thus looking for rational points on a b b 1 hyperbola (see the figure above). Notice that x2 gives an element of the f irst column of the table. Thus they seem to have picked points rational points on the hyperbola in increasing order. How do you locate such a point? We note that y 2 x2 = (y x)(y + x) (we will discuss what this might n have meant to the Babylonians soon). We write y + x = m , y x = m then n 1 m n 1 m n m2 +n2 m2 n2 y = 2 n + m and x = 2 n m . Thus y = 2mn and x = 2mn . This suggests that we take a = m2 n2 , b = 2mn and c = m2 + n2 . If m and n are positive integers then you can check easily that this assignment generates a Pythagorean triple. There is an algebraic proof of Fibonacci that this method generates all Pythagorean triples that have no common factor. Andr Weil (1906-1999) has pointed out that there is a geometric argument in Euclid Book X, Lemma 1,2 in preparation for Proposition 29 that proves that this method gives all such triples that are relatively prime (in fact a bit more than this). We will come back to this later. Consider the Pythagorian triple 3, 4, 5. We will nd numbers m, n as above. The method above says take y = 5 and x = 3 . Then y + x = 2 and y x = 1 . 4 4 2 This suggests take m = 2 and n = 1. We can check that this works m2 1 = 3, 2mn = 4 and m2 + n2 = 5. We will now discuss a probable meaning for the formula y 2 x2 = (y x)(y + x). The formula y 2 x2 is geometrically the area of the gure gotten by removing a square of side x from one of side y. If you take the smaller square
12
out of the lower right corner then in the lower left corner one has a rectangle of side x and base y x. If we cut this rectangle o and rotate it 90o then we can attach it to what is left of the big square and get a rectangle of side y x and base y + x.
The two problems above are similar to the word problems of high school algebra and were probably used in the same way as we use them now. That is, to hone the skills of a student learning basic algebra. Plimpton 322 is another matter. It contains number theoretic relationships at a sophisticated level. Imagine a line of reasoning similar to the one in the previous paragraph without any algebraic notation and without even the notion of a fraction. 1.2.6 Exercises.
1. Problem 26 on the Rhind papyrus is: A quantity whose fourth part is added to it becomes 15. Use the Egyptian method to solve the problem. 2. Use the Babylonian approximation to calculate 2. (Suggestion: Start with a = 4 so that b = 2 . Can you improve on this?) 3 9 3. A problem on a Babylonian tablet says: I have added 7 times the side of my square to 11 times the area and have 6; 15. Find the side. Use the Babylonian method to solve this problem. 4. Find m, n so that a = m2 n2 , b = 2mn and c = m2 + n2 for the Pythagorean triples 119, 120, 169 and 5, 12, 13.
1.3
Some number theory taken from Euclid.
We now jump about 1500 years to about 300BC and the time of the school of Euclid in Alexandria. We will examine parts of Books VII,VIII,IX of his Elements that deal with numbers. We will have more to say about the other books at appropriate places in this work. We will use the translation of Sir Thomas Heath for our discussion. 13
1.3.1
Def initions
Euclid begins Book VII with 22 def initions that set up basic rules for what we have been calling the primitive number system. We will see in the next chapter that Euclid did not think of numbers in this sense. He rather thought of numbers as intervals. If we have two intervals AB and CD and if we lay out AB a certain number of times an this covers CD exactly then AB is said to measure CD. 1. An unit is that by virtue of which each of the things that exist is called one. This doesnt make too much sense but it is basically establishing that there is a unit for measurement.. We have been denoting this by |. 2. A number is a multitude composed of units. Thus ||| is a number as before. However, Euclid thinks of it as an interval that is exactly covered by three unit intervals.Be warned that the unit is not considered to be a number. 3. A number is a part of a number, the less of the greater, when it measures the greater; Thus the greater, ||||||, is measured by the less |||. 4. but parts when it does not measure it. ||||| is not measured by |||. 5. The greater number is a multiple of the less when it is measured by the less. Notice that the def initions are beginning to be more accessible. Here we measure |||||| by two of the |||. This thus |||||| is ||| multiplied by ||. 6. An even number is that which is divisible into two equal parts. 7. An odd number is that which is not divisible into two equal parts, or that which differs by a unit from an even number. 8.,9.,10. talk about multiplication of odd and even numbers. (e.g. an odd by an even is an even). 11. A prime number is that which is measured by a unit alone. Thus |||||| is measured by |, ||, ||| so is not prime. ||||| is only measured by |. 12. Numbers prime to one another are those which are measured by an unit as a common measure. |||| is measured by |, || ||||||||| is measured by |, ||| thus the only common measure is |. Thus |||| and ||||||||| are prime to one another. 13., 14. are about numbers that are not prime (to each other). A number that is not prime is composite. 14
In 15. he describes multiplication as we did (repeated addition). 16. And when two numbers having multiplied one another make some number, the number so produced is called plane, and its sides are the numbers which have been multiplied. Here Euclid seems to want to think of the operation of multiplication in geometric terms: an area. In 17 the product of three numbers is looked upon as a solid. 18,.19. def ine a square and a cube as we do. We will study these concepts in the next chapter. 20. Numbers are proportional when the f irst is the same multiple, or the same part, or the same parts, of the second that the third is to the fourth. ||| |||||| and |||| |||||||| are proportional. This is a relationship between two pairs of numbers. It is essentially our way of looking at rational numbers. In 21. there is a discussion of similar plane and solid numbers. 22. A perfect number is one which is equal to its parts. The parts of |||||| are |, ||, ||| and | + || + ||| = ||||||. So it is perfect. |||| is not. To us this is not a very basic concept. Perfect numbers are intriguing (28 is one,what is the next one?) but it is hard to see any practical reason for their study. We shall see that Euclid gave a method for generating perfect numbers.
1.3.2
Some Propositions
Having disposed of the def initions, Books VII,VIII,IX consist of a series of Propositions. Number one is: B_____________F __A D________G__C __E Two unequal numbers being set out, and the less being continually subtracted in turn from the greater, if the number which is left never measures the one before it until an unit is left, the original numbers will be prime to one another. Let us try this out. Take 27 for the larger and 8 for the smaller. Subtract 8 from 27 and get 19, subtract 8 and get 11, subtract 8 and get 3, subtract 3 from 8 and get 5 subtract 3 from 5 and get 2 subtract 2 from 3 and get 1. Thus the numbers are relatively prime. We will now describe the Euclidian proof. The numbers are denoted AB and CD and Euclid draws them as vertical intervals. He assume on the contrary that AB and CD are not prime to each other. Then there would be a number E that measures both of them. We now come to the crux of the matter: Let 15
CD measuring BF leaving F A less than itself. (Here it is understood that BF + F A = BA and that BF is evenly divisible by CD) This assertion is now called the Euclidean algorithm. It says that if m, n are whole numbers with m < n then we can write n = dm or n = dm + q with q a whole number strictly less than m. For some reason he feels that this assertion needs no proof. To Euclid this is evident. If n is measured by m it is obvious. If it is not then subtract m from n and get q1 if q1 < m we are done. q1 cannot be measured by m hence q1 6= m and so q1 > m. We now subtract m from q1 and get q2 . If q2 < m then we are done otherwise as before q2 > m. Subtract m again. This process must eventually lead to a subtractend less than m since if not then after n steps we would be able to subtract nm from n so mn < n. But this is impossible since m > 1 so mn = n + n + ... + n (m times). Hence we are asserting n > mn n + n. Since it is obvious that n + n > n we see that the process must give the desired conclusion after less than n steps. We now continue the proof. Let AF measuring DG leaving GC less than itself. E measures CD hence BF and E measures AB so E measures F A. Similarly, E measures GC. Since the procedure described in the proposition now applies to AF and GC, we eventually see that E will eventually measure a unit. Since E has been assumed to be a number (that is made up of more that one unit) we see that this is impossible. In Euclid this unbounded procedure (f inite for each example) is only done three times. Throughout his arguments he does the case of three steps to represent the outcome of many steps. The second proposition is an algorithm for calculating the greatest common divisor (greatest common measure in to Euclid). Given two numbers not prime to one another, to f ind the greatest common measure. Given AB and CD not prime to one another then and CD the smaller then if CD measures AB then it is clear that CD is the greatest common measure. If not consider AB CD, CD. There are now two possibilities. The f irst is that AB CD is smaller than CD. In this case if AB CD measures CD then it must measure AB and so is the greatest common measure. In the second case CD is the smaller and if CD measures AB CD then it must measure AB and so it is the greatest common measure. If not the previous proposition implies that if we continually subtract the smaller from the larger then we will eventually come to the situation when the smaller measures the larger. We thus have the following procedure: we continually subtract the smaller from the larger stopping when the smaller measures the larger. Proposition 1 implies that the procedure has the desired end. Here is an example of proposition 2. Consider 51 and 21. Then 51 21 = 30 (30,21) , 30 21 = 9(21,9), 21 9 = 12 (12,9), 12 9 = 3 (9,3) so the greatest common divisor is 3. Why is it so important to understand the greatest common divisor? One important reason is that it is the basis of understanding fractions or rational numbers. Suppose that we are looking at the fraction 21 . Then we have seen 51 16
that the greatest common divisor of 21 and 51 is 3. Dividing both 21 and 51 by 7 3 we see that the fraction is the same as 17 . This expression is in lowest terms 7 21 42 and is unique. 17 = 51 = 102 = ... We will emphasize his discussion of divisibility and skip to Proposition 31. Any composite number is divisible by some prime number. We will directly quote Euclid. Let A be a composite number; I say that A is measured by some prime number. For since A is composite, some number will measure it. Let a number measure it, and let it be B. Now, if B is prime then we are done. If it is composite then some number will measure it. Let a number measure it and call it C. Since C measures B and B measures A, C measures A. If C is prime then we are done. But if it is composite then some number will measure it. Thus, if the investigation is continued in this way, some prime number will be found which will measure the number before it, which will also measure A. For if it were not found an inf inite series of numbers will measure the number, A, which is impossible in numbers. Notice that numbers are treated more abstractly as single symbols A, B, C and not as intervals. (Although they are still pictured as intervals.) More important is the inf inite series of divisors of A. No real indication is given about why this is impossible for numbers. However, we can understand that Euclid considered this point obvious. If D is a divisor of A and not equal to A then D is less than A. There are only a f inite number of numbers less than a given number n, 1, 2, ..., n 1. This argument uses a version of what is now called mathematical induction which we will call the method of descent. Suppose we have statements Pn labeled by 1, 2, 3, .... If whenever Pn is assumed false we can show that there is an m with 1 < m < n with Pm false then Pn is always true. The proof that this method works is that if the assertion for some n were false then there would be 1 < m1 < n for which Pm1 is false. But then there would be 1 < m2 < m1 for which Pm2 is false and this procedure would go on forever. Getting numbers m1 > m2 > ... > mn > ... with all the numbers bigger than 1. Let us try in out. The assertion Pn is that if n is not a unit then n is divisible by some prime. If Pn is false that n is not a prime and not a unit. Hence it is composite so it is a product of two numbers a, b neither of which is a unit and both less than n. If Pa were true then a would be divisible by some prime. But that would imply that n is divisible by some prime. This is contrary to our assumption. Thus if Pn is false then Pa is false with 1 < a < n. The method of descent now implies that Pn is true for all n. We will now jump to Book IX and Proposition 20. Prime numbers are more than any assigned multitude of prime numbers.
17
Here we will paraphrase the argument. Start with distinct primes A,B,C. Let D be the least common multiple of A, B, C (this has been discussed in Propositions 18 and 19 of Book IX). In modern language we would multiply them together. Now consider D + 1. If D + 1 were composite then there would be a prime E dividing it. If E were one of A,B,C then E would divide 1. Notice that we are back with three taking the place of arbitrarily large. The modern interpretation of this Proposition is that there are an inf inite number of primes. What is really meant is that if the only primes are p1 ,...,pk then we have a contradiction since p1 pk + 1 is not divisible by any of the primes and this contradicts the previous proposition. After Book IX, Proposition 20 there are Propositions 21-27 that deal with combining even and odd numbers and seem to be preparatory to Euclids method of generating perfect numbers. For example, Proposition 27 (in modern language) says that if you subtract an even number from an odd number then the result is an odd number. Here one must be careful and also prove that if you subtract an odd number from an even number you get an odd number (Proposition 25). We would say that the two statements are essentially the same since one follows from the other by multiplication by 1. However, since negative numbers were not in use in the time of Euclid Proposition 25 and 27 are independent. We now record one implication of Proposition 31 (and Proposition 30 which is discussed below) that is not explicit in Euclid (we will see why in the course of our argument). This Theorem is usually called the fundamental theorem of arithmetic. If A is a number (hence is not a unit) then A can be written uniquely (up to order) in the form pe1 pe2 per with p1 , ..., pr distinct primes and e1 , ..., er r 1 2 numbers (here B m is B multiplied by itself m times). We rst show that any number is a product of primes using a technique analogous to the method of Euclid in his proof of Proposition 31. If A is prime then we are done. Otherwise A is composite hence by Proposition 31 A = q1 A1 with A1 not a unit and q1 a prime. If A1 is a prime then we are done. Otherwise, A1 = q2 A2 with q2 a prime and A2 not a unit. If A2 is prime we are done since then A = q1 q2 A2 . Otherwise we continue this procedure and either we are done in an a nite number of steps or we have A1 > A2 > ... > An > ... an innite sequence of positive numbers. This is impossible for numbers according to Euclid. We have mentioned in our discussion of Proposition Let us show how the principle of descent can be used to prove the assertion that every number is a product of primes. Let Pn be the assertion that if n is not the unit then is a product of primes. If Pn is false then n is not a unit and not prime so n is composite. Hence n = ab with neither a nor b a unit. If both Pa and Pb were true then a is a product of primes and b is a product of primes so ab is a product of primes. Thus one of Pa or Pb would be false. If
18
Pa is false set m = a otherwise Pb is false and set m = b. Then 1 < m < n and Pm is false. The principle implies that Pn is true for all n. This principle can be made into a direct statement which we call the principle of mathematical induction.. The idea is as follows if S1 , S2 , ... are assertions and if S1 is true and if the truth of Sm for all 1 < m < n implies that Sn is true then Sn is true for all n. This is intuitively clear since starting with S1 which we have shown is true we have. S1 is true so S1 and S2 are true so S3 is true, etc. For example suppose that the statement Sn is the assertion 1 + 2 + ... + n = n(n + 1) . 2
Then S1 says that 1 = 1 which is true. We now assume that Sm is true for all 1 < m < n. Then 1 + 2 + ... + (n 1) + n = (1 + 2 + ... + (n 1)) + n = n(n + 1) n(n 1) 2n + = 2 2 2 which is the assertion Sn . Let us see how the method of descent implies the principle of mathematical induction. Suppose we have a statement Sn for n = 1, 2, ... and suppose that we know that S1 is true and whenever we assume Sm is true for 1 < m < n the Sn is true. Assume that Sn is false. Then n cannot be the unit. If Sm were true for all 1 < m < n then we would know that Sn were true. Since we are assuming the contrary we must have Sm is false for some m with 1 < m < n. Thus the method of descent implies that Sn is always true. One can show that the two principles are equivalent but we have traversed to far away from Euclid already. Returning to the fundamental theorem of arithmetic we have shown that if A is not the unit then A can be written as q1 q2 qn with qi a prime for i = 1, ..., n. Since the qi are not necessarily distinct we can take p1 , ..., pr to be the distinct ones and group those together to get A = pe1 pe2 per (here e1 is the number r 1 2 of i such that p1 = qi , e2 is the number of i such that p2 = qi ,...). We are now ready to prove the uniqueness. The crux of the matter and is Proposition 30 of Book VII which says: If two numbers by multiplying one another make some number, and any prime number measure the product, it will measure one of the original numbers. Let us see how this proposition implies our assertion about uniqueness. We will prove it using the principle of mathematical induction. The assertion Pn is that if n is not one then up to order there is only one expression of the desired form. Notice that P1 doesnt say anything so it is true (by default). 19 n(n 1) +n= 2
Suppose we have proved Pn for 1 m < n. Assume that n = pe1 pe2 per and r 1 2 f f f n = q11 q22 qs s . with p1 , ..., pr distinct primes and q1 , ..., qs distinct primes. Then we must show that r = s and we can reorder q1 , ..., qr so that qi = pi and fi = ei for all i = 1, ..., r. Since p1 divides n we must have p1 divides f f f f f f q1 (q11 1 q22 qs s ). Thus p1 divides q1 or q11 1 q22 qs s by Proposition 30. f If it divides q1 it is equal to q1 . Otherwise since it cannot divide q11 1 it divides f2 fs q2 qs . Proceeding in this way we eventually see that there must be an index i so that p1 = qi . Relabel so that i = 1. Then we see that if m = n/p1 then f f e f m = p11 1 pe2 per , and m = q11 1 q22 qs s . If m = 1 then n = p1 = q1 . r 2 Otherwise 1 < m < n so Pm is true. Hence s = r and f1 1 = e1 1 and the other qi can be rearranged to get the conclusion qi = pi and fi = ei for i = 2, ..., r. So to complete the discussion of our Proposition we need only give a proof of Proposition 30 Book VII. This proposition rests on his theory of proportions (now rational numbers). We will give an argument which uses negative numbers (jumping at least 1500 years in our story). We will assume here that the reader is conversant with integers (0, 1, 2, ...). Our argument is based on Propositions 1 and 2 Book VII given in the following form: If x,y are numbers that are relatively prime (prime to each other) then there exist integers a, b such that ax + by = 1. We follow the procedure in the argument that demonstrates Propositions 1 and 2 of Book VII . If x > y then the f irst step is x y. We assert that at each stage of this subtraction of the lesser from the greater we have a pair of numbers ux+vy and zx+wy with u, v, z, w integers. At step one this is clear. So suppose this is so at some step we show that it is so at the next step. So if ux + vy and zx + wy are what we have at some step then if (say) ux + vy > zx + wy then at the next step we have (ux + vy) (zx + wy) and zx + wy. That is (u z)x + (v w)y and zx + wy. According to Propositions 1 and 2 Book VII this will eventually yield 1. We will now demonstrate Proposition 30 Book VII. Suppose that p is a prime, a, b are numbers and p divides ab, but p does not divide a. Then p and a are relatively prime. Thus there exist integers u, v so that up + va = 1. Now b = upb + vab since and ab = pc we see that b = ubp + vcp = (ub + vc)p. We will also describe how Euclid proves Proposition 30. Let C be the product of A and B and assume that D is a prime dividing C then C is the product of D and E. Now assume that A and D are prime to each other (since D is prime this means that D does not measure A). Then D, A and B, E are in the same proportion. Since D is prime and A all pairs in the same proportion to D, A are given as multiples F D, F A (this is a combination of Propositions 20 and 21 in Book VII) thus D measures B.
20
1.3.3
Exercises.
1. Use the method of Propositions 1 and 2 of Book VII to calculate the greatest common measure of 315 and 240 and of 273 and 56.. 2. Read the original proof of Proposition 30 Book VII. Explain how it differs from the argument given here. Also explain in what sense the two proofs are the same. 3. Use the principal of mathematical induction to show (a) 1 + 4 + 9 + .... + n2 = n(n+1)(2n+1) . 6 (b) 1 + 2 + 4 + ... + 2n = 2n+1 1. 4. Use the material of this section to show that if a is a fraction then it can b c be written uniquely in the form d with c, d in lowest terms (relatively prime). In other words complete the discussion of the proof of Proposition 30 Book VII). 5. Assume that 1 + 2m + ... + nm = pm (n) with pm a polynomial of degree m + 1 in n. Set up a formula of the form of (a),(b) for the sum of cubes. Prove it by induction. Why do you think that the assertion about pm is true? 6. Use the method of descent to prove that there is no rational number a b 2 so that a = 2. Hint: Let Pn be the statement that there is no m such that b n2 = 2m2 . Use Proposition 30 show that if n2 = 2m2 then n is even. Use this to show that if Pn is false then Pm is false for m such that n2 = 2m2 .
1.4
1.4.1
Perfect numbers and primes.
The result in Euclid.
Perfect numbers are not a central topic in mathematics. However, their study has led to some important consequences. As we saw Euclid devoted one of his precious 22 def initions in Book VII to this concept. We recall that a perfect number is a number that has the property that the sum of its divisors (including 1 but not itself) is equal to itself. Thus 1 has as divisor 1 which is itself so it is not perfect. 2 has divisor 1 other than itself as does 3 and 5 so 2,3,5 are not perfect. Four has divisors 1,2 other than itself so it is not perfect. 6 has divisors 1,2,3 other than itself so it is perfect. Thus the smallest perfect number is 6. One can go on like this the next is 28 whose factors other than itself 1,2,4,7,14. It is still not known if there are only a f inite number of perfect numbers. Euclid in Proposition 36 Book IX gave a method that generates perfect numbers. Let us quote the proposition. If as many numbers as we please beginning from an unit be set out continuously in double proportion, until the sum becomes prime, and if the sum multiplied by the last make some number, the product will be perfect. This says that if a = 1 + 2 + 4 + ... + 2n is prime then 2n a is perfect. Notice that as Euclid gives the result it allows us to discover perfect numbers if we know that certain numbers are prime. We will now try it out. Euclid does not think of 1 as prime. 1 + 2 = 3 is prime. 2 3 = 6 is thus perfect. 1 + 2 + 4 = 7 is 21
prime so 4 7 = 28 is perfect. 1 + 2+ 4 + 8 = 15 not prime. 1 + 2 +4 + 8 + 16 = 31 is prime so 16 31 = 496 is perfect. We now check this because it tells us why the proposition is true. Write out the prime factorization of 496 (which we have seen is unique in the last section as 24 31. Thus the divisors of 496 other than itself are 1, 2, 22 = 4, 23 = 8, 24 = 16, 31, 2 31 = 62, 22 31 = 124, 23 31 = 248. Add them up and we see that Euclid was correct. The example of 496 almost tells us how to demonstrate this assertion of Euclid. If a = 1+2+...+2n is prime then the factors of 2n a are 1, 2, ..., 2n , a, 2a, ..., 2n1 a. So the sum of the factors is 1 + 2 + ... + 2n + a + 2a + ... + 2n1 a. This is equal to a + (1 + 2 + ... + 2n1 )a. Now we observe that Exercise 2 (c) of section 1.4 implies that 1 + 2 + ... + 2n1 = 2n 1. Thus the sum of the factors is a + (2n 1)a = a + 2n a a = 2n a. 1.4.2 Some examples.
This proposition is beautiful in its simplicity and we will see that the Swiss mathematician Leonhard Euler (1707-1783) proved that every even perfect number is deducible from this Proposition. The catch is that we have to know how to test whether a number is prime. We have noted that 1 + 2 + 4 + ... + 2n = 2n+1 1. Thus we are looking for numbers of the form 2m 1 that are prime. Let us make an observation about this point. If m = 2k were even then 2m 1 = 22k 1 = (2k + 1)(2k 1). If k = 1 then we have written 3 = 3 1 so if m = 2, 2m 1 is prime. If k > 1 then 2k 1 > 1 and 2k + 1 > 1 so the number is not prime. We therefore see that if 2m 1 is prime and m > 2 then m must be odd. To get 496 we used 25 1 = 31. The next number to check is 27 1 = 127. We now check whether it is prime. We note that if a = bc and b c then b2 a. This is so because if b c then b2 bc = a. Thus we need only check whether 127 is divisible by 2,3,5,7,11 (since 122 = 144 > 127). Since it is not we have another perfect number 127 64 = 8128. Our next candidate is 29 1 = 511 = 7 73. We see that 22 1, 23 1, 25 1, 27 1 are prime but 29 1 is not. One might guess from this that if 2m 1 is prime then m must be prime. Obviously we are guessing on the basis of very little information. However, this is the way mathematics is actually done. So suppose that m = ab, a > 1, b > 1 we wish to see if we can show that 2ab 1 is composite. Set x = 2a then our number is xb 1. We assert that xb 1 = (x 1)(1 + x + ... + xb1 ). One way to do this is to remember long division of polynomials the other is to multiply out (x 1)(1 + x + ... + xb1 ) = x + x2 + ... + xb 1 x ... xb1 . Then notice that x, x2 , ..., xb1 subtract out and we have xb 1 left. Armed with this observation we can show the following proposition. If p = 2m 1 is prime then m is prime. If m = ab and a > 1, b > 1 then setting x = 2a we see that p = xb 1 = (x 1)(1 + x + ... + xb1 ) = cd, c = x 1 > 1 and d = 1 + x + x2 + ... + xb1 > 1. 22
Our next candidate is 11. But 211 1 = 2047 = 23 89. Using Mathematica (or any program that allows one to do high precision arithmetic) one can see that among the primes less than or equal to 61, 2p 1 is prime for exactly p = 2, 3, 5, 7, 13, 17, 19, 31, 61. Notice the last yields a prime 261 1 = 2305843009213693951. We note that at this writing (2002) the largest known prime of the form 2p 1 is 213466917 1(Michael Cameron, 2001 with the help of GIMPS -Great Internet Mersenne Prime Search). 1.4.3 A theorem of Euler.
We give the theorem of Euler that shows that if a is a perfect even number then a is given by the method in Euclid. Write a = 2m r with r > 1 odd. Suppose that r is not prime. Let 1 < a1 < ... < as < r be the factors of r. Then the sum of the factors of a other than a is (1 + 2 + ... + 2m ) + (1 + ... + 2m )a1 + ... + (1 + ... + 2m )as + (1 + ...2m1 )r = 2m+1 1 + (2m+1 1)a1 + ... + (2m+1 1)as + (2m 1)r. Since we are assuming that a is perfect this expression is equal to a. Thus (2m+1 1)(1 + a1 + ... + as ) + (2m 1)r = 2m r. We therefore have the equation (2m+1 1)(1 + a1 + ... + as ) = r. From this we conclude that 1 + a1 + ... + as is a factor of r other than r. But then 1 + a1 + ... + as as . This is ridiculous.. So the only option is that r is prime. Now we have (2m+1 1) + (2m 1)r = 2m r. So as before, r = 2m+1 1. This is the assertion of the proposition. 1.4.4 The Sieve of Eratosthenes.
In light of these results of Euler and Euclid, the search for even perfect numbers involves searching for primes p with 2p 1 a prime. So how can we tabulate primes? The most obvious way is to make a table of numbers 2, ..., n and check each of these numbers to see if it is divisible by an earlier number on the list. This soon becomes very unwieldy. However, we can simplify our problem by observing that we can cross off all even numbers, we can then cross off all numbers of the form 3 n then 5 n then 7 n, etc. This leads to the Sieve of Eratosthenes (230 BC) 1 2 3 4 5 6 7 8 9 10 11 12 13 ... 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30...
23
3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60... We cross out all numbers in the f irst row that are in the second or third row and have 1 5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 49 53 55 59 61 ... The next sequence of numbers to check is the multiples of 5 5 10 15 20 25 30 35 40 45 50 55 60 65 ... This reduces the f irst row to 1 7 11 13 17 19 23 29 31 37 41 43 47 49 53 59 61 ... Now the number to check is 7 7 14 21 28 35 42 49 56 63 ... Deleting these gives 1 11 13 17 19 23 29 31 37 41 43 53 59 61 ... The next to check is thus 11 11 22 33 44 55 66 ... We note that 112 = 121 > 61. Thus we see that the primes less than or equal to 61 are 2,3,5,7,11,13,17,19,23,29,31,37,41,43,53,59,61. The main modern use of the Sieve of Eratosthenes is as a benchmark to compare the speed of different digital computer systems. Most computational mathematics programs keep immense tables of primes and this allows them to factor relatively large numbers. For example to test that 231 1 is prime one notes that this number is of the order of magnitude of 2.4 109 thus we need only check whether it is divisible by primes less then or equal to about 5104 so if our table went to 50,000, the test would be almost instantaneous. However, for 261 1 which is of the order of magnitude of 2.4 1019 the table would have to contain the primes less than or equal to about 50 billion. This is not reasonable for the foreseeable future. Thus other methods of testing primality are necessary. Certainly, computer algebra systems use other methods since, say Mathematica, can tell that 261 1 is a prime in a few seconds. Using Mathematica one can tell that 289 1 = 618970019642690137449562111 is a prime. This gives the next perfect number 288 (289 1) which is (base 10) 191561942608236107294793378084303638130997321548169216. We will come back to the question of how to produce large primes and factoring large numbers later. In the next section we will give a method of testing if a number is a prime. We will see that the understanding of big primes has led to practical applications such as public key codes which today play an important role in protecting information that is transmitted over open computer networks. 1.4.5 Exercises.
1. Use the Sieve of Eratosthenes to list the primes less than or equal to 1000. 2. Write a program in your favorite language to store an array of the primes less than or equal to 500,000. Use this to check that 231 1 is prime. 3. The great mathematician Pierre Fermat (1601-1665) considered primes of the form 2m + 1. Show that if 2m + 1 is prime then m = 2k . (Hint: If m = ab 24
with a > 1 and odd then set x = 2b . 2m +1 = xa +1. Now ((x)a 1) = xa +1 since a is odd. Use the above material to show that 2m + 1 factors.) This gives 21 + 1 = 3, 22 + 1 = 5, 24 + 1 = 17, 28 + 1 = 257,... (so far so good). Fermat m guessed that a number of the form 22 +1 is prime. Use a mathematics program (e.g. Mathematica) to show that Fermat was wrong. 4. The modern mathematician George Polya gave an argument for the proof that there are an inf inite number of primes using the Fermat numbers Fn = n 22 + 1. We sketch the argument and leave the details as this exercise. He asserted that if n 6= m then Fn and Fm are relatively prime. To see this he observes that x2r 1 = (xr 1)(xr + 1) and so x2 1 = (x2
k k1
1)(x2
k1
+ 1) = (x2
k2
1)(x2
k2
+ 1)(x2
k1
+ 1)
= ... = (x2 1)(x2 + 1)(x4 + 1) (x2 If n > m then n = m + k so 2n = 2m 2k . Hence
m k
k1
+ 1) =
k1
(x 1)(x1 + 1)(x2 + 1)(x4 + 1)(x8 + 1) (x2 Fn = (22 )2 + 1.
+ 1).
So (setting x = 22 ) Fn 2 = (x + 1)K with K = (x1 + 1)(x2 + 1)(x4 + 1)(x8 + 1) (x2
k1
m
+ 1).
Thus Fn 2 = Fm K. So if p is a prime dividing Fn and Fm then p must divide 2. But Fn is odd. So there are no common factors. Now each Fn must have at least one prime factor, pn . We have p1 , p2 , ..., pn , ... all distinct. 5. We say that a number, n, is k perfect if the sum of all of its factors (1, ..., n) is kn. Thus a perfect number is 2-perfect. There are 6 known 3perfect numbers. Can you nd one?
1.5
The Fermat Little Theorem.
In the last section we saw how the problem of determining perfect numbers leads almost immediately to the question of testing if a large number is a prime. The most obvious way of testing if a number a is prime is to look at the numbers b with 1 < b2 a and check if b divides a. If one is found then a is not a prime. It doesnt take much thought to see that this is a very time consuming method of a is really big. One modern method for testing if a is not a prime goes back to a theorem of Fermat. The following Theorem is known as the Fermat Little Theorem.
25
1.5.1
The theorem.
If p is a prime and if a is a number that is not divisible by p then ap1 1 is divisible by p. Let us look at some examples of this theorem. If p = 2 and a is not divisible by 2 then a is odd. Hence ap1 1 = a 1 is even so divisible by 2. If p = 3 and a is not divisible by p then a = kp + 1 or a = kp + 2 by the Euclidean algorithm. Thus ap1 1 is either of the form (3k +1)2 1 or (3k +2)2 1. In the f irst case if we square out we get 9k2 + 6k + 1 1 = 3(3k 2 + 2). In the second case we have 9k 2 + 12k + 4 1 = 3(3k2 + 4k + 1). We have thus checked the theorem for the f irst 2 primes (2,3). Obviously, one cannot check the truth of this theorem by looking at the primes one at a time (we have seen that Euclid has demonstrated that there are an inf inite number of primes). Thus to prove the theorem we must do something clever. That is demonstrate divisibility among a pair of numbers about which we are almost completely ignorant. 1.5.2 A proof.
We now give such an argument. If a is not divisible by p then ia is not divisible by p for i = 1, ..., p 1 (Euclid, Proposition 30 Book VII). Thus the Euclidean algorithm implies that if 1 i p 1 then ia = di p + ri with 1 ri p 1. If i > j and ri = rj then ia ja = di p + ri dj p rj = (di dj )p. So (i j)a is divisible by p. Since we know that this is not true (1 ij p1) we conclude that if i 6= j then ri 6= rj . This implies that r1 , ..., rp1 is just a rearrangement of 1, ..., p 1. Before we continue the proof let us give some examples of the rearrangements. We look at a = 2, p = 3. Then a = 0 3 + 2, 2 a = 4 = 1 3 + 1. Thus r1 = 2, r2 = 1. Next we look at a = 3 and p = 5. Then 3 = 0 5 + 3, 6 = 1 5 + 1, 9 = 1 5 + 4, 12 = 2 5 + 2. Thus r1 = 3, r2 = 1, r3 = 4, r4 = 2. We can now complete the argument. Let us denote by sj for 1 j p 1 numbers given by the rule that rsj = j. Thus in the case a = 2, p = 3, s1 = 2, s2 = 1. In the case a = 3, p = 5 we have s1 = 2, s2 = 4, s3 = 1, s4 = 3. Then we consider a (2 a) (3 a) ((p 1) a). We can write this in two ways. One is 1 2 (p 1) ap1 . The second is (ds1 p + 1) (ds2 p + 2) (dsp1 p + (p 1)). If we multiply this out we will get many terms but by inspection we can see that the product will be of the form 1 2 (p 1) + c p.
26
We are getting close! This implies that 1 2 (p 1) ap1 = 1 2 (p 1) + c p. If we bring the term 1 2 (p 1) to the left hand side and combine terms we have 1 2 (p 1) (ap1 1) = c p. Thus p divides the left hand side. Since p cant divide any one of 1,2,...,p 1, we conclude that p divides ap1 1. 1.5.3 The tests.
This leads to our test. If b is an odd number and 2b1 1 is not divisible b then b is not prime. If b is odd and 2b1 1 is divisible by b then we will call b a pseudo prime to base 2. It is certain that if b is odd and not a pseudo prime to base 2 then b is not a prime. The aspect that is amazing about this test is that if we show that b does not divide 2b1 1 then there must be a number c with 1 < c < b that divides b about which we are completely ignorant! On the other hand this test might seem ridiculous.. We are interested in testing whether a number b is prime. So what we do is look at the (generally) very much bigger number 2b1 1 and see if b divides it or not. This seems weird until you think a bit. In principal to check that a number is prime we must look at all numbers a > 1 with a2 b and check whether they divide b. Our pseudo prime test involves long division of two numbers that we already know. That is the good news. The bad news is that the smallest pseudo prime to base 2 that is not a prime is 341 = 11 31 and it can be shown that if b is a pseudo prime to base 2 then so is 2b 1. Thus there are an innite number of pseudo primes to base 2. For example if p is prime then 2p 1 is also a pseudo prime to base 2 (see Exercise 4 below). Note that we could add to the test as follows. If b is odd and 2b1 1 is divisible by b we only know that b is a pseudo prime. We could then check whether 3 divides b and if it does we would know it is not a prime. If it doesnt we could check whether b divides 3b1 1. The smallest number that is not a prime but passes both tests is 1105 = 5 13 17. One can then do the same thing with 5. We note that if we do this test for 2, 3, 5 the non-primes less than 10, 000 that pass the test are {1729, 2821, 6601, 8911}. This leads to a rened test that was rst suggested by Miller-Rabin. Choose at random a number a between 1 and b 1. If the greatest common divisor of a and b is not one then b is not prime. If a and b are relatively prime but ab1 1 is not divisible by b then b is not prime. If one repeats this k times and the test for being composite fails then the probability of b being composite is less than or equal to 21 . Thus if k is 20 the probability is less than one in k a million. Obviously if we check all elements a less than b then we can forget about the Fermat part of the test. The point is that the number b is very big and if we do 40 of these tests we have a probability of better than 1 in 1012 that we have a prime. 27
A number, p, that is not a prime but satises the conclusion of Fermats theorem for all choices of a that are relatively prime to p is called a Charmichael number the smallest such is 561. Notice that 561 = 3 11.17. One further sharper test(the probabilities go to 0 faster and have strictly less failures than the Miller-Rabin test) is the Solovay-Strassen probabilistic test. We can base it on the proof we gave of Fermats Little Theorem. Suppose that a and b are relatively prime and b is odd and bigger than 1. For each 1 j < b we write ja = mj b + rj with 0 rj < b. We note that rj cant be zero since then b divides ja. Since b has no prime factures in common with a this implies that b divides j. This is not possible since 0 < j < b. Thus, as before the numbers r1 , ..., rb1 form a reordering of 1, 2, ..., b 1. We denote by the number that is gotten by multiplying together the numbers rj ri with j > i and j < b. Then since r1 , ..., rb1 is just a rearrangement of 1, ..., b 1 we see that is just 1 times the number we would get without a rearrangement. We write J(a, b) for 1 if the products are the same and 1 if not. We now consider the product ja ia) for j > i and 1 j < b then as we argued above we see that this number is + cb with c a number. range then This says that if is the product of j i over the same a
(b1)(b2) 2
= J(a, b) + cb.
(b1)(b2)
2 This implies that if b is prime that a J(a, b) is divisible by b. Now n 2 is odd so J(a, b) = J(a, b)n2 . Hence we see that if b is prime then
a
b1 2
J(a, b) = db
for some number d. This leads to the test. We say that a number a between 2 and b 1 is a witness that b is not prime if a and d are not relatively prime or b1 a 2 J(a, b) is not divisible by b. One can show that if there are no witnesses then b is prime. One can also prove that if b is not prime then more than half of the numbers a between 2 and b 1 are witnesses. The test is choose a number a between 2 and b 1 at random. If a is a not a witness that b is not prime then the probability is strictly less than 1 that b is composite. Repeating the 2 test say 100 times and not nding a witness will allow us to believe with high probability that b is prime. The point of these statistical tests is that if we dene log2 (n) to be the number of digits of n in base 2 then the prime number theorem (J. Hadamard and de Valle Poussin 1896we will talk about this later) implies that if N is a large number then there is with high probability a prime between N and log2 (N ). For example, N = 56475747478568 then log2 (N ) = 45 and 56475747478601 is a prime. Thus to search for a prime with high probability with say 256 digits base 2 choose one such number (at random), N , then use the statistical tests on the numbers between N and N + 256. 28
The reader who has managed to go through all of this might complain that the amount of calculation indicated in these tests is immense. When we talk about modular arithmetic we will see that this is not so. In fact these tests can be implements very rapidly. As a preview we consider the amount of calculation to test that a number b is a is a 2-pseudo prime. We calculate 2b1 as follows: We write out b 1 in base 2 say b 1 = c1 2 + c2 4 + ... + cn 2n with ci either 0 or 1. We then 2 2b1 = (22 )c1 (24 )c2 (22 )cn . As we compute the products indicated we note that if m is one of the intermediate products and if we apply division with remainder we have m = ub + r with 0 r < b. In the test we can ignore multiples of b. Also we use the m+1 m m = (22 )2 . And the 22 can be replaced by its remainder after fact that 22 m division by b. Let rm be the remainder for 22 . Thus if we have multiplied the rst k terms and reduced to a number less than b using division with remainder to have the number say s if ck+1 = 1 we multiply s by rk and then take the remainder after division by b. We therefore see that there are at most n operations of division with remainder by b and never multiply numbers as big as b. We will see that a computer can do such a calculation very fast even if b has say 200 binary digits. We give an example of this kind of calculation consider the number n = 65878161. Then 2n1 is an immense number but if we follow the method described we have the binary digits of n 1 (written with the powers of 2 in increasing order) are {0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1}. The method says that each time we multiply we take only the remainder after division by n. We thereby get for the powers {2, 4, 16, 256, 65536, 12886831, 1746169, 1372837, 38998681, 33519007, 56142118, / 28510813, 45544273, 49636387, 27234547, 48428395, 5425393, 65722522, 46213234, 3252220, 64423528, 16511530, 46189534, 45356743, 15046267, 47993272}. Now the intermediate products are (we include the terms where the digit is 0) {1, 1, 1, 65536, 65536, 65536, 46555867, 46555867, 46555867, 46555867, 18206659, 42503458, 24542662, 24542662, 24542662, 54699517, 54699517, 29732113, 728509, 728509, 38913619, 36121177, 1794964, 30837667, 23401021}. The last number is not one so the number n is not a prime. This seems like a lot of computation but most modern personal computers can do these calculations instantly. It turns out the n = 7919 8319. So nding a factor by trial and error would have involved more computations. We also observe that the same m+1 m method can be used for any choice of a using a2 = (a2 )2 . 29
1.5.4
Exercises.
1. Make a large list of pseudo primes base 2 less than or equal to 1000. Compare this with a list of primes less than or equal to 1000. (You will want to use a computer for this.) 2. If n is any positive integer show that there exists a consecutive list of composite integers of length n. (Hint: If we set (n + 1)! = (n + 1)n(n 1) 2 then (n + 1)! + 2, (n + 1)! + 3, ..., (n + 1)! + n + 1 are all composite.) For each n = 2, 3, 4, 5, 6, 7, 8, 9 nd the consecutive list of primes that starts with the smallest number (for example if n = 3 the answer is 8, 9, 10). Why do we need to only check n odd? 3. Calculate the rearrangement of 1,2,...,6 that corresponds to a = 2 and p = 7 as in the proof of the Little theorem. Use this to calculate J(2, 7). 4. Given a and p as in Fermats Little theorem and r1 , ..., rp1 and s1 , ..., sp1 show that if 1 < a < p then r1 = a and s1 r1 = u p + 1 with u a whole number. 5. Show that if p is a pseudo prime to base 2 then so is 2p 1. (Hint: If q = 2p 1 then p p1 2q1 1 = 22 2 1 = 22(2 1) 1. Now p divides 2p1 1. So 2p1 1 = cp. Thus 2q1 1 = 22cp 1 = x2c 1
with x = 2p .) 6. We note that 1 + 22 + 1 = 6 (so divisible by 3), 1 + 24 + 34 + 44 + 1 = 355 (so divisible by 5). Show more generally that if p is prime then 1p1 + 2p1 + ... + (p 1)p1 + 1 is divisible by p. It has been shown that if p satises this condition (that it divides the above sum) then it has been shown by Giuga(Giuga, G. Su una presumibile propertiet caratteristica dei numeri primi. Ist. Lombardo Sci. Lett. Rend. A 83, 511-528, 1950) that p is a Charmichael number. He also conjectured that such a number must, in fact be prime. This has been checked for p < 1013800 (Borwein, D.; Borwein, J. M.; Borwein, P. B.; and Girgensohn, R. Giugas Conjecture on Primality. Amer. Math. Monthly 103, 40-50, 1996). 7. Use a package like Mathematica or Maple to show that 341 is a pseudo prime to base 2 and that 21104 1 and 31104 1 are both divisible by 1105. 8. To do this problem you should use a computer mathematics system. Calculate the remainder of dividing 2n1 by n for n = 57983379007789301526343247109869421887549849487685892237103881017000 7677183040183965105133072849587678042834295677745661721093871. Use the outgrowth of the calculation to deduce that n is not a prime.
1.6
Large primes and cryptography.
In the last section we saw that large primes appear naturally in the unnatural problem of f inding perfect numbers. Large primes have also become an important part of secure transmission of data. Most modern cryptographic systems 30
involve two keys one to be used to encode and the other to decode messages. Public key systems have a novel aspect in that the information necessary to encode a message is in principle known to everyone. But the information to decode the message is only known to the person the intended recipient of the message. In other words, even if you know how to encode a message you still do not know how to decode a dierent message encoded by that method. Alternatively, even if you find a method of deciphering one message deciphering another is not easier. This is a seeming contradiction and although most believe that the methods now in use have this contradictory property there is no mathematical proof that this is so. This type of cryptography was f irst described by W.Die and M.E.Helman New directions in cryptography, IEEE Transactions in Information Theory IT-22 (1976),644-654. One of the f irst practical implementations was due to Rivest, Shamir and Adelman (1978) and is called RSA. It is based on the hypothesis that the factoring of large numbers is much harder than multiplying large numbers. We will discuss this point and describe the implementation of RSA later in this section. 1.6.1 A problem equivalent to factorization.
In the RSA system a person (usually called Alice) chooses (or is assigned) two very large primes p and q. Alice calculates n = pq and makes n public. She also chooses a number e (for encode) that has greatest common divisor 1 with the number m = (p 1)(q 1) and such that 1 < e < m. This number is also made public. The rest of the system involves enciphering messages using these two numbers (n, e). The point of the methods of enciphering is that to decode the message one must know a number 1 < d < m (for decode) such that ed = rm+1 for some integer r (note that the form of Proposition 1 Book VII in Euclid tells us that d exists. It is hypothesized that one cannot nd d without knowing m. There are also probabilistic arguments that indicate that with high probability if we know d then we know m. The main point is thus the following Proposition: If we know the number m then it is easy to factor n. Before we demonstrate this we will interpret the line of thought. This assertion then says that with a high probability, deciphering the RSA cipher is at the same level of di iculty as factoring n. Since we have hypothesized that this is impractically hard we have implemented a public key system. As for the Proposition, if we know m then we know (p1)(q1) = pqpq+ 1. Since we know n = pq we therefore know p + q. Now, (p + q)2 2pq = (p q)2 we see that we know (p q)2 at the same level of di iculty as squaring (which the ancient Egyptians thought was relatively easy) that we have hypothesized is much easier than factoring. The last step is to see that there is an easy method of recovering a if we know a2 . We will see that this is so below. Thus with little di iculty we have calculated p + q and p q. We can recover p and q by adding and subtracting and dividing by 2.
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1.6.2
What do we mean by hard and easy?
Before we describe an implementation of RSA we will give a working explanation of the terms hard and easy. In what follows we will use the notation log2 (n) to mean the smallest k such that 2k is greater than or equal to n. In other words log2 (n) is the number of operations necessary to write the number n in base 2. We will say that a procedure depending on integers N1 , ..., Nd is easy if the there is a method for implementation (an algorithm) that takes a time to complete that is proportional to a xed power (depending on the procedure) of (log2 (N1 ) + ... + log2 (Nd )). If an operation is not easy then we say that it is hard. The study of hard and easy belongs to complexity theory. It is a formalism that is useful for testing whether good computational methods exist (or dont exist). We will just touch the surface. As our rst example we consider the problem of comparing two numbers M and N . We assert that this takes at most 4(log2 (N ) + log2 (M )) operations. We will go through most of the (gruesome details for this case since it is the simplest. The reader should have patience). Indeed, it takes log2 (N ) + log2 (M ) operations to write the two numbers. Once we have done this we know log2 (N ) and log2 (M ). We prove by induction on r = log2 (N ) + log2 (M ) that it now takes at most 4(log2 (M ) + log2 (N )) operations to test whether N is bigger than M is smaller than N or is equal to N . If r 1 then all we must do is look at the two indicated numbers which are 0 or 1. Assume for r s (the induction hypothesis). We now show that it is true for s. We rst check that if log2 (N ) > log2 (M ) (or log2 (N ) < log2 (M )) then N > M or (N < M ). This by the induction hypothesis we need at most 4(log2 (log2 (N )) + log2 (log2 (M )) steps to check this. If we have strict comparison of the logs we are done in 2(log2 (log2 (N ))+log2 (log2 (M )) steps. Otherwise we now know that log2 (N ) = log2 (M ) we now check the digits one by one from the top and look for the rst place with one of M or N having a 1 and the other a 0 the one with the 1 is the larger. If we do the full number of steps we have equality. Thus we have done the comparison in at most (log2 (N ) + log2 (M )) additional steps. Now we observe that log2 (n) n . If n 2. If n = 2 this says that 1 1. If it is 2 true for n and if n 6= 2k 1 then log2 (n + 1) = log2 (n) n < n+1 . Otherwise, 2 2 n = 2k 1. So log2 (n + 1) = k + 1. We are left with observing that 2k k + 1, for k = 1, 2, ... For k = 1 we have equality. If 2k k + 1 then 2k+1 = 2(2k ) 2(k + 1) = 2k + 2 k + 2. This implies that 2(log2 (log2 (N )) + log2 (log2 (M )) log2 (N )) + log2 (M ) So the total number of steps is at most (log2 (N ) + log2 (M )) + 2(log2 (N )) + log2 (M )) + (log2 (N ) + log2 (N )) the rst term for writing the two numbers, the second for comparing the number of digits and the third for the main comparison. Thus comparison in easy (as we should guess). 32
We now look at addition. We have numbers a and b write out the numbers in base 2 assume that a is the larger and ll out the digits of b by 0 (easy). This involves 2log2 (a) operations. We write n = log2 (a). Now add the lowest digits, if one is 0, then put the other digit in the lowest position of the answer otherwise both are 1 so put a 0 in the lowest position and then look at the next digit of a if it is 0 change it to 1 if it is 1 change it to 0 then do the same operation on the next digit continue until you get to a 0 digit of a or to the top one which must be 1 and we would change it to 0 and add one more digit to a. This happens only if all the digits of a are 1 in this case a = 2n 1. So to add a and b you need only change the lowest digit of b to 0 and then add 2n which involves at most 3 steps. This implies that we are either done in 3 steps or we need at most n operations to add the lowest digits. We then go to the next digit. We see that if we are adding at the rth digit we will need to at most the larger of 3 and n r easy operations. Thus the number of operations is at most n + (n 1) + ... + 1 = n(n+1) easy operations. So addition is easy. 2 The next case is that division with remainder is easy. To see this we look at M and N and we wish to divide M into N . Comparison is easy. So if M > N the division yields 0 with remainder N . If M = N we get division 1 and remainder 0. Thus we nay assume M < N . Let m be the number of digits of M and n that of N . If n = m then the division is 1 with remainder N M (subtraction is easy, you will do this in an exercise). Thus we can assume that n > m. Now multiply M by 2nm and (this just means putting n m zeros at the end of the base two expansion of m) subtract this from N . Getting N1 with less than n digits. If N1 M we are done otherwise do the operation again. After at most n of these steps we are done. Thus we must do at most n easy operations. So division with remainder is easy. We also note that similar considerations imply that addition, subtraction and multiplication are easy. Consider Euclidian method of calculating the greatest common divisor (g.c.d.)of two numbers n > m > 1. f irst subtract m from n repeatedly until one has m or one has a number that is less than m. If the number is m then the g.c.d. is m. If not put n1 = m and m1 equal to the number we have gotten and repeat. If m1 = 1 then we know that the g.c.d. is 1. Thus The initial step involves about n/m subtractions. It also involves one division with remainder. If n is not divisible by m then m1 is the remainder after division. Thus, if we use division rather than subtraction each step involves one division with remainder. Since each step reduces the bigger number to a number less than or equal to one half its size we see that the number of such operations is at most log2 (n). Thus it takes no more than log2 (n) times the amount of time necessary to calculate the division with remainder of n by m. By a hard operation on n or on n > m we will mean an operation that involves more than a multiple of log2 (n)k steps for each k = 1, 2, 3, ... (the multiple could depend on k). Thus calculating the g.c.d. is easy. To complete the line of reasoning in the previous subsection we show that if a is a positive integer then the calculation of the positive integer b such that b2 a < (b + 1)2 is easy b is called the integer square root of a.. The idea is 33
to write out a to base 2. If the number of digits is 4 or less look it up in a table. If the number of digits is n which is odd n = 2k + 1 then take as the f irst approximation to b the number 2k if this satisf ies the upper inequality we are done otherwise try 2k + 2k1 if it satisf ies both inequalities we are done otherwise if it doesnt satisfy the lower one replace by 2k + 2k2 and continue the same testing to see if we leave the bit on or not. The involves calculating 2n squares so since n 1 = log2 (a) and we have decided that squaring is easy we have shown that in this case calculating b is easy. If n = 2k is even then look at the f irst 2 bits of a (the coe icients of the highest and next highest power of 2) then start with 2k1 and use the same procedure. Is anything hard? The implementation of RSA assumes that factoring a large number is hard. There is no proof of this assertion, but the best known methods of factorization take the order of magnitude of 2C(log2 (N )) 3 steps. 1.6.3 An implementation of RSA.
1
Suppose that you are shopping on the internet and you must transmit your credit card number, C, to the merchant. You know that it is possible that Joe Hacker is watching for exactly this sort of transaction. Obviously, you would like to transmit the number in such a way that only the merchant can read it. Here is an RSA type method that might accomplish this task. The merchant chooses two big primes p and q (so big that they are both bigger than any credit card number) then forms the numbers n = pq and m = (p 1)(q 1). He also chooses e randomly between 1 and m that has greatest common divisor. 1 with m. He transmits the numbers n and e to your computer (and probably Joes computer). Your computer then calculates the remainder that is gotten when C e is divided by n. Call this number S. Your computer sends S to the merchant. This is what Joe sees. The merchant calculates the number d that has the property that de = 1 + mk for some k. He then calculates the remainder after division by n of S d and has C we will explain this in the next paragraph. If Joe can calculate d then he also knows C. However, if the primes are very large we have seen that this is very improbable. We now explain why S d = C + nh for some h. Neither p nor q divides C since it is too small. By denition of S, C e = S + ng for some g. Thus S = C e ng. We therefore have S d C de = (C e ng)d C de . One can check the formula xd y d = (x y)(xd1 + xd2 y + ... + xy d2 + y d1 ). 34
by direct multiplication. (x y)(xd1 + xd2 y + ... + xy d2 + y d1 ) = x(xd1 + xd2 y + ... + xy d2 + y d1 ) y(xd1 + xd2 y + ... + xy d2 + y d1 ) = xd + xd1 y + ... + x2 y d2 + xy d1 xd1 y ... x2 y d2 xy d1 y d = xd y d . If we make the replacement x = C e ng and y = C e in this formula we nd that S d C de is divisible a multiple of x y = ng and is thus divisible by n. Thus the remainder after dividing by n of S d and C de is the same. We note that (C e )d = C de = C 1+mk = C(C m )k . Now m = (p1)(q1) and (C k(q1) )p1 = 1+ap by the Fermat Little Theorem. Similarly, C mk = 1 + bq. Thus C mk 1 is divisible by both p and q hence by n. (See Exercise 2 below.) Thus S d = C(1 + cn) = C + un for some whole number u. We will now do an example of this but with smaller numbers than those that would be in a practical implementation.. We take p = 71 and q = 97. Then n = 6887 and m = 6720. Choose e = 533. Then the decoder is d = 2132. If C = 45 then the remainder after division by n of C e is 116. We note that 116d has remainder 45 after division by n. 1.6.4 Fermat factorization.
The reverse is also true, that is, if 1 s t and if n = t2 s2 then if a = t s and b = t + s then n = ab. This leads to a method. Start with the number n let g be its integer square root. if g 2 = n we have factored the number into two smaller factors. Otherwise try t = g + 1 and calculate t2 n if this number is a perfect square s2 then apply the above observation Otherwise replace t by t + 1 and try again. Keep this up until t2 n = s2 . This is practical only if n has two factors that are very close together. This tells us that for the sake of security of RSA one must choose p and q far apart. We will try this factorization out for the example we used above n = 6887 then the integral square root is 82. 822 = 6724. 832 n = 2, 842 n = 169 = 132 . 35
RSA is based on the assumption that factoring big numbers is hard. How would we go about doing a factorization of a big number. If we knew that the number came from RSA we would then know that it has only two prime factors. Does this make the problem easier? Fortunately for the internet this doesnt seem to be the case. We will, however, look at a pretty good method of factoring now. Suppose that n is an odd number and that n = ab with 1 < a < b. Set t = a+b and s = ba . Note that a and b are odd so t and s are whole numbers. 2 2 We have t2 s2 = ab.
So taking a = 84 13 = 71 and b = 84 + 13 = 97 weve found our original p = a, q = b. There are many variants of this method that involve signicant improvements in the number of operations necessary to do the factorization. However, the best known methods are hard in the sense of this section. In the next section we will show how a change in the rules allows for an easy factorization algorithm. 1.6.5 More approaches to factorization.
In 1994, Peter Shor published a proof that if a computer that obeys the rules of quantum mechanics could be built then it would be possible to factor large numbers easily. The subject of quantum computing would take us too far aeld. However, one of the ingredients of Shors approach can be explained here. We start with a large number, N . Choose a number y randomly. We calculate the remainder of division by N of y x for x = 0, 1, 2, ... and call that number f (x). Then there is a minimal number 1 < T < N such that f (x + T ) = f (x) for all T x. We call T the smallest period. If T is even we assert that y 2 + 1 and N have a common factor larger than 1. We can thus use the Euclidean algorithm (which is easy in our sense above) to nd a factor of N . Before we demonstrate that this works consider N = 30 and y = 11. Then f (0) = 1, f (1) = 11, 112 = 121 = 1 + 4 30, so f (2) = 1 = f (0). Thus T = 2. Now 111 + 1 = 12. The greatest common divisor of 12 and 30 is 6. We will next check that this assertion about y, T, N is correct. We rst note that T T (y 2 + 1)2 = y T + 2y 2 + 1. But y T = 1 + m N by the denition of T . Thus after division by N one gets T T the same remainder for (y 2 + 1)2 and for 2(y 2 + 1). This implies that (y 2 + 1)2 2(y 2 + 1) is evenly divisible by N . Thus so is T T T T (y 2 + 1) 2 y 2 + 1 = (y 2 1 y 2 + 1 . Thus if y 2 + 1 and N have no common factor then y 2 1 is evenly divisible by N . This would imply that T which is smaller than T satises 2 f (x + T ) = f (x). 2
T T T T
This contradicts the choice of T as the minimal period. There are several problems with this method. The most obvious is what happens if the minimal period is odd? It can be shown that the probability is small that one would make many consecutive choices of y with odd period. Thus the method is probabilistic. However, if you could decode RSA with 36
probability, say, .6 , then you would be able to decode about 60% of the secure internet commerce. There, however, is a much more serious problem. There is no easy algorithm for computation of such periods. The standard ways of nding the T above are as dicult as the factoring algorithms. This is where quantum computing comes in. Shors contribution was to assume that his computer allowed for superpositions (be patient we will know what this means later. For now if you dont know what this means read quantum mechanical operations.) of digits and that these superpositions obeyed the rules of quantum mechanics. Under these assumptions he proved that he could nd the period easily. 1.6.6 Exercises.
1. Why are subtraction, multiplication and division with remainder easy (in the sense above)? 2. Show that if p, q are distinct primes that if p and q divide a then pq divides a. 3. Use Fermat factorization to factor each of the following numbers into a product of two factors 3819, 8051, 11921. 4. Suppose that you have intercepted a message that has been encoded in the following variant of RSA. Each letter in the message is translated into a number between 1 and 26. We will ignore case and all punctuation but spaces and a space is assigned 27. So a and A become 1 , z and Z become 26. Thus we would write NUMBER as 14, 21, 13, 2, 5, 18. We think of this as a number base in base 28. (Here this number is 14+2128+13282 +2283 +5284 +18285 = 312 914 602. We expand the number and write it to base 60. getting 22, 43, 40, 8, 24. We then encode each digit using RSA with n = 8051 and e = 1979. This gives 269, 294, 7640, 652, 198. Suppose that you know that 402, 2832 was coded in this way. What did the original message say? (Even for relatively small numbers such as these you will almost certainly need a computer algebra package to do the arithmetic.) 5. A form of RSA is the standard method of sending secure information on the internet. Do you agree that it is secure? 6. Consider all y between 10 and 20 and N = 30. Calculate the periods, T in the sense of the Shor algorithm (see the previous section).
37
2
2.1
2.1.1
The concept of geometry.
Early geometry.
Babylonian areas.
In section 1.3 we alluded to the fact that Euclid did not look upon arithmetic as an outgrowth of simple counting. He rather looked upon it as arising from measurement of intervals with respect to a unit. The word geometry when analyzed has two parts geo for earth and metry for measurement. The earliest known record of geometry can be found in Babylonian tablets dated from about 3000 B.C. These tablets are concerned with calculating areas. One starts (as did Euclid) by measuring intervals with respect to a unit interval. The subject of these tablets was the calculation of areas bounded by four straight lines. If we think a bit about this question and decide that a square with side given by the chosen unit has unit area and if we take two of them on put then side by side (ore one on top of the other) then we have a rectangle with sides 2 and 1. It is reasonable to think that this rectangle has area 2.
Similarly we can put six such unit squares together and make a rectangle of sides 2 and 3 which has area 6. Thus if we have a rectangle of sides a and b then the area should be a b (square units). Obviously, not every area is as regular as a rectangle and the Babylonians concerned themselves with four sided gures that could be determined by 2,3 or 4 measurements.. Thus a rectangle of sides a and b is determined by two measurements. What about 3 measurements? Here imagine a rectangle of sides a, b and on one of the sides b an distance c from the side of length a is marked. One then joins the marked point with the endpoint of the other side of length b. One now has a gure that is sometimes called a rectangular trapezoid. Let us deduce the corresponding area.
38
The gure has sides labeled by a, b, c and the diagonal if we fold it over as in the picture above the two trapezoids t together to make a rectangle of sides a and b + c. Thus the trapezoid is half of that rectangle and so we have shown that its area is 1 (b + c)a. This is the Babylonian formula. 2 It is still a subject of debate as to what the Babylonians meant by a gure determined by four measurements.. However, what seems to be agreed is that the formula that was used for the area does not jibe with any general notion of four measures since if the measurements are a, b, c, d then the formula they give is (a+c)(b+d) . This seems to be what they thought was the area of a general 4 four sided gure with sides of lengths a, b, c, and d. 2.1.2 Right triangles.
2.1.3
As we saw the Babylonians understood Pythagorean triples. They in fact seemed to be aware of what we call the Pythagorean Theorem. In 1916 the German historian of mathematics Ernst Weidner translated a tablet from 2000 BC that contained the assertion that if a right triangle has legs a and b then the other side has length b2 c=a+ . 2a This is not correct, in general, however we should recall that the Babylonians used the approximation p v u2 + v = u + . 2u If we apply this formula we nd that they are using p c = a2 + b2 . Some Egyptian Geometry.
In the Moscow Papyrus (approximately 1700 BC) there is the following problem The area of a rectangle is 12 , and the width is three quarters of the length, what are the dimensions? The solution was given in the following way. If we attach a rectangle of side one third of the smaller to the longer side to make the gure into a square then
39
the area of the square would 16. Thus the longer side must be 4 and the shorter 3. This method is what we now call completing the square. This example indicates that the Egyptians understood rectilinear areas. However in the same Papyrus there is the following: If you are told: A truncated pyramid of 6 for the vertical height by 4 on the base by 2 on the top. You are to square this 4, result 16. You are to double 4 result 8. You are to square 2 result 4. You are to add the 16, the 8, and the 4, result 28. You are to take one third of 6, result 2. You are to take 28 twice result 56. You will nd it right. Here the scribe clearly has in mind a truncated pyramid of the type that we know that the Egyptians built. That is the base is square and the top is parallel to the base and centered over it.. If we write u for the height (6) and a for the side of the base (4) and b for the side of the top (2). Then the scribe has written: 2 b + ab + a2 ( u ). Which is the correct formula. We will just indicate how it 3 follows from the formula for the volume of a pyramid of base a and height h, ha2 3 . Consider the picture below
Then the total volume is (u+v)a . Now the theory of similar triangles (see Thales 3 fourth Theorem below or Proposition 4 Book VI in Euclid) we have u+v v = . a b The desired volume is thus va3 vb3 (u + v)a2 vb2 = . 3 3 3b 3b We now rewrite the identity just used as v u v + = a a b 40
2
that is
u =v a
1 1 b a v=
=
(a b)v ab
this gives b . ab Substituting this into the formula we have for the desired area yields u a3 b3 . 3(a b)
We now note that this implies the Egyptian formula once it is understood that (a b)(a2 + ab + b2 ) = a3 b3 . The consensus is that the Egyptians were aware of this identity.. The later Egyptian geometry seems to have been inuenced by the Babylonians since on the tomb of Ptolomy XI who died in 51 BC the inscription contained the incorrect formula for the area of a quadrilateral of sides a, b, c and d (a + c)(b + d) . 4 The Babylonians and the Egyptians also had an understanding of the geometry of circles. 2.1.4 Exercises.
1. Can you nd any quadrilaterals have area in accordance with the Babylonian formula? 2. What is the area of a rectangular trapezoid with dimensions 2,4,3? 3. A problem on the Moscow Papyrus says: One leg of a right triangle is two and a half times the other and the area 20. What are its dimensions? Use the Egyptian method of completing to a rectangle to solve the problem (this is the way it was done on the papyrus). 4. Calculate the volume of a right pyramid (not truncated) of base 9 and height 15.
2.2
Thales and Pythagorus.
The geometry of the ancient civilizations is important but pales next to the developments in early Greece. Perhaps one reason why we are so aware of Greek mathematics is because of their rich literature and historical writing dating from the earliest eras of their civilization. The rst Olympic games were held in 776 BC (a documented historic event). The works of Homer and Hesiod (still read) predate this event. During the sixth century BC there is a record
41
of two great mathematicians Thales and Pythagorus. Their individual achievements are only documented by secondary sources which, perhaps, exaggerate the accomplishments of these two mathematicians. The Greek world in 600 BC had spread from its original boundaries of the Aegean and Ionian seas to scattered settlements along the Black and Mediterranean Seas. Most of the mathematics that has been recorded comes from these outskirts. One possible reason for this is that they interacted with the older cultures of the Babylonians and the Egyptians. Thales of Milatus (624-548 BC) and Pythagorus of Samos (580-500 BC) were known to have travelled to the ancient centers of Babylonia and Egypt to study their mathematics. 2.2.1 Some theorems of Thales.
Eudemus of Rhodes (320 BC a student of Aristotle) wrote a history of mathematics that is now lost but a summary of this history (also lost) was incorporated by Proclus (410-485 AD) in his early pages of commentary on the rst book of the Elements by Euclid. Proclus reports as follows: ... (Thales) rst went to Egypt and thence introduced this study to Greece. He discovered many propositions himself and instructed his successors in the principles underlying many others, his methods of attack being in some cases more general in others more empirical. Later quoting (the quote of) Eudemus he attributes that following ve theorems (found in the Elements) to Thales. 1. A circle is bisected by its diameter. 2. The base angles of an equilateral triangle are equal. 3. If two lines intersect the two opposite angles are equal. 4. If two triangles have all their angles equal then the corresponding sides are in proportion. 5. If two triangles have one side and the two adjacent angles equal then they are equal. We will consider these theorems in our discussion if Euclidean geometry. Thales was a practical man whose motto according to Proclus was know thyself. 2.2.2 Pythagorus.
Pythagorus, on the other hand, was a mystic and a prophet. His motto (and that of the Pythagoreans) was all is number. As with Thales, only secondary sources still exist (Aristotle was known to have written a biography of Pythagorus). The Pythagoreans were a vegetarian sect since they believed (possibly inuenced by a trip of Pythagorus to India) in the migration of souls. The term mathematics comes from Pythagorus and literally means that which is to be learned. Proclus, in his introduction to the books of Euclid says: 42
Pythagorus, who comes after him [Thales], transformed this science into a liberal form of education, examining its principles from the beginning and probing the theorems in an immaterial and intellectual manner[meaning abstract]. He described the theory of proportions and the construction of cosmic gures. Johann Stevin Kepler (1571-1630) wrote: Geometry has two great treasures: one is the Theorem of Pythagorus; the other, the division of a line into extreme and mean ratio. The rst we may compare to the measure of gold the second we may name a precious jewel. The meaning of this quotation will be clearer after reading the next section. 2.2.3 The golden ratio.
First we look at a square of side a with its diagonals drawn.
This picture is quite similar to one on the Babylonian tablet Yale 7289. We note that if we use the method the Babylonians each of the triangles with legs 2 b,b and hypotenuse a has area b2 since 4 of them make up the square of area 2 a2 we see that a2 = 4 b2 = 2b2 . Note that the obvious symmetry implies that all of the angles in the center are equal and so each must be a right angle. The drawing is therefore an elegant proof of the Pythagorean theorem in this case. It is reasonable to ask what happens for a pentagon? Consider the two gures
43
If we rotated the pentagon so that a vertex would go to a vertex then the gure would look exactly the same. This says that all the segments labeled by an a are equal to the same value which we will call a. Similarly for the ones marked b and c. Now each of the triangles with base c and two sides a full diagonal (a + b + a) rotate one on the other. For example, AEC and ABD. Each of the triangles with base b and sides a (for example, E 0 A0 C) is similar to the of the triangles with base c and sides a + b + a. So a 2a + b = . c b We note that the line BD is parallel to AE and that AC is parallel to ED. This 0 implies that AE has the same length as ED. So c = a + b. We therefore have 2a + b a = . a+b b Cross multiplying gives 2ab + b2 = a2 + ab. Hence b2 + ab = a2 44
If we divide both sides of this equation by a2 and set x = becomes x2 + x = 1.
b a
then the equation
b Thus the ratio a satises the above equation. This ratio was called the golden ratio by the Pythagoreans. It is this division that Kepler called a precious jewel. The ancients believed that a rectangle whose sides are in this ratio was the most pleasant to the eye. The Greeks designed the Parthenon so that its 1 sides conformed to this ratio. The number y = x is called the golden section and satises y 2 = 1 + y.
A rectangle whose smaller side is in the proportion of the golden ratio to the larger is called a golden rectangle. It has the property that if we take a golden rectangle ABDF as in the picture below and if we We mark the point C so that the length of BC equals the length of the shorter side AB. Then one has a subdivision into a square ABCH and a rectangle CDF H. The rectangle is another golden rectangle:
To see this we observe that if b is the length of AB and if a is that of BD then CD has length a b and F D has length b. We assert that ab b = . b a To see this cross multiply we are trying to see if a(a b) = b2 . That is if b a2 = ab + b2 . Which is just the assertion that a is the golden ratio. The point here is we can now look at the new golden rectangle as a square and a golden rectangle. In fact, we can continue this forever. If in each of the squares we draw the part of the circle of radius equal to the length of a side starting at the far corner (relative to our labeling) we have a spiral. The arcs seem to t smoothly. They do (but not as smoothly as the picture indicates) and well understand why after we discuss the innitesimal calculus.
45
2.2.4
Relation with the Fibonacci sequence.
If we recall the problem of Fibonacci: A rabbit takes one month from birth to become sexually mature. Each month a mature pair gives birth to two (assume a male and a female) rabbits. If you have a pair of newborn rabbits how many pairs rabbits will you have in a year? We start with 1 pair after a month they have just gotten mature so there is still only one pair. They give birth at the end of the next month so there are then 2 pair. In one month the original pair will give birth again but the second pair will have just gotten sexually mature. So there are 3 pairs. 2 of the pairs sexually mature and 1 newborn. The next month the two pairs of mature rabbits give birth and the newborn matures there are now 5 pairs of rabbits 3 mature and 2 newborn. Next month there will be 5 mature and 3 newborn. The pattern (rst apparently pointed out by Kepler) is if the number of rabbits at the beginning of month k is denoted Fk with Nk newborn and Mk mature then Mk+1 = Fk (every rabbit that existed at the beginning of month k is mature in one month) and Nk+1 = Mk (only the mature give birth in one month). Since Fk+1 = Mk+1 + Nk+1 , we see that Fk+1 = Fk + Mk = Fk + Fk1 . Viz.. F0 = N0 = 1 (this is where we start), F1 = M1 = 1, M2 = 1,N2 = 1 so F2 = 2. Now F3 = F2 + F1 = 3. Similarly, F4 = F3 + F2 = 3 + 2 = 5. Continuing in this way we have the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144....
If we start with two squares of side one. Put a square of side two on top of them. Then a square of side 3 to the left. After that a square of side 5 below, etc. (as in the picture) .If we draw circles as in the case of the golden spiral we nd that we have a spiral that is almost identical. This leads us to consider the ratios FFk we have k+1 1, .5, .666..., .6, .625, .6153846, .61904 76, .6176471, .61818..., .6179775, .618055... for the rst 12 ratios to at least an accuracy of 7 digits. We note that the golden ratio is .61803398875 to 11 decimal places. This seems to indicate that if we us the notation for the golden ratio and if we write Qk = FFk then k+1 46
1. Q2k1 < < Q2k for k = 1, 2, 3, ... 2. Q2k Q2k1 becomes arbitrarily small with k. 3. Q2k1 < Q2k+1 , Q2k > Q2k+2 . If these observations are true then the Fibonacci sequence gives an eective way of calculating the golden ratio to arbitrary precision. These observations are true (as we shall soon see) but a much more surprising relationship is true. We rst observe that the quadratic formula implies that 51 = . 2 We also note that if we use the notation for the golden section then 5+1 = . 2 A theorem of J.P.M.Binet proved in 1843 says Fk = k+1 ()k+1 , k = 0, 1, 2, ... 5
Let us check this for some small k. If k = 0 then the numerator is 5+1 + 2 51 = 5. So the formula is correct. If k = 1 then the numerator is 2 ( 5+1 )2 ( 51 )2 = 5. To prove the formula for all k we will use mathemat2 2 ical induction. The assertion Sn is that the formula is true for all k between 0 and n. We know that S0 and S1 are true. Let us assume that Sn is true. We must show that Sn+1 is true. To do this we need only show that the formula is correct for Fn+1 and we may assume that n + 1 2. Thus Fn+1 = Fn + Fn1 . Our assumption implies that Fn = n+1 ()n+1 n ()n , Fn1 = . 5 5
If we add these two terms together we nd that Fn + Fn1 = n ( + 1) ()n (1 ) . 5 5
We now observe that 2 = 1 and 2 = + 1, So n 2 ()n 2 n+2 ()n+2 Fn + Fn1 = = . 5 5 5 This is the desired formula for Fn+1 . Notice that we have given no indication as to why we thought that such a theorem might be true. The method of mathematical induction can only be used to prove assertions that we have guessed 47
in advance or, perhaps, we can derive in using deeper insight. We will describe an alternate more direct approach when we study matrix theory. The formula of Binet easily implies the 3 assertions above. We have (using 1 = ) that if k is even than Qk = if k is odd then Qk = Also Q2k Q2k1 = + 2k+3 > 1 2k+4 2k+3 < . 1 + 2k+4 4k ( + 23 + 5 ) . (1 + 4k+2 )(1 4k+4 )
This implies 2. We leave 3. to the reader. We note that Binets formula can be used to prove many results about the Fibonacci sequence. One very nice formula is
2 Fn+1 Fn1 Fn = (1)n+1 .
To check this we do the substitution of Binets formula in the left hand side of the equation:
2 Fn+1 Fn1 Fn = (( n+2 ()n+2 )( n ()n ) ( n+1 ()n+1 )2 )/5.
If we multiply out the terms in the braces the left hand side of this equation is equal to n+2 ()n ()n+2 n + 2 n+1 ()n+1 . 5 We now use the facts that = 1 and 2 + 2 = 3. So the above display is indeed (1)n+1 . 2.2.5 Phyllotaxies.
In this subsection we will discuss an apparent relationship between the Fibonacci numbers and the spiraling that occurs in plants. It has been observed that the number of petals of a specic type of ower is usually a Fibonacci number. Lilies have 3, buttercups 5, marigolds 13, asters 21 most daisies 34,55, or 89. The head of a ower (like a sunower or a daisy) can be seen to have two families of interlaced spirals, one winding clockwise and the other counterclockwise. The pair of numbers is (see the gure below) 34 and 55 or 55 and 89 for the sunower. Another such phenomenon is the spiralling of pine cones. Among the pine cones found in a cursory look in Del Mar, California one can nd 5,8 and 8,13 pine cones.
48
There are many attempts at an explanation as to why the Fibonacci numbers appear in so many ways in nature. The most convincing are related to the assertion that the golden ratio is the most irrational number. We will give one explanation of this statement in terms of continued fractions. First we will explain the idea of a continued fraction. If we have a number 0 < a < 1 then an expression for a as a continued fraction is a= 1 a1 +
1 a2 + a
1 3 +...
With a1 , a2 , ... positive integers. This means that we should consider 1 1 , a1 a1 +
1 a2
,
1 a1 +
1 1 a2 + a
3
,
1 a1 +
1 a2 +
1 a3 + 1 a4
, ...
as better and better approximations to a. These rational numbers are called the convergents (here we have the rst,second,third and fourth convergent). 1 Here is the method for nding a1 , a2 , .... Dene r1 = a , and a1 to be the largest integer less than or equal to r1 . In general, assuming rn and an have been dened and rn > an then dene rn+1 = 1 rn an
and an+1 to be the largest integer less than or equal to rn+1 . If rn = an . Then the nth convergent is equal to a and we stop. If a is irrational this procedure will never stop. If a 1 then we set a0 equal to the largest integer less than or equal to a and we write a0 + a1 + 1 1 for the continued fraction and the
a2 + 1 a3 +...
49
convergents are a0 + 1 1 , a0 + a1 a1 +
1 a2
, a0 +
1 a1 +
1 1 a2 + a
3
, a0 +
1 a1 +
1 a2 +
1 a3 + 1 a4
, ...
n n Then one can show that if pn is the nth convergent then pn is in lowest terms q q and is closer to a then any fraction in lowest terms with denominator at most 2 qn . If a is the Golden Mean then r1 = 51 = 5+1 . Thus 1 < r1 < 2 so 2 a1 = 1. 1 2 r2 = = 5+1 51 1 2
thus r2 = r1 so a2 = a1 = 1. We note that this goes on forever, rn+1 = rn for n = 1, 2, ... and thus an+1 = an = ... = a1 = 1. Thus the partial fraction expansion of the Golden Mean is 1 1+ The convergents are 1 2 3 5 8 1, , , , , , ... 2 3 5 8 13
1 1+
1 1+ 1 1+....
.
n1 Which we recognize as FFn for n = 1, 2, 3, ... Recent explanations of phyllotaxies involve this irrationality and these convergents. The suggested theory is that if blobs of expanding matter that radiate from a central source are such that as they radiate out they repel then they should be propelled initially at a slope that is badly approximated rationally thus allowing the most space for the radiated initial blobs which are all assumed to be the same size. The fact that the golden ratio is so badly approximated makes it a likely candidate for this angle of radiation. The corresponding ratio of the counts of spirals then give a rational approximation to this number. One of the rst observations of this phenomenon is in the work of 1837 Auguste and Louis Bravais who observed this angle in the ratio of the left and right spiraling of leaves on many trees. In 1872 P.G. Tait extended the work of the Barvais brothers to the spiralling we have been discussing. A controlled experiment was performed by Stpane Douady and Yves Couder in 1993(La Reserche, 24 (1993), 26-25) which conrms these ideas. In their experiment they had a medium of liquid silicon on a disk and from the center of the disk they shot blobs of magnetized liguid. On the edge of the disk they had a strong magnetic source which would cause the blobs to radiate. They found that the count of the spiraling depended on the rate of radiation. The most likely count was a pair of consecutive Fibanocci numbers. However, by changing the rate they found other sequences such as 1, 3, 4, 7, 11, .... This sequence satises the same recursion as the Fibonacci sequence. An amusing discussion of this work can be found in the Mathematical Recreations column of Ian Stewart in the January 1996 Scientic American.
50
2.2.6
Exercises.
1. Join the vertices of a regular six sided gure. Can you see any interesting ratios, etc.? 2. Explain why the golden spiral looks smooth. 3. Show that Qk+1 = Qk1+1 . Use this to show assertion 3. above directly. Suppose that we have a sequence ak of positive rational numbers satisfying 1 ak+1 = 1+ak . Show that if limk ak exist then the limit is . 4. Use mathematical induction to show that n+1 = Fn +Fn1 , n = 1, 2, .... What is the analogous formula for ? 5. We dene a sequence E0 = 1, E1 = 1 and En+1 = E0 + E1 + ... + En . What can you say about this sequence? 6. Consider the sequence dened by the following rules, A0 = 3, A1 = 0, A2 = 2, An+1 = An1 + An2 . This sequence is called the Perrin sequence. In 1991 Steven Arno proved that if n is prime then n divides An . (A3 = 3, A4 = 2, A5 = 5, A6 = 5, A7 = 7, A8 = 10, A9 = 12, A10 = 17, A11 = 22, ...). It has been shown that calculating the remainder of the division of An by n is easy (in the sense of section 7 of Chapter 1). Devise a primality test based on this result.
2 7. Use the formula Fn+1 Fn1 Fn = (1)n+1 to show that consecutive Fibonacci numbers are relatively prime.
8. Find as many examples (or counter examples) to the phenomenon described in the above section (phyllotaxies). 9. Show that the nth convergent of the Golden Mean is
Fn1 Fn . 22 7
10. Let a = . Show that the 0th convergent is 3 and the rst is can use 3.1416 as an approximation for ).
(you
2.3
The Geometry of Euclid.
When we think of the work of Euclid we think about his Thirteen Books of the Elements and plane geometry. We have already seen that this is a misconception. Books VII,IX and XI are concerned with number theory. Solid geometry also appears in several places Books X,XI,XII and XIII. He also wrote books on other topics. Some of his work still exists including his Optics and a book called Phenomena which is a treatise on spherical geometry as it applies to astronomy. He also wrote The Elements of Music which is unfortunately lost. However, his book Sectio Canonis on the Pythagorean theory of music still exists. Without a doubt, his reputation rests on his masterpiece: The Elements. Since the geometry in the elements is much better known than the number theory, we will make an even less complete study of it than we did of the number theory. As in the case of the number theory Book I begins with denitions 23 in this case. There are then 5 Postulates and 5 Common Notions. 51
2.3.1
The denitions.
1. A point is that which has no part. Like his rst few denitions in Book VII this denition must be taken with a grain of salt. He seems to mean that points are the smallest objects that we will consider. 2. A line is breadthless length. As we shall see a line is not necessarily a straight line. In fact, we will see an attempt in Denition 3 to dene a straight line. In modern terminology Euclids line would be a curve. (Denition 15 denes a circle as a part of a line.) 3. The extremities of a line are points. 4. A straight line is a line which lies evenly with the points on itself. This is Euclids expression for a line as we know it. It seems clear that he is asking us to picture a straight line and is just saying that our picture is correct. In a nutshell, a straight line is a line that has some sort of uniformity that should imply straightness. 5 denes a surface, 6 says that the extremities of a surface are lines and 7 denes a plane surface. These denitions are completely analogous to what he does for lines and straight lines. 8. A plane angle is the inclination to one another of two lines that meet each other and do not lie on a straight line. Here he is giving us the notion of an angle between (what we would call two curves). He doesnt seem to think that a denition of inclination is necessary. Furthermore he must be thinking of lines that have exactly one point in common (where they meet) but both do not lie on the same straight line. This is a bit confusing since the lines are not necessarily straight. We can conceive of curves that are partially in a straight line and partially o of it. With the use of the methods of Calculus one can give a notion of angle between two curves. But these curves must be well approximated by straight lines near the point where they meet. 9. And when the lines containing the angle are straight, the angle is called rectilineal. This is dening what we usually mean by an angle (that is between two straight lines). Next he denes a right angle. 10. When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right and the straight line standing on the other is called a perpendicular to that on which it stands. Here we are asked to know what it means for two angles to be equal. Euclid seems to have no need to dene such a concept. It seems clear that he feels that 52
he must introduce some terminology but that all he is doing is describing objects with which we are already familiar. The next denitions dene obtuse angle to be one greater than a right angle and acute angle to be one less than a right angle. Euclid does not seem to feel that he has any need to explain the meaning of the terms less than or greater than in the context of angles. Denitions 13 and 14 are concern boundary and gure. A boundary is dened to be an extremity but an extremity in this context is not dened. Although there is an indication in Denition 3. What he seems to mean is that the boundary is swept out by extreme points of lines. A gure is that which is contained in a boundary. 15. A circle is a plane gure contained by one line such that all straight lines falling upon it from one point among those lying within the gure are equal to one another. 16. And the point is the center of the circle. So a circle is contained by one line. So a line is really what we think of as a curve. There is a point so that if we take a straight line with one extremity at this point and the other on the circle getting a straight line L then do the same for another point on the circle getting a straight line M and if we lie the two lines one on top of the other they are the same. Denitions 17 and 18 dene diameter and semicircle. We note that one of the parts of the denition of diameter is the rst Theorem that Eudemus attributed to Thales. We should also note that Euclid felt no need to prove this part of the denition. 19,20,21,22 dene various types of gures using the terminology with which we are all familiar. Denition 23 involves a concept that is needed in the statement of the fth Postulate. 23. Parallel straight lines are straight lines which, being in the same plane and being produced indenitely in both directions, do not meet one another in either direction. The point of the denitions seems to be to attach names to concepts that we already know. Euclids denitions are not denitions as they are understood in modern mathematics. 2.3.2 The Postulates.
Here Euclid describes assumptions that he feels must be made as the basis of geometry. These are of two types. The rst 3 describe constructions that are possible. 1. To draw a straight line from any point to any point. In other words if we have two points there is always a straight line that joins them. 2. To produce a nite straight line continuously in a straight line. This can mean several things. He seems to want it to mean that we can choose any point on a straight line and have that point be one of the endpoints of a straight line of xed length. 53
3. To describe a circle with any center and distance. We can draw a circle with any center and any radius (in any plane). The next two are assertions about angles. 4. That all right angles are equal. 5. That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indenitely, meet on the side on which are the angles less than the two right angles. (In the picture below the angles in question are a and b.
This is the famous parallel postulate. It seems obviously true and is conrmed by every picture we draw. We will see that it was the subject of intense speculation into the nineteenth century. The brunt of the study was to see if it could be deduced from the other 4 using the denitions and the common notions that we will now describe. This work on the parallel postulate will be studied in greater detail after we have developed a more sophisticated groundwork for our analysis. 2.3.3 The Common Notions.
These are the basic axioms for equality and inequality. 1. Things that are equal to the same thing are equal to each other. 2. If equals be added to equals the wholes are equal. This common notion is a geometric assertion. It applies to areas, geometric gures and numbers (as in the denitions before Book VII). The next common notion should be interpreted in this way also. 3. If equals be subtracted from equals the remainders are equal. 4. Things which coincide with one another are equal. This is the basic method of showing that things are equal in the Elements. The proofs devise a method of laying one object onto another object in such a way that they coincide. That is they t together perfectly. This can be seen graphically in Proposition 4 of Book I which shows that if two triangles have to
54
pairs of equal sides and the included angles are equal then if you lay the angle made by the corresponding sides of one triangle onto that for the other the two triangles coincide. 5. The whole is greater than the part. 2.3.4 Some Propositions.
Euclid is now in business. All terms he will need in Book I are dened (we should assume to his satisfaction also other books such as book II will dene more terms). The rest of Book I involves basic plane geometry. We give the avor of the proofs by looking at two examples, in detail. Proposition 1 and Proposition 47 (The Pythagorean Theorem) in Book I. We will rst look at Proposition 1.
On a given nite straight line construct an equilateral triangle. This means that we are asked to show that if we are given a nite straight line (an interval) we can construct an equilateral triangle with one side equal to the given one. We will now give the proof as given in Euclid The argument is as follows. We have the line AB. We use Postulate 3 twice to make the two circles shown the rst with center A the second with center B and both with distance AB. Let C be the intersection of the two circles. Then AC = AB by the denition of circle (Denition 15) and BC = AB for the same reason (that AC and BC exist is Postulate 1). Thus AC = BC by 1. in the Common Notions. The triangle thus has all of its sides equal. This is ne except for one assertion that Euclid does not deem necessary to be proved: That the circles intersect. This is more serious than the lack of the need to prove what we called the Euclidean property in section 1.4. A proof of the existence of this intersection involves more sophisticated mathematics. At a minimum it involves real denitions of some of the terms. The crux of the matter has to do with the fact that a circle has an inside and an outside and
55
that a line (or a circle) that contains a point in the inside of the circle and a point on the outside must have a point on the circle itself. Proposition 5 is the assertion that the base angles of an isosceles (legs equal) triangle are equal. This is a strengthening of Thales second Theorem as we quoted from Eudomus. Proposition 15 is the third Theorem of Thales that we quoted. Proposition 26 is the fth Theorem of Thales in our list. The other Proposition that we will analyze in detail is number 47 in Book I. We call it the Pythagorean theorem. The proof below seems to be original to the Elements (in other words most of the other proofs are transcriptions of other peoples arguments). In right-angled triangles the square of the side subtending the right angle is equal to the squares on the sides containing the right angle.
The basic idea is to show that the triangles ABD and F BC are equal as are the triangles AEC and BCK. To see how this proves the theorem we note that since the triangle ABD has base BD and height DL as does the rectangle with sides BD and DL (Euclid simply calls it the parallelogram BL). We conclude (as did the Egyptians, Babylonians and Proposition 41, Book I) that the rectangle BL is twice the triangle ABD. The same argument shows that the square ABF G is twice the triangle BCF . Hence since doubles of equals are equal to each other (this is a statement in braces without any further reference) this implies that the square ABF G is equal to the rectangle BL. Similarly, the square ACKH is equal to the rectangle CL. Since, BL and CL make up the square BCED the Proposition follows. (Euclid says: Therefore etc. Q.E.D.). We are left with the assertion about the triangles. We will consider the rst pair notice that AB = BF , BD = BC thus in light of Proposition 4 Book I
56
(Thales fth theorem in our list above) we need only show that the angles ABD and F BC are equal. To see this we observe that the angle DBA is the sum of DBC and ABC. The angle F BC is the sum of ABF and ABC. Since all right angles are equal ABF = DBC. So the assertion about the angles follows from Common Notions 2. 2.3.5 Exercises.
1. Prove the converse of the Pythagorean theorem. That is, if the square of one side of a triangle is equal to the sum of the squares of the other two sides then the angle opposite this side is a right angle. (This is Proposition 48 in Book I. Explain the proof in the Elements and give a proof using, say, trigonometry). 2. In Proposition 11 of Book II of the elements show that Euclid is showing that one can construct the Golden ratio.
2.4
Archimedes.
Archimedes lived during the period 287-212 BC. He was a citizen of Syracuse. In his youth he is thought to have traveled to Egypt and while there he invented the water screw as a way of lifting large amounts of water. He developed a theory of levers and made the famous boast : Give me a place to stand on and I can move the Earth. He is said to have backed this up by raising a ship out of the water using one arm. He was also a military engineer who invented many weapons during the defense of Syracuse against the Romans. He is said to have used giant lenses to focus the sunlight to burn down the Roman eet. The history of his practical inventions is largely second hand since he wrote commentary on only one of these (On sphere making which is lost). Most of Archimedes writings on mathematics have been preserved. He wrote his work in the form of letters to his friends: Conon of Samos and Eratosthenes. After Conon died he sent his letters to Conons student Dositheus of Pelusium. When the Romans eventually invaded Syracuse in 212 BC their general, Marcellus, ordered that Archimedes and his household be spared in the ensuing massacre. However, when a soldier went to escort Archimedes to an audience with Marcellus, Archimedes was concentrating on a geometric problem. He told the soldier that he would come once he solved the problem. The soldier was furious and killed Archimedes: perhaps the greatest mathematician who ever lived. At this point we will discuss two of Archimedes works: The Sand-Reckoner and Measurement of the Circle. The rst is in the nature of a study of very large numbers. The second is the genesis of the elegant approximation 22 of 7 . In later chapters we will be looking at Archimedes work on what we call calculus (although the work alluded to above on the calculation of involves ideas that are usually associated with calculus.
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2.4.1
The Sand-Reckoner.
This paper begins with an introduction written in the form of a letter to King Gelon of Syracuse and ends with a conclusion also addressed to Gelon. Let us quote from the initial material in the paper (as translated by Heath). There are some, King Gelon, who think that the number of the sand is innite in multitude; and I mean by sand not only that which exists about Syracuse and the rest of Sicily but also that which is found in every region whether inhabited or uninhabited. Again there are some who, without regarding it as innite, yet think that no number has been named that is great enough to exceed its multitude...But I will try to show you by geometrical proofs, which you will be able to follow, that, of the numbers named by me and given in the work I sent to Zeuxippus, some exceed ... that of the mass equal to the magnitude of the universe. [ Here the point that will be critical is the phrase no number has been named that is great enough...] He then discusses the various possibilities for the size of the universe with the idea that whatever sizes are believed he will always take one bigger. Included in the models that he considers is that of Aristarchus of Samos of which Archimides says: His hypothesis are that the xed stars and the Sun remain unmoved, that the Earth revolves about the Sun in the circumference of a circle, the Sun lying in the middle of the orbit, and that the sphere of the xed stars, situated above the same center as the sun, is so great that the circle in which he supposes the Earth to revolve bears such a proportion to the distance of the xed stars as the center of the sphere bears to its surface. Archimedes goes on to discount this theory for technical reasons. However, his point is not to establish a theory of the universe but just to get an upper bound on its size. Now comes the point of the whole exercise. There was no known notation or theory of big numbers. Recall that the Egyptians really didnt get past 10 million. The Romans would be constantly inventing new symbols and would eventually run out of letters in the alphabet. The biggest number that the Greeks used was a myriad which is 10,000. Archimedes considers what happens if we multiply two myriads. One then has a myriad myriads. Then he proposes to take a myriad myriads and treat it as a basic unit (a number of the rst order) then he can multiply it by a myriad myriads. One can continue this way a myriad number of times and get a number that Archimedes called P (probably but we reserve this symbol for something else) a number of the second order. In modern notation P = (100000000)100000000 . He then observes that he can continue this process by taking P to be a number of the rst order and consecutively multiply 100000000 P by itself P times getting P P = (100000000100000000 )100000000 . This new number can now be treated as a number of the rst order and the process repeated once again. He then gives a reasonable argument that one of his new found immense numbers is big enough. He in fact argues that the number of particles in the universe is less than 1063 (much smaller that P ). The modern estimates are somewhat nearer to 10100 . 58
In modern mathematics we the ideas of this paper lead to the Archimedian Property that is given any number (we mean here a rational or real number) there is an integer that is strictly bigger. 2.4.2 Exercises.
1. Prove the following is a theorem in The Sand-Reckoner by induction (this is the way Archimedes proved it): If there be any number of terms in continued proportion say A1 , A2 , ..., An , ...and if the rst is 1 the second is 10 [so the third is 100] and if the mth term is multiplied by the nth term the distance [i.e. the number of terms between them] from this term to An is the same as the distance from 1 to Am . 2. In Archimedes paper he takes as the diameter of the Sun 30 times the diameter of the moon. Do you agree with this? (He quotes Euxedus, his own father Pheidias and Aristarchus for estimates of 9, 12 and 20 times. Thus he was estimating higher than anyone else at the time.) 2.4.3 Archimedes calculation of .
We will next look at Archimedes study of the number . He is the rst to prove that 22 is a remarkably good approximation to the ratio of the circumference of 7 a circle to its diameter. In his paper Measurement of a Circle he in shows that 3 10 1 <<3 . 71 7
His method (as we shall see) could yield to arbitrary precision. The important point to note is that he has lower and upper bounds of (in decimal notation) 3.1408 and 3.1429 thus is 3.14.. to an accuracy of at least 0.002. After we study this remarkable result we will look at various ramications of Archimedes work that span about 2200 years. Before we begin his we will discuss the understanding of before Archimedes did his work. The Babylonians routinely used the value 3 for the ratio of the circumference to the diameter of a circle however in some tablets the other values closer to and perhaps including 22 were 7 indicated . In the Rhind Papyrus the value was taken to be 3 1 and sometimes 6 16 2 = 3.16... At least once in the Bible (Revised Standard Version 1952 the 9 King James Bible is a bit more poetic but has the same meaning) in 1 Kings 7-23 it says: Then he made the molten sea; it was round, ten cubits from brim to brim, and ve cubits high, and a line of thirty cubits measured its circumference. The ten cubits from brim to brim is the diameter and the circumference is thirty so the ratio is 3. One can argue that the Bible wouldnt bother with fractions. But even so 31 or 32 would be much closer to the correct value. It is interesting that the rst convergent of the partial fraction expansion of is 22 7 59
(see section 2.2.5). This means that there is no better rational approximation with denominator less than of equal to 7. We will now give a discussion of Archimedes method. The rst proposition involves the following diagram.
The area of any circle is equal to a right-angled triangle in which one of the sides about the right angle is equal to the radius, and the other to the circumference of the circle. This proposition says the if a circle has radius r and circumference c then the area is rc . We know this in a dierent way. We know that c = 2r so the 2 proposition says that the area is r2 . However, this result allows us to calculate the area without knowing . The argument is truly ingenious. Let K denote the area of the triangle and let a be the area inside of the circle. Archimedes observes that there are three possibilities K < a, K > a, K = a. The point is to show that the rst two possibilities cannot occur. He rst assumes a > K and shows that this leads to a contradiction. He draws the inscribed square ABCD. He then bisects the arcs AB, BC, CD and DA and draw the lines from the center of the circle through the bisectors. If necessary bisect again and continue until the area of the inscribed gure is greater than K. To see that (under the hypothesis a > K) this is possible since all we need do is take the subdivision so ne that the sum of the maximal distances from the sides of the gure to the circle is less than a K. That this can be done is obvious from the picture and although the Archimedesmethod of determination of the subdivision involves an assertion equivalent with the desired one that is unproved. He now observes that it is easily seen that the area of each of the polygonal gures is less than K.. In fact the area is the sum of the triangles whose vertices are 60
consecutive vertices of the gure and the center. The height of each of these triangles is less than r and the sum of all the sides is less than c. Thus the area is less than K. So the case a > K is impossible. To show that the case a < K is impossible he argues as above using circumscribed polygons (see the picture). As we have pointed out there are still a few points in this argument that have not been proved (these are easily checked using trigonometry). However, if m is the area of any of the inscribed polygons as in the argument and if M is the area of any of the circumscribed polygons then we have M >K>m and M > a > m. To prove the result we must observe that for each (small) E there is a subdivision such that M m < E. This is basically what Archimedes is asserting. Notice how close to modern calculus this is.
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Archimedes next proves the upper bound for using the above gure (the part OAC is to be thought of as part of the hexagon above it). In this gure the line BA is part of a diagonal of the circle. The line AC is tangent to the circle at A. He starts by taking the angle AOC equal to 1 of a right angle. He then 3 observes that OA > 265 . Fortunately, another Greek mathematician named AC 153 Eutocius inserted an explanation of this inequality. the actual ratio is 3. So 265 2 we must just check that 153 < 3. One checks that the square is 2.9999 to 4 decimal places. To see where the 3 comes from consider the following picture (the curve AN should be an arc of the circle of radius OA)
The angles OAC and OM N are right angles. The angle AOC is 1 of a right 3 angle so the angles ON M and OCA are each 2 of a right angle. This implies 3 that N M = 1 (ON = OA since both are radii of the same circle). Now ON 2 ON 2 = N M 2 + OM 2 so OM 2 = 3 ON 2 = 3 OA2 . Using the fact that OAC 4 4 3 OM OM and OM N are similar triangles we see that MN = OA . Since MN = 2 = 3. 1 AC The assertion follows. Similarly, OC = 2 = 306 . Next we bisect the angle AOC CA 153 which yields the line OD in the picture above. Now Proposition 3 in Book VI of the Elements implies that CO = CD (see Exercise 1 in 2.4.4). Now we OA DA CA have CO+OA = CO + 1 = CD + 1 = CD+DA = DA . Now multiplying the OA OA DA DA OA two ends of this string of equations by OA we have CO+OA = OD . We now CA CA OA 265 OC 306 OA use the inequalities AC > 153 and CA = 2 = 153 . Thus OD = CO+OA > CA 265 306 571 2 2 2 153 + 153 = 153 . He now applies the Pythagorean theorem OD = AD + AO . 2 2 2 2 2 +AO +153 450 > 5711532 = 349409 (1532 = 23 409). Now Archimedes So OD2 = ADAD2 AD 23 apparently guesses another very good lower bound for a square root ((591+1/8)2 591 1 = 3494 28.7 66...) yielding OD > 1538 . The point here is that he now has a OA good lower bound for the ratio OD instead of OC . He can now bisect the angle DA CA AOD getting the point E in the main diagram above and argue as before getting 1172 1 a lower bound on OE > 1532 . He then bisects again and yet again. At his EA 1 time he has an angle the size of 48 of a right angle and at this fourth bisection he only does half the argument and gets an inequality
OA OG
2
>
4673 1 8 153
. Now the
62
diameter of the circle is 2OA and HG = 2AG. Thus the ratio of the diameter of the circle to of the circumference of this 96 sided circumscribed regular gure 4673 1 2 is at least 15396 . The reciprocally of this then gives an upper bound for of = 3 + 467321 < 3 + 467221 = 3 + 1 . 7 2 2 The next task is to derive a lower bound for . Archimedes does this by starting with a regular inscribed hexagon and bisecting 4 times just as he did for the upper bound.
14688 4673 1 2 667 1 667 1
He starts with the above picture with the angle equal BAC to one third of a AC right angle. As before, CB = 3. This time Archimedes needs an upper 2 bound for 3. He chose 1351 since 1351 = 3.0000016... As before he bisects 780 780 the angle BAC getting the straight line AD. He observes that AD intersects BD at the point d. We note that the angles at C and D are right angles. Thus since dAC and BAD are the two halves of the angle just bisected we see that AD the triangles ADB, ACd and dBD are similar. Thus DB = BD = AC . Now Dd Cd CA Cd we observe (see Exercise 1 below) that AB = dB . Thus AC = AB (this implies Cd dB AD that (AC)(dB) = (AB)(Cd) which we will use in a moment). So DB = AB . Bd We also note that (AB)(Bd) + (AB)(Cd) = (AB)(Bd) + (AC)(Bd) (see the parenthetic remark). Thus AB = AB+AC (cross multiply). The denominator Bd Bd+Cd of the right hand side of this equation is Bd + Cd = BC. We therefor have AB AB+AC AB+AC AD . The outgrowth of all of this is DB = AB+AC . Bd = Bd+Cd = BC BC AC 1351 BA 1560 AD We now note that BC < 780 and BC = 2 = 780 . So DB < 2911 . We 780 now use the Pythagorean theorem for the right triangle BDA. Finding that 2911 2 2 2 2 AB 2 +AD2 AB 2 = BD2 + AD2 . So BD2 = BDBD2 < 780 + 1 = 2911 +780 . As 7802 before Archimedes must approximate a square root and he takes an upper bound 3013 3 AB 4 BD < 780 . He now bisects the angle BAD getting the line AE and proceeds AB in exactly the same way to get an upper bound for BE . He bisects two more times getting the line AG. Using the same technique he gets the estimate
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66 < 66 4 . So GB > 2017 1 . Since GB is a side of a regular inscribed polygon AB 4 with 96 sides we see that the ratio of the perimeter of the polygon to the radius of the circle is greater than 6696 > 3 10 . 71 2017 1 4 There are many theories as to how Archimedes found his accurate upper and lower bound for 3 one that is very convincing can be found in A.Weil, Number Theory, Birkhuser, Boston, 1984. He suggests that Archimedes was applying the formula (5x + 9y)2 3(3x + 5y)2 = 2(x2 3y 3 ).
AB BG
2017 1
Then according to Weil he started with x = 1 and y = 0 and since 52 3 32 = 2. We have 2 5 2 =3 . 3 9 5x + 9y 3x + 5y , 2 2
The iteration involves two parts (x, y) then 2 1 The rst iteration (x = 5, y = 3) yields 262 3(15)2 = 1, that is 26 = 3+ 152 . 15 In the second (x = 26, y = 15) one has 526+915 = 265, 326+515 = 153 2 2 and (265)2 3(153)2 = 2 thus 265 = 3 (153)2 . This gives Archimedes 153 lower bound. The upper bound is obtained by putting x = 265 and y = 153 into the rst formula. Getting the pair 5 265 + 9 153 3 265 + 5 153 = 1351, = 780 2 2 the upper bound used by Archimedes. 2.4.4 Exercises. (x, y) (5x + 9y, 3x + 5y) .
1. Consider the diagram (the curve AN should be an arc of the circle of radius OA)
64
with OD the bisector of the angle AOC show that OC = CD . (Hint: Use OA DA trigonometry. Let be the angle AOC. Then we are asked to show that tan tan 1 2 = . cos tan 2 Do this by using the usual trigonometry identities sin = 2 sin cos and 2 2 cos = (cos )2 (sin )2 .) 2 2 2. In the calculation of the upper bound for Archimedes replaced the 667 1 better estimate 3 + 467321 by 3 1 . Why do you think he did it? 7
2
3. The Babylonians preferred the upper bound 3 1 over 3 1 for . Can you 6 7 give a reason for this? 4. Do the indicated iterations at the end of this section for sharper and sharper approximations to 3. What would you do to get good approximations of 5? 2.4.5 The iteration in Archimedes calculation.
It is clear that Archimedes could have in principle continued his bisection procedure indenitely. However, the calculations become more and more complicated and even powerful arithmetician (as Archimedes obviously was) would be stymied by the calculation after two more bisections. Furthermore, little would have been gained since his initial choices of square roots of 3 limited him to about 4 digits of accuracy. We now live in an age of cheap high speed calculation power and can therefore implement many more iterations of Archimedes method. Let is rst abstract the iteration that Archimedes does 4 times for the upper bound and 4 times for the lower.
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We will use the diagram of Exercise 1. above we take the angle AOC to be such that 2m times it makes exactly one rotation. Let denote that angle. Then 2AC is a side of the regular m-gon circumscribed on the circle of radius OA and 2N M is the side of the regular m-gon inscribed in the same circle. Thus the circumference of the circumscribes m-gon is 2mAC and the circumference of the inscribed is 2mN M . We observe that AC = tan and NM = sin . Thus OA OA the circumscribed circumference divided by the diameter (2OA) is a = m tan and the circumference of the inscribed divided by the diameter is b = m sin . We now bisect the angle AOC getting AOD. Then using the same argument we nd that the circumference of the circumscribed 2m-gon divided by the diameter is a0 = 2m tan and the corresponding ratio for the inscribed 2m-gon 2 is b0 = 2m sin . The key point is 2 a0 = b0 = 2ab , a+b a0 b.
Let us check this with standard trigonometry. We will use sin = 2 sin cos 2 2 and cos = (cos )2 (sin )2 . If we use the rst identity we nd that 2 2 a0 b = 2m sin (2m sin cos ) 2 2 2 = 4m2 (sin )2 = (b0 )2 . 2 cos 2
This shows that the second identity is true. As for the rst 2m2 (sin ) 4m sin cos 2ab 2m sin cos 2 2 = . = = sin a+b 1 + cos m( cos + sin ) 1 + (cos )2 (sin )2 2 2 We now use the fact that 1 = (cos )2 + (sin )2 so the denominator of the 2 2 last expression is 2(cos )2 . Substituting we nd that the last expression is 2 = 2m tan . 2 We will now use these observations to set up the implied iteration in Archimedes. We start with m = 6 then Archimedes has shown that tan = 33 and sin = 1 . 2 Thus the corresponding ratios (which we denote by a0 and b0 ) are a0 = 2 3 and b0 = 3. If we do the rst bisection then we have a1 and b1 with a1 = and b1 = 2a0 b0 a0 + b0 p a1 b0 . 2an bn an + bn
4m sin cos 2 2(cos )2 2
2 2
In general we have after n + 1 bisections an+1 =
66
and bn+1 =
Archimedes is using the upper and lower bounds b4 < < a4 and extremely clever choices of approximate square roots. If you do the calculation using a computer you nd that a4 = 3.14271... and b4 = 3.14103... whereas Archimedes estimates are 22 = 3.142857... and 3 10 = 3.140845... The esti7 71 mates of Archimedes are therefore truly remarkable. We note that a10 = 3.141592930... and b10 = 3.141592519... Thus 4 iterations gives 2 decimal place accuracy and 10 gives 6 decimal place accuracy. One nds that after 16 iterations a16 and b16 agree to 8 decimal places. This predicts that 2k iterations should give an accuracy of k decimal places (indeed a calculation shows that one has 50 digit accuracy after 100 iterations). This can be (essentially) proved as follows. We note that s 2an bn 2an b2 n an+1 bn+1 = = an + bn an + bn 2an bn (an bn ). (an + bn )( 2an + an + bn ) One can check that note that for all n 0
2an bn (an +bn )( 2an + an +bn )
p an+1 bn .
1 2+ 2
< 1 . To see this we rst 3
bn < an indeed if n = 0 this is the assertion that 2 3 > 3. Assuming this for n we note that the iteration implies that bn = Thus bn+1 = s an+1 (an + bn ) . 2an r an + bn < an+1 2an
a2 (an + bn ) n+1 = an+1 2an
since an > bn . Thus the principle of mathematical induction proves the assertion. In 2an bn (an + bn )( 2an + an + bn ) we divide the numerator and denominator by 2an bn and get We now observe that than
1 1 2(1+ 1 ) an bn
1 q . n 1 + an 1 + an +bn b 2an q
an +bn 2an
> 1 and
>
1 . 2
Thus the expression is less
=
1 . 2+ 2
67
There is something very odd about this recursion that is that the expression for bn+1 involves an+1 . Consider the following change in the recursion: an+1 bn+1 2an bn an + bn p = an bn . =
One checks that with this recurrence starting with the same initial values we have agreement between a5 and b5 to 50 decimal places. There is only one problem with replacing Archimedes iteration with this one. It converges to 3.219546022.... Which could be an interesting number but it is not . What is it? The next number will unravel this mystery and lead to a method of determining to very high orders of accuracy. Before we go on to these developments we will make one general observation about the above iteration we rst note that. The iteration implies that we have p an bn bn+1 an+1 = (an + bn 2 an bn ) = (an + bn ) 2 an bn p bn an . (an + bn ) This implies that if n > 1 then bn > an . We note that bn an bn + an = bn an . We therefore have an bn 2 bn+1 an+1 = 2 (bn an ) . (an + bn ) bn + an We estimate the expression
an bn 2. (an +bn )( bn + an )
For simplicity we assume
that a0 1 and b0 1 we assert that an 1 and bn 1 for all n. If n = 0 this is our assumption. If we assume this assertion for n then an bn an and an bn bn so 2an bn > an + bn hence an+1 1 also an bn 1 implies that an bn 1 so 2 bn+1 1. We have already seen that an + bn 2 an bn = bn an so an + bn 2 an bn . Thus we have 1 an bn an + bn 2 2 bn + an (1 + 1)2 = 4. Thus and since an 1 and bn 1 we see that 1 an bn 2 8 . (an + bn ) bn + an This implies that in the modied iteration we have bn+1 an+1 68 1 (bn an )2 8
for n 0 if a0 and b0 are both at least 1. Actually one has a similar estimate if we only assume that a0 and b0 are bigger than 0 (we will see why in the next section when we relate this iteration to the arithmetic-geometric mean iteration.. This accounts for the rapid convergence. Starting with a0 = 2 3, b0 = 3. Then b1 a1 1 (a0 b0 )2 = 0.0269238... Now 8 b2 a2 b3 a3 2.4.6 1 1 (b1 a1 )2 3 (b0 a0 )4 = 0.00009063..., 8 8
n
1 1 (b0 a0 )2 , ... (b2 a2 )2 = 7 (b0 a0 )8 , ..., bn an 8 8 82n+1
The arithmetic-geometric mean iteration of Gauss.
Recall the new iteration of the previous subsection: an+1 bn+1 2an bn an + bn p = an bn . =
With a0 , b0 positive real numbers. If we write an = have the recursion un+1 vn+1 un + vn 2 = un vn . =
1 un
and bn =
1 vn
then we
The rst is the arithmetic mean of un and vn and the second is their geometric mean. If u and v are positive then their mean or arithmetic mean is their average a(u, v) = u+v their geometric mean or multiplicative average is 2 m(u, v) = (uv) 2 . We note that a(u, v)2 m(u, v)2 = (uv) 0. Thus if we 4 start the iteration with u0 , v0 0 then un vn for all n 1. This iteration was discovered independently by J.L.Lagrange (1736-1813) and C.F.Gauss (1777-1855). Lagrange alluded to it in 1785 and Gauss studied it in 1790 (when he was about 14). We attach the name of Gauss since he did the most profound work on it in particular answering the question we asked at the end of the last section. The iteration is called the AGM On May 30, 1799 Gauss wrote (in his diary) that if we start the iteration with u0 = 1 and v0 = 2 then u1 and v1 n n are equal to Z dt 2 1 0 1 t4 to at least 11 decimal places for n large. Notice that he is predicting the value of the original variant of the Archimedean iteration. He was absolutely certain that the limit of the sequences was in fact this number. In his diary he said that should this be true then it will surely open a whole new eld of analysis. The area of analysis that was opened is the theory of elliptic functions which is still one of the most important areas of mathematics that has permeated every 69
1 2
aspect of the science and about which we shall hear much more later. Before giving Gausss solution to the general problem we will explain a possible reason why he might believe that the limit in the above case might be given by an integral of the above sort. We rst return to the Archimedean iteration of the previous section an+1 bn+1 2an bn an + bn p = an+1 bn =
with a0 = tan , b0 = sin and 0 < < . Then as above we see that there is a 2 number L such that bn < L < an and that an bn can be made as small as we wish by increasing n. The amazing fact is that the number L is . Thus for 1 example if we start with a0 = 1, b0 = 2 then = . This could have lead him 4 to look at his iteration multiplied by . He was no doubt certain that integrals 4 of the type of his projected formula for his limit could not be calculated using elementary methods (e.g. modern Freshman or Sophomore Calculus). The integral Z 1 dt 1 t4 0 is one of the simplest of the type that he would have studied. He therefore would have known that it was about 1.311028777. From this and his no doubt 2 very accurate approximation to he could have easily come up with the ap8 proximation in his diary entry of 1799. Gauss later derived a formula for the limit of the AGM which can be found in volume 3 of his collected works. The solution is given in terms of a completed elliptic integral. We will just quote the formula (a very nice discussion can be found in Borwein and Borwein, Pi and the AGM. Consider the AGM then p since xa+xb = x a+b and (xa)(xb) = x ab for x, a, b > 0 we see that if we 2 2 denote be M (a, b) the limit of the AGM with u0 = a, v0 = b. Then if a, b > 0, b M (a, b) = aM (1, a ). It is therefore enough to calculate M (1, x) for x > 0. Here is the formula Z 1 2 2 d q = . M (1, x) 0 1 (1 x2 ) sin2 2.4.7 Exercises. 1. Consider the following iteration an + bn 2 2an bn . bn+1 = an + bn With a0 > 0, b0 > 0. Show by induction that an bn = a0 b0 . Also that an+1 = an+1 bn+1 = 70 (an bn )2 . 2(an + bn )
Use these observations to derive a very fast method of calculating square roots. (Note that if we start with a0 = 3 , b0 = 2 then a2 = 97 , b2 = 168 and 2 56 97 a3 = 18817 , b3 = 32592 , further (b3 )2 = 2.999999992... 10864 18817 2. Use the method in section 2.4.5 of the derivation of the Archimedean iteration to show that if = m then the Archimedean iteration (the main iteration in 2.4.5) starting with a0 = tan , and b0 = sin eyelids in the limit. 3. Make the appropriate change of variables to show that the value of M (1, 2) using Gausss general formula agrees with the one he predicted in 1799.
2.4.8
A short history of calculations of .
As we have seen, Archimedes is the author of the famous approximation 22 for 7 . We have also seen that most ancient peoples who were aware of used 3. The Babylonians used the somewhat better approximation 3 1 . We will end 6 this short history with a method of approximation based on the AGM. After Archimedes the iteration described above was used to nd approximations to until the 17th century.perhaps the best usage (and perhaps the last) was by Ludolph van Ceulen (1540-1610) who used the method to calculate 34 digits of . This method converges relatively slowly so the computational overhead overwhelms hand computation. It wasnt until the advent of calculus that more precise approximations were found using more rapidly converging sequences. Until the middle of the twentieth century the approaches involved clever uses of a formula attributed to James Gregory (1638-1675) which says that x2 x4 x6 arctan(x) = x + + ... 3 5 7 If, in this series, we set x = 1 then we have 1 1 1 = 1 + + ... 4 3 5 7 p Edmond Halley (1656-1743) used x = 1/3 in Gregorys series to produce the series 1 1 1 1 = (1 + 2 3 + ...). 6 33 3 5 3 7 3
He used this to nd to 71 decimal places to do this he needed to sum at least p 143 terms of the series. The approximation is q3 with p = 2975751933403306660762781037851580732457179725218341337 8517664256040092164338566715216074032725294059375304662800 8784897242437840350479660097317139924948757850741584109975 3560486485710547648 and
71
q = 5468726754975023858173190008331026443083349550027969750 06063504744927456329014146000945985504325020793071970588029 48449190349218434866194124401527196795946520854577134466195 5929457343724625 This is an amazing achievement using hand calculation. Later variants of these methods are related to the formula of John Machin (1680-1752). He observed that 1 1 = 4 arctan( ) arctan( ) 4 5 239 the point here is that the rst term is easily calculated and the second is an alternation of terms that become very small rapidly. Machin used his formula to nd 100 digits of . In 1961 using an IBM mainframe D. Shanks and J.W.Wrench produced 100,000 digits of using two 3 term variants of Machins formula. (one to check the other). Using the same method one million digits were computed in 1973 by Guillard and Bouyer. Further precision has for the most part been based on the AGM. As of 1999 the record is held by Kanada, Takahashi 1999 with 206158430000 digits. We include here an iterative scheme for calculating that was discovered by Borwein and Borwein in the 1980s that is derived from the AGM. The iteration is as follows: 1 xn + 1 xn+1 = n0 2 xn yn xn + 1 yn+1 = n1 xn (1 + yn ) xn + 1 n1 n = n1 yn + 1 with x0 = 2, 0 = 2 + 2, y1 = 4 2. One can show that if n 2 102
n+1
n 0.
This says that after 10 iterations we have to the accuracy of over 2000 decimal digits. After 20 we heve over 2000000 digits.
2.4.9
Exercise.
1. Use a computer algebra system to check that the iteration does indeed give the asserted accuracy for n = 2, 3, 4, 5 (asserted 8, 16, 32, 64 digits). Devise an algorithm combining 2.4.7 Exercise 1 with the above iteration to get a high precision algorithm for with no square roots.
72
3
The emergence of algebra
As we saw in chapter one the ancient Babylonians and Egyptians had an understanding of much of what we would now call Algebra. For the most part their algebra comes down to us in the form of problems that are not very dierent from those that are assigned to modern students. The ancients certainly had an understanding of how to solve linear and quadratic equations in one variable. The Babylonians with the use of copious tables could solve some cubic equations. But they were hampered in their lack of two basic formalisms that we take for granted. The rst is that they had no concept of negative numbers and they had no notation such as our modern algebra which allowed them to handle an unknown quantity as if it were a number. There was still a basic distinction between the role of numbers for counting and numbers for measurement. As we shall see, the nal synthesis involves the identication of the two notions of number.
3.1
Algebra in Euclids Elements.
In Euclids elements Book II can be considered to be devoted to algebra. For example, Proposition 4 book II says: If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments.
We will not go through Euclids proof here (which is surprisingly long). We will just point out that what it says is that the square ADEB is made up of the two squares CBIG, DF GH and the two equal rectangles ACGH and F EIG. In more modern notation the side of the big square is AB = x + y. The side of the square CBIG is y that of DF GH is x and the two adjacent sides of the two rectangles are x, y. Thus the content of the Proposition is (x + y)2 = x2 + 2xy + y 2 . We will see that until the time of Descartes, the part of mathematics that we consider to be algebra was consistently phrased in geometric terms. In Euclids number theory a number was a concatenation of unit intervals. He 73
only considers whole numbers. In his geometric algebra (Book II) he considers lengths and areas but gives no direct relationship with the concept of number which comes later (Book VII). Thus intervals, squares and rectangles, cubes are dealt with as if they are what we consider to be numbers. The addition and subtraction meant putting together gures as in the one above. The amazing aspect of all of this is that within these constraints mathematicians were able to do serious work in algebra such as solving a polynomial of degree3 or 4. The constraints were broken by the seventeenth century French mathematicians.
3.2
The Arabian notation.
In chapter 1 we studied the methods that were used by several early cultures to represent numbers. We also looked at our own decimal system. This positional system was used by the peoples of the middle east and comes to us under the name Arabic notation. This notation when it appeared in Europe was very similar to our modern notation (however it is likely that it had its genesis in India). One of the most important and earliest western advocates of this system was Leonardo of Pisa (alias Fibonacci) who used the system in his book Liber Abaci (published in 1228). In this book he used the arabic notation for everything but fractions. For fractions he used sexagesimal, Egyptian fractions and common fractions (that is a/b in lowest terms). He preferred the latter two types. We have seen that he devised an algorithm to convert common fractions to Egyptian fractions. We also observed that he gave a complete characterization of Pythagorean triples. 3.2.1 The completion of the characterization of Pythagorean triples.
The following argument involves the understanding of squares of integers. Fibonacci was so enamored of squares that he wrote a book Liber Quadratorum (Book of Squares) which contained the proof of the following theorem (see also Euclid Book X, Lemmas 1,2 before Proposition 29): If a, b, c are positive integers such that a2 +b2 = c2 then one of a and b, say, b must be even and there exist numbers m, n, x such that a = x(m2 n2 ), b = 2xmn, c = x(m2 + n2 ). If a, b are relatively prime we can take x = 1. To prove this assertion we rst show that one of a or b must be even. Suppose not. Then a = 2r+1, b = 2s+1 and so c2 = a2 +b2 = 4r2 +4r+1+4s2 +4s+1 = 4(r2 + r + s2 + s) + 2. We can thus conclude that c2 is even. This can only be so if c is even. But then c = 2t. We conclude that 4t2 = 4(r2 +r +s2 +s) +2. This leads to the conclusion that 4 divides 2. We can thus assume that b is even. To complete the hard part of the argument we rst observe that the last assertion implies the main assertion. Indeed, if x divides a and b then x2 divides c2 so x divides c. (You should be starting to see why the book had its title.). Thus if a, b, c is a Pythagorian triple and if x is the greatest common divisor of a
74
a b c and b then x , x , x is a Pythagorian triple. So we are left with showing the last assertion. We thus may assume a and b are relatively prime and that b is even. Since a2 + b2 = c2 it follows that c2 a2 = b2 . So (c a)(c + a) = b2 . We notice that if c were even then a must be even. But then a and b would have 2 as a common factor. Thus a and c are odd. If y is odd and divides both c + a and c a then y divides their sum and dierence which are 2c and 2a. But then x2 divides b2 . So x divides b which is contrary to our assumption. Thus if p is an odd prime so that pr divides b then p divides exactly one of c + a and c a thus p2r divides one of the factors. Hence if b = 2t pr1 pru is a prime u 1 factorization of b then we can reorder the indices so that c + a = 2v p2r1 p2rs s 1 2rs+1 and c a = 2w ps+1 p2ru with v + w = t. If v and w were both bigger than u 1 then 4 would divide both 2c and 2a. Since both are at least one we see that one of w and v must be one. This implies that the other must be odd. If v = 1 then c + a = 2m2 and c a = 2n2 . It is clear that if w = 1 we come to the same conclusion. Thus b2 = 4m2 n2 so b = 2mn. Also 2c = (c+a)+(ca) = 2(m2 +n2 ) and 2a = (c + a) (c a) = 2(m2 n2 ). So a, b, c are of the desired form.
3.2.2
Exercises.
1. Observe that if m = 2, n = 1 then m2 n2 = 3, 2mn = 4, and m2 + n2 = 5. If m = 3, n = 1 then m2 n2 = 8, 2mn = 6 and m2 + n2 = 10. Thus aside from the factor of 2 and the order the two give the same Pythagorian triple. Show that if m, n are relatively prime then the greatest common divisor of any pair of the Pythagorian triple is either 1 or 2. 2. Show that if x, y are rational numbers such that x2 + y 2 = 1 then there exist integers m, n such that either (x, y) = ( or (x, y) = ( 2mn m2 n2 , ) m2 + n2 m2 + n2
2 a2 (Hint: y = 1 x2 . If x = a in lowest terms then 1 x2 = c c2 . If the c square root is rational then c2 a2 must be the square of an integer b. Thus a2 + b2 = c2 .) 3. What is the overlap between the sets described by the two formulas in problem 2? 4. If we divide the numerators and denominators of the rst expression in problem 2 by n2 and write t = m then we have n (x.y) = ( 2t 1 t2 , ). 1 + t2 1 + t2
2mn m2 n2 , ). m2 + n2 m2 + n2
Show that the only pair of real numbers (x, y) not covered by a value of t is (1, 0). This is called the rational parametrization of the circle. 75
5. Look at Euclid Book X. Lemmas 1,2 before Proposition 29 and write out what the assertions mean algebraically. 3.2.3 Polynomials of higher degree.
Not as well known is the fact that Fibonacci studied cubic equations. Studying algebraic equations was quite dicult in his time due to a lack of appropriate notation and since the algebra of Fibonacci was still the geometric algebra of Euclid. Furthermore, cube roots were not constructed in the plane geometry so they were somewhat more mysterious. In his book the Flos (1225) he studied some cubic equations in particular x3 + 2x2 + 10x = 20. He proved that this equation has no rational roots and even no roots of the form u + v with u and v rational. He also gave an approximate solution to the equation in sexagesimal (1;22,7,42,33,4,40). The Middle Eastern mathematicians had methods of calculating roots of polynomials to arbitrary precision. Most notable is the work of Abul Kamil (850-930). Also Omar Khayyam (1050-1130) had interpretations of roots of certain cubics as intersections of conic sections. One reason for the slow progress in general methods of solution of polynomial equations was the lack of good notation (which persisted into the seventeenth century) and a lack of the ability to manipulate unknowns and indeterminates. For example, the unknown quantity x made sense to them (even to the ancients) as a quantity one could also say a cube a side and a face none of which are known. Then a description of a cubic equation could be given as a cube added to 2 times a face added to 6 times as side is equal to 12. We would write this as x3 + 2x2 + 6x = 12. The mathematicians developed clever short hand notations for such expressions. However, they did not go to the next stage and replace the 2, 6, 12 by indeterminates a, b, c thus getting x3 + ax2 + bx = c. Rather, they dealt with the specic equation with explicit coecients and used techniques that could work with many other coecients. We have seen this approach in Euclid. It persisted into the seventeenth century and the work of Vite and Descartes. 3.2.4 Exercises.
1. Show that there are no rational solutions to the equation x3 + 2x2 + 10x = 20.
76
(Hint: If x =
a b
in lowest terms then a3 + 2a2 b + 10ab2 = 20b3 .
Thus every prime divisor of b divides a. Hence b = 1. Now conclude a divides 20 and check all of the cases.) 2. Convert Fibonaccis approximation to decimal and check that it is a good approximation.
3.3
3.3.1
The solution of the cubic and quartic
The Tartaglia, Cardano approach to the cubic.
In spite of the fact that modern algebraic notation did not exist in the sixteenth century the general solution to the cubic (degree 3) and to the quartic (degree 4) was deduced by the Italian mathematicians Niccolo Tartaglia (1500-1577) and Gironomo Cardano (1501-1576) for the cubic and Ludovico Ferrari (15221565) for quartic. The history of that endeavor is not the most savory in the annals of mathematics and in fact it is almost certain that the solution to the cubic is in fact due to Scipione del Ferro (1465-1526) but unpublished. It seems that neither Tartaglia nor Cardano were morally as strong as they were mathematicians. It also seems that Tartaglias role in the solution of the cubic was much more substantial than that of Cardano, although he seems to have been inuenced by the rumors that a solution by Ferro existed.. We will leave these historical questions aside and just point out that there is an English translation of the Ars Magna published by the M.I.T Press (1968), translated by T.R.Witner, also the book A History of Mathematics by C.B.Boyer, et al has an interesting discussion of this history and further references. What seems to be well documented is that Tartaglia could solve a general cubic of the form x3 + ax = b. It is quite conceivable that Cardanos contribution was to reduce the general cubic x3 + cx2 + dx = e to this form. In modern notation this is a fairly simple task. Set x = y u. Then the equation says y 3 3uy 2 + 3u2 y u3 + cy 2 2cuy + cu2 + dy du = e. So if u =
c 3
then the equation is given in y as y 3 + ay = b
with a = d c3 and b = dc + e 2c . This step seems to be truly minor in 3 27 our notation. But in the sixteenth century the methods used to derive such formulas were completely geometric. Recall that cubes had to be interpreted 77
2
3
as volumes and squares as areas. Also the formulas used had to be given as geometric properties of areas of geometric gures. The nal problem was that negative numbers were still not allowed and so there were many variants of the equations that needed to be analyzed. For example x3 + ax + c = 0 was not seen as the same x3 + ax = b with b = c. Similarly, the term ax might be on the right hand side of the equation. In addition to all of these complications, there was still no direct way of dealing with general quantities such as a and b above (the x was better understood). Thus rather than write the equation above Cardano would consider (say) x3 + 3x = 4. He would than say: Let the cube plus 3 times the side equal 4. The 3 and the 4 would take the place of the a and the b. He would then go through an equivalent geometric discussion to the one below with the special values of a and b. With these provisos we will now derive a solution to the above reduced form of the cubic. The critical idea is to write x = u v . Then substituting in the equation we have u3 3u2 v + 3uv 2 v 3 + a(u v) = b. u3 3uv(u v) v 3 + a(u v) = b. u3 v 3 = b. Now v =
a 3u .
That is
If we take 3uv = a then the equation becomes
So upon substitution we have a 1 u3 ( )3 3 = b. 3 u
Multiply through by u3 and we have a u6 bu3 ( )3 = 0. 3 Apply the quadratic formula to solve for u3 and get r p b b2 + 4( a )3 b a b 3 3 u = = ( )2 + ( )3 . 2 2 2 3 Notice that Cardano must choose the plus sign. Now v 3 = have (at least as a possibility) sr sr 3 3 b 2 a 3 b b a ( ) +( ) + ( )2 + ( ) 3 x= 2 3 2 2 3 78 u3 b. Thus we b . 2
r !3 r !3 q q 3 3 b 2 a 3 b b 2 a 3 b Thus x + ax = (2) + (3) + 2 (2) + (3) 2 = b.
3
This is one of Cardanos solutions (depending on various signs as we have pointed out). Notice that in the course of this development we have made choices. However, if we assume that a > 0 and that the only cube roots we can have are positive then we can reverse the steps sr sr 3 b 2 a 3 b 3 b a b ( ) +( ) + ( )2 + ( ) 3 2 3 2 2 3 2 r r b 2 a 3 b 2 a a 3 ( ) + ( ) ( ) = 2 ( )3 = . = 2 3 2 3 3
3.3.2
Some examples.
First let us give some examples of Cardanos formula. Consider the equation x3 + x2 + x = 14. The rst step is to eliminate the x2 . According to the recipe above, taking c = 1, d = 1, e = 14 we are reduced to y 3 + ay = b with a =
2 3
and b =
385 27 .
We can no plug into the formula
Observe that this expression involves only square roots of positive numbers so at least it makes sense geometrically. If you do the calculation indicated you are looking at r r 385 385 3 17 3 17 57 + 57 y= . 18 54 18 54 We now know that y 1 is a solution to our original equation. If you use 3 a calculator and evaluate this expression numerically you will nd that y 1 3 is approximately 2 and if you substitute 2 into the original equation you will nd that 2 is indeed a solution. This indicates that there could be serious diculties in the use of the elegant formula above. We will look at several other such diculties in the exercises. Cardano and his contemporaries were much more worried about another problem. Which we will now describe. First we must consider the form that was necessary for they must use to write the solution to x3 = ax + b. 79
vs vs u u 2 2 3 3 u 3 3 385 385 u 385 385 2 2 t t + + + . y= 9 54 54 9 54 54
without recourse to negative numbers. We would replace a by a and use the previous solution. Cardano did something equivalent using the substitution x = u + v. This gave rise to the solution s s r r 3 b 3 b b 2 a 3 b a x= + ( ) ( ) + ( )2 ( ) 3 . 2 2 3 2 2 3 If the term under the square root sign was non-negative then he had no trouble understanding the solution. However, if we consider (as Cardano did) x3 = 15x + 4. The formula yields q q 3 3 2 + 121 + 2 121
which made no sense to Cardano. If you ask a mathematical software package to evaluate this expression numerically then it yields 4.0. Direct check shows that x = 4 is indeed a solution. The mathematics package would be hard pressed to see that this expression is exactly 4. (In Maple V version 4 the simplify operation doesnt yield 4. However, the f actor operation does. Mathematica 4 (but not 3)actually returns 4 when it encounters this expression.) We note that x3 15x4 = (x4)(x2 4x+1). This implies that the equation has three distinct roots: 4 and the roots corresponding to the quadratic factor that involve square roots of positive numbers. Getting ahead of ourselves, we will see that if all three roots of a cubic are real and distinct Cardanos formula always involves a square root of a negative number (see exercise 3 below) q q 1. Show that the number 3 17 57 + 385 3 17 57 385 1 is equal to 2 18 54 18 54 3 using Cardanos formula. (Show that the only real root of the corresponding equation is 2.) 2. Observe that if that if x is real then 3 x = 3 x. Use this to see that the choice made in the derivation of Cardanos formula didnt change the outcome. 3. Consider the equation x3 2x = 5. Calculate the solution given by an appropriate variant of Cardanos formula. Next use a calculator or a computer to do Newtons iteration to derive an approximate solution (Newton actually did this calculation to 5 decimal places in 1669.) The Newton method is to guess a solution x0 . The iteration is xn+1 = xn f (xn ) . f 0 (xn ) 3.3.3 Exercises.
Here f (x) = x3 2x 5 and f 0 (x) = 3x2 2. Thus if we start with the approximate root 2 then 1 x1 = 2 + . 10 80
Newton showed that x2 was accurate to 5 decimal places. 4. Let f (x) = (x u)(x v)(x w) with u, v, w distinct and real. Assume that u + v + w = 0. Show that f (x) = x3 ax b with a = u2 + uv + v 2 , b b = (uv)(u + v). Show that ( 2 )2 ( a )3 is always negative. Thus if there are 3 3 real roots then the formula cannot be written directly in terms of real numbers. 3.3.4 The early attempts to explain the paradox. q q 3 3 2 + 121 + 2 121
This strange expression
that must be 4 was a thorn in the side of the remarkable achievement of solving the cubic (and relatively soon the quartic). The resolution of this paradox that the expression must be 4 but involves meaningless objects would not be fully resolved for about 400 years. As we shall see it goes to the heart of what we understand of numbers. Cardano had in earlier studies encountered other types of equations with solutions had a form (u + v) + (u v) and he knew that 2u was indeed a solution. Such numbers are now called conjugate complex numbers and we know that they do indeed add up to a number closer to the sense of Cardano and his contemporaries. Rafael Bombelli (1526-1573) made a proposal the explain the paradox. He suggested the following wild thought. Suppose the cube roots p a pair of conof jugate numbers were conjugate? That is suppose we could write 3 2 + 121 = p u + v and 3 2 121 = u v. Then the irksome sum would be 2u. Since he knew that by all rights 2u should be 4 he chose u = 2. He then comp puted (2 + v)3 = 8 6v + (12 v)2 v. Thus if Bombellis wild thought is to work he must have 8 6v = 2. He was therefore forced to have v = 1 and he found (probably to his own amazement) that 2 + 121 = (2 + 1)3 and 2 121 = (2 1)3 . He now felt justied to plug his newfound cube roots into the Cardano formula and found that with his interpretation the Cardano solution was indeed equal to 4. This brilliant analysis was no doubt very convincing at the time. However, from our perspective it leaves open more questions than it answers. However, before we begin to attempt to study the larger issues we will need to bite the bullet and understand what is meant by numbers. This will be begun in the next section when we discuss analytic geometry. First we will give a short discussion of Ferraris solution of the quartic and another related problem that arises from that result.
81
3.3.5
The solution of the quartic.
We rst describe Ferraris reduction of the solution of the quartic to the cubic. We are considering x4 + ux3 + vx2 + w = mx. Cardanos technique for eliminating the square term in the cubic can be used to eliminate the cube. That is replace x by x u . Then the equation is in the 4 form x4 + ax2 + b = cx. The idea of Ferrari is to complete x4 + ax2 to a square by adding sides. We are thus looking at a a2 (x2 + )2 = ( b) + cx. 2 4 The critical step is to throw in another parameter (say) y in the left hand side and to observe that (x2 + a + y)2 2 a a = (x2 + )2 + 2(x2 + )y + y 2 2 2 a2 = ( b) + cx + 2x2 y + ay + y 2 . 4
2
a2 4
to both
For the last equation we have substituted ( a b) + cx for (x2 + a )2 . This 4 2 term can be written in the form Ax2 + Bx + C. With A = 2y, B = c, C = 2 ( a b) + ay + y 2 . We solve for y so that the quadratic equation has exactly 4 one root. That so that B 2 4AC = 0 (i.e. we eliminate the term in the quadratic formula). Substituting the values of A, B, C we have B 2 4AC = c2 8y(( a2 b) + ay + y 2 ). 4
This is a cubic equation in y. Let u be a root of this equation (which we presumably can nd using Cardanos formula). Then for this value we have (x2 + a + u)2 2 a2 b) + cx + 2x2 u + au + u2 4 B 2 = Ax2 + Bx + C = A(x ( )) 2A c 2 = 2u(x + ) . 4u = (
a c 2 + u)2 = 2u(x + ) . 2 4u This says that to nd a solution x we need only solve (x2 + x2 + a c + u = 2u(x + ). 2 4u 82
That is
To do this we can apply the quadratic formula. The point of this is that in light of Cardanos formula we can write a solution to the quartic as an algebraic expression that involves arithmetic operations on square roots and cube roots of arithmetic operations on the coecients of the equation. This is also true for the cubic and the quadratic formula does the same for degree 2. Of course, we must take into account the same provisos as we did for the cubic. When we study analytic functions of a complex variable we will come back to the sense in which Cardanos solution to the cubic (and thereby Ferraris of the quadric) is actually a well dened solution. In spite of these possible misgivings these results came at least 4000 years after the Babylonians understood how to solve the quadratic equation. The next natural problem was to nd a solution of the quintic (fth degree polynomial) in terms if arithmetic operations (addition, subtraction, multiplication, and division) and square roots, cube roots and fth roots (radicals). The greatest mathematicians of that time and in fact for about the next 200 years could not nd any clever method that would solve this problem. The answer to this problem was given by two of the most tragic cases in the history of mathematics. We will rst discuss the solution of the problem for the quintic. 3.3.6 The quintic
The success of the Italian algebraists of the sixteenth century was extraordinary. The next step would be the quintic and then, of course, equations of higher degree. To the surprise of the mathematical community, there were no clever methods that they could nd to reduce the solution of the quintic to that of the quartic, cubic and quadratic and extraction of fth roots (or for that matter roots of any order). The prevailing idea had always been that one should be able to nd the roots of any polynomial by doing arithmetic operations and extraction of roots. However, in 1799, Paolo Runi (1765-1822) published his two volume treatise Teorie Generale delle Equazioni in which he included an argument to show that there was no such method of solving the quintic. As happens in the history of mathematics, announced proofs of major new results are often incomplete or even wrong. Runi, in fact was on the right track but wrong in detail. One can imagine the scrutiny to which this treatise was subjected. The proof of the impossibility for the quintic was given a rigorous proof by Nicolas Abel (1802-1829) at the age of 19 (notice that he lived at most 27 years!). He published his proof in the form of a pamphlet at his own expense in 1824. Due to his limited funds, he had to keep the pamphlet brief and for that reason it was extraordinarily dicult to understand. He later proved a theorem that applied to all equations of degree 5 and higher. It states If n 5 then there is no formula for a solution involving arithmetic operations and extraction of roots on the coecients of the equation xn + a1 xn1 + a2 xn2 + ... + an = 0. This assertion is now known as the Abel-Runi theorem. A great deal of mathematics ocurred in the time that intervened between the work of Cardano, 83
et. al. and the work of Abel. Most notably, the algebraic notation, which we now take for granted, was invented. Also the understanding and general usefulness if negative numbers was nally a standard part of mathematics. Another major development in the interim was the invention of complex numbers. We will make a rst (relatively geometric) attempt at explaining complex numbers in this chapter and will approach this concept more analytically in the next chapter. These numbers were used in so-called conjugate pairs in Bombellis solution of the apparent paradox in the solution of the cubic. Within the system of complex numbers Carl Friedrich Gauss (1777-1855) proved the fundamental theorem of algebra which states Within the complex numbers every equation xn + a1 xn1 + a2 xn2 + ... + an = 0 with n > 0 has a root. Here we have two seemingly contradictory theorems. Abel asserts that there is no way of writing a formula (involving extraction of roots) for a solution if n 5 and Gauss asserts that even so there is a solution. We will encounter such apparently paradoxical situations throughout our investigations. It should also be pointed out that Abel sent his pamphlet to Gauss who was acknowledged to be the most important mathematician of his time. Gauss was furious with the brevity of the work and made scathing remarks about it. Abels article on this subject was eventually published in Crelles Journal (founded in 1826) in the rst issue which in addition contained 21 other articles by Abel. The theorem of Gauss will be studied in more detail in the later chapters (however, we will discuss an important special case later in this chapter). This theorem appeared in the thesis of Gauss and he went on to give numerous alternative proofs. The most notable aspect of the theorem is that it is not really a theorem in algebra. The reason for this is that, as we shall see, complex numbers (and in fact real numbers) are of a very dierent nature from integers and rational numbers. This dierence is the basis of mathematical analysis (modern calculus). Before we leave this subject in order to learn enough mathematics to discuss it in more depth, we should point out that the theorem of Abel only proves that there is no general formula. Obviously there are equations that we can solve by radicals. For example, x6 2 = 0. It is thus reasonable to ask the question: What equations can be solved by radicals? Here we mean that even though we cant nd a formula we can still write out a solution in terms of the coecients of the equation using arithmetic (that is addition, multiplication and division) and extraction of roots (square roots, cube roots, fth roots, etc.). Abel studied this problem also and laid the partial foundation of the mathematical theory of groups (terms like abelian groups are in the honor of Abels work). However, the solution of this problem was completed by another teenager, Evariste Galois (1811-1832). Galois gave a complete criterion as to when an equation can be solved by radicals. But we are now well ahead of ourselves in this story. We now return to the situation at hand 84
and begin the development of a broader notion of number that would at least include the expressions described (recall Bombellis analysis). In this broader formulation Gausss fundamental theorem of algebra will hold. We will next need to introduce the theory of Galois (now called appropriately Galois Theory) which lays the basis of the theory of groups. The latter will be studied in later chapters. To begin the analysis of the rst problem we must step back to the seventeenth century and study the work if Descartes.
3.4
Analytic Geometry.
Ren Descartes (1596-1650) is now mainly known for the notorious x, y axis and Cartesian coordinates (which we will see he never directly used) and for the quotation I think therefore I am. Both are oversimplications of what he actually did and what he actually meant. We will discuss his geometry which was an important beginning to what we now take for granted in algebra. The work that we will discuss is The Geometry which was published in 1637 as an addendum to his treatise The Discourse on Method. The Geometry consists of three books (we would probably call them sections). The rst establishes a basis for the meaning of number in terms of geometry and establishes the notation that we still use today for polynomials with indeterminate coecients. In the second he shows how one can use his algebraic methods to analyze plane gures in terms of polynomial equations. The third analyzes 3 and higher dimensions. It relates his notation with the earlier works of Cardano, et. al. and for example writes out Cardanos formula in exactly the same way we do. There were several people whose work predated Descartes who understood the idea of independent variable and dependent variable. Nicole dOresme (1323-1382) actually did graphing of data much the way we do today (he did explicitly use what we call Cartesian coordinates thus a more accurate but cumbersome name might be Oresmian coordinates). We will have more to say about the work of this amazing man in the next chapter. Also, Francoise Vite (15401603) established our formalism of unknown quantities that we manipulate in the same way as if they were known numbers. Bombelli, in addition to his work on the mysteries of the cubic and the foundations of complex numbers wrote a treatise on algebra in which he also handled unknowns algebraically. Bombelli did not label his unknowns by letters, instead he invented a symbol for an unknown (not unlike our "frowning face" -( rotated ninety degrees). Fortunately that notation didnt catch on. 3.4.1 Descartes notation and interpretation of numbers.
He begins his rst book with the following assertion (we will use the translation given by David Eugene Smith and Marcia Latham): Any problem in geometry can easily be reduced to such terms that a knowledge of the lengths of certain straight lines is sucient for its construction. Just as arithmetic consists of only four or ve operations, namely, addition,
85
subtraction, multiplication and the extraction of roots, which may be considered a form of division, so in geometry, to nd required lines it is merely necessary to add or subtract other lines; or else, taking one line which I shall call unity in order to relate it as closely as possible to numbers, and which can in general be chosen arbitrarily, and having given two other lines, to nd a fourth which shall be to one of the given lines as the other is to unity (which is equivalent to multiplication);... Before going on let us see what he means here. First he chooses a line segment which he calls unity and in the diagram below (which is essentially the same picture that occurs on page one in The Geometry) is denoted AB on the inclined
line he measures BC (the rst line) and on the line containing AB he measures BD. He then joins the points A and C. From D he draws the parallel line to AC which intersects the line containing BC at E. He then says that if we consider the ratios of BC and BD to AB then the ratio of BE to AB is the product of the corresponding ratios. Let us demonstrate the correctness of this assertion (Descartes feels no need to explain any more than what we have already said). The triangles ABC and DBE are similar. Thus the corresponding sides are all in the same proportion (Euclid, Elements, Book VI, Proposition 10). Thus BE BC BD = AB . If we think of BE as c times a unit, BC as a times a unit, BD as b times a unit and AB as 1 times a unit then the assertion is just that a b = c. Descartes also had a method for doing division geometrically (we will give it as an exercise). We now come to an important point. Often it is not necessary thus to draw lines on paper, but it is sucient to designate each by a single letter. Thus to add the lines BD and GH, I call one a and the other b and write a + b. Then a b will indicate that the line b is subtracted from a; ab is the line a multiplied by b;... Here he is saying that a+b is just the line corresponding to hooking together BD and GH on the same line. ab denotes the geometric operation of multiplication. The symbol a is a bit more abstract than a line since it corresponds to a line measured by a unit (shades of Euclid!). He writes division as we do and aa is a2 and if this is multiplied by a then we have a3 , etc. The square 86
root of a2 + b2 is denoted a2 + b2 . The cube root of a3 b3 + ab2 is written 3 a3 b3 + ab2 and as he says similarly for other roots. Notice that so far he has only taken square roots of combinations of squares and cube roots of combinations of cubes or rectangular solids. Now comes the crux: Here it should be observed that by a2 , b3 and similar expressions, I ordinarily mean only simple lines, which however I name squares, cubes, etc., so that I may make use of the terms employed by algebra. This means that even though we might be looking at a line segment of length 2 times the unit and then considering a cube that has one edge that segment we can think of the corresponding volume as a unitless number 8. This is obvious to us but it was not standard at the time. Further, it leads to our modern approach to numbers being produced by geometry. Descartes goes on to study the variants of the quadratic formula that are formed by changing the signs of the coecients. Again his approach is quite modern and he writes such things as x2 = ax + b2 . To him negative numbers have a right to existence. However, since he still interprets the solution of the equation geometrically he looks three cases the above, x2 = ax + b2 and x2 = ax b2 . We show how Descartes handles the rst two. Consider the following gure:
He takes LM N to be a right triangle with LM = b, LN = a . Now prolong M N 2 to M Q a distance equal to N L. Then x = M Q is the solution. If on the other hand we were considering the equation x2 = ax + b then we would use the same gure from the point N we lay o N P on the line N M with the length of N P = N L. This time x = P M is the answer. The last case is perhaps more interesting we are looking at x2 = ax b2 . For this we consider the following gure:
87
Here LM is of length b, LN is of length a perpendicular to LM and the circle 2 is of radius N L. The line through M is parallel to N L. There are three possibilities. The rst is that the circle cuts the line through M in 2 points R, Q and both M Q and M R are solutions. The second is that the circle touches in one point, say Q, then M Q is the solution. The third is that b > a that is the 2 circle doesnt touch the line through M . Then he asserts that the equation has no solution. The main distinction between Descartes and his contemporaries is that his reason for the geometric constructions is the quadratic formula. So in the rst example x = M Q = QN +N M. QN = a and the Pythagorian theorem 2 q q p 2 2 says that N M = QN 2 + LM 2 = a + b2 so x = a + a + b2 . Which is 4 2 4 the positive solution given by the quadratic formula. We will leave the other cases to the reader in the exercises. The point here is that in Descartes formalism numbers and their units have been separated. Vite had allowed for the handling of unknowns as numbers but he still considered a product of two numbers to be an area, of three a volume. Thus x2 + 2x + 1 makes sense to him as making an area that is a disjoint combination of a square of side x, a rectangle of side x and side 2 and a gure of area 1. For Descartes this "homogenization" is unnecessary. Even Descartes makes a distinction between positive solutions (actual) and negative (false). In the third case he has a third possibility no solution. In the The Geometry he has several results about counting actual solutions or converting false solutions into actual ones. Thus although he did not believe that negative numbers could be actual solutions to geometric problems he was aware of their existence in his algebraic formalism. Book 2 of The Geometry is a study of curves in the plane. Although, the familiar x, y axes of analytic geometry do not appear explicitly in Descartes work they are certainly implicit. We will come back to these ideas in the next chapter. In the next subsection we will consider his approach to what we now call analytic geometry. 3.4.2 Exercises.
1. Show how to divide using the same diagram as Descartes used to give a geometric interpretation of multiplication. 88
2. Consider the gure below. We quote Descartes: If the square root of GH is desired, I add F G equal to unity; then bisecting F H at K, I describe the circle with radius F K with center K, and draw from G the perpendicular and extend it to I, and GI is the required root. Show that this is indeed a geometric interpretation of the square root of GH.
3. Show that Descartes geometric method does indeed describe the solutions to the quadratic equations described above. 3.4.3 Conics and beyond.
The second book of The Geometry contains the meat of the Cartesian method of algebraic geometry it has the title On the nature of Curved lines. The opening paragraph says: The ancients were familiar with the fact that the problems of geometry may be divided into three classes, namely plane, solid, and linear problems. This is equivalent to saying that some problems require only circles and straight lines for their construction, while others require a conic section and still others require more complex curves. I am surprised, however, that they did not go further, and distinguish between dierent degrees of these more complex curves ... The chapter ends with (the perhaps unwarranted) paragraph: And so, I think I have omitted nothing essential to an understanding of curved lines. Descartes begins by rejecting the study of certain curves such as spirals by saying that they really belong only to mechanics. We will study one example from that chapter that rst shows how congurations involving two lines (we will make this more precise) yield conic sections which can be described by quadratic equations and that if one if one allows a line to be a conic section then one has a cubic equation. We will also show how to use Descartes method to derive an equation for an ellipse. We start with the following picture.
89
The angles at A, B, L are right angles. Descartes looks at the curve traced out as follows (this is just a paraphrase of what he actually says the algebra, however is the same as his). The points G and A are xed. The dimensions of the gure KN L are xed and the side KL is slid up and down the line AB. As it slides we look at the path that the point of intersection, C, of the line joining G to L and the line extending KN . He then observes that the two quantities BC and AB determine the point C. Since they are unknown he uses the notation y for BC and x for AB. (This is as close as he gets to the x, y axes.) The quantities that are known (or xed) are AG which he calls a, KL which he calls b and BK LN which he calls c. He then uses similar triangles to observe that KL = BC. . NL b BL+b b Thus c = y . That is, BL = c y b. He uses similar triangles again to y a see that GA = BC . That is, x+BL = BL . This gives aBL = y(x + BL). So AL BL ab b b 2 c c y ab = yx + y( c y b) = yx + c y by. Multiplying through by b we have c 2 ay ac = b yx + y cy. We therefore have c y 2 = (a + c)y xy ac. b Which Descartes observes is the equation of a hyperbola. He then observes that if the gure to be slid were say a hyperbola then one would have gotten a higher order equation. Let us do one more example. We will follow his method to calculate an equation for the locus of points so that the sum of the distances from two xed points is xed (we now call the gure an ellipse). Consider the following diagram
90
P is the point on the ellipse. The two xed points are A and B. C is the midpoint of the line segment AB joining A and B. The line P D is perpendicular to AB. We will call the constant value of the sum of the distances from P to A and to B, 2a. We will also denote by c the length of the segment AC. Then AP + P B = 2a. We set x = CD and y = P D. Then the Pythagorian theorem says that AP 2 = (c + x)2 + y 2 , P B 2 = (c x)2 + y 2 . Thus AP 2 P B 2 Now AP 2 P B 2 = (AP P B)(AP + P B) = 2a(AP P B). We therefore have 2xc . a Since AP P B + AP + P B = 2AP and by the above AP P B + AP + P B = 2xc a + 2a we have xc AP = a + . a Similarly, AP + P B (AP P B) = 2P B. Thus AP P B = PB = a We have a This gives xc 2 a xc . a
x2 c2 a2
= (c + x)2 (c x)2 = (c2 + 2cx + x2 ) (c2 2cx + x2 ) = 4xc.
= y 2 +(cx)2 . Hence a2 2xc + x2 + y 2 = a2 c2 + c2 2 x . a2
= y 2 +c2 2xc+x2 .
91
It is convenient to write b2 = a2 c2 . Then we have x2 + y 2 = b2 + This implies that b2 2 x + y 2 = b2 . a2 Dividing both sides of this equation by b2 we have y2 x2 + 2 = 1. 2 a b Notice that here the x, y axes are not explicitly drawn but they are used implicitly. 3.4.4 Solution to higher degree equations in The Geometry. a2 b2 2 b2 x = b2 + x2 2 x2 . a2 a
The Third Book of The Geometry has the title On the Construction of Solid and Supersolid Problems. This chapter establishes our normal notation for higher degree equations. It also lays the foundation for polynomial algebra. Descartes rst order of business is to make clear that only positive roots of equations are true. We quote: It often happens, however, that some of the roots are false or less than nothing. He gives the example, rst considering (x2)(x3)(x4) = x3 9x2 + 26x 24 with roots 2, 3, 4. He then multiplies by x + 5 that has false root 5 (notice a false root in his sense is still described by a positive number and labeled as false). Thus he has x4 4x3 19x2 + 106x 120 which has three true roots 2, 3, 4 and one false root 5. He uses this as an example for his celebrated Law of Signs. Which says (the assertions refer to the signs occurring before the coecients the coecient of x4 should be taken as +1). An equation has as many true roots as it contains changes of sign from + to - or from - to +; and as many false roots as the number of times two + or two - are found in succession. We note that 0 should be ignored. Thus the quartic example above the signs are +, , , +, thus it has 3 changes + + and two in succession so the theorem asserts that the number of true roots (i.e positive) is 3 and the number of false (negative) is 1. Thus the theorem is correct without any interpretation in this case. In general, there are some caveats. First we must count a root with multiplicity since x2 2x + 1 = 0 has signs +,-,+. Hence 2 sign changes. But it only has one root 1. The method that he nds the example by successively multiplying seems to be all he feels is necessary for a proof of his assertion. In fact, proofs are noticeably absent from Descartes 92
book. However, there are several detailed derivations of formulas. In this case one can easily see that the above assertion is false as stated. If we consider x2 2x + 2 = 0 then the sequence of signs is +,-,+ so he predicts two true roots and no false ones. However there are no real roots of this equation by the quadratic formula. We must therefore interpret the result to be about equations with all roots true or false. The next really substantive part of this chapter involves what happens when one starts with a polynomial in an unknown x and substitutes x = y a (in fact he uses the example of a = 3 in a variant of the polynomial he has been studying. He says that he has increased by 3 every true root and decreased by 3 every false root that is greater than 3. He then gives a full discussion of how to carry out the substitution on a specic quartic. He then considers the same calculation but this time diminishing by a that is substituting x = y + a (again a = 3 and he looks at a specic quartic). These calculations take up a full half page of this extremely terse book. But then we come to the point he says: ...we can remove the second term of an equation by diminishing its true roots by the known quantity of the second term divided by the number of dimensions of the rst term, if these terms have opposite signs; or if they have like signs by increasing the roots by the same quantity. In other words the reduction used in Cardanos Ars Magna. Descartes formulas look just like ours but his text is still far from our approach of considering negative numbers as true objects. It is also important to note that it is here that he considers equations with indeterminate coecients (we know them but they are arbitrary). He later applies these considerations to the cubic and for all practical purposes gives a derivation of Cardanos formula for the solution of the cubic geometrically but writes the formula in exactly the same way that we do. We note that Descartes attributes it to Ferro (he in fact says that ...the rule, attributed by Cardan to one Scipio Ferreus ...). In the rest of the book he considers equations of higher degree mainly 6 and gives methods of solving specic equations. This small book lays the foundation of a synthesis of algebra (polynomials) and geometry. It lays out the power of a notational scheme that has lasted through our time. 3.4.5 Exercises.
1. Prove Descartes rule of signs for polynomials of degree 1,2,3,4 with only real roots. 2. In the derivation of the equation for the ellipse it is necessary to have a > c. However, if we had a < c then we would have had to write b2 = a2 c2 . Follow 93
the line of argument from there to see that y2 x2 2 = 1. a2 b Can you make any sense out of this?
3.5
Higher order equations.
As we indicated Abel proved that, in general, an equation of degree 5 or higher cannot be solved using algebraic operations combined with extraction of roots (i.e. solvable by root extraction). Galois gave a method of determining which equations could be solved. Before we study what these two prodigies actually did in later chapters. In this chapter we will content ourselves with a better understanding of their accomplishment by improving our understanding of the concept of number. We will also resolve some ambiguities that arise in Cardano formula when we include (as we must) complex numbers. We will not attempt, as yet, to be completely rigorous (perhaps we never will) with this concept but will build on Descartes ideas. We rst introduce the concept of complex number more carefully than we did when we studied Bombellis explanation of Cardanos strange example.
3.5.1
Complex numbers.
To Descartes, once a unit square is chosen the square of a number a, a2 must be considered to be the area of the square of side a. Thus the square of a number can never be negative. However, in the Cardano formula we must include the possibility of taking the square root of a negative number. We note that if we wish to allow for this possibility we must only nd a meaning for 1 since if a < 0 then a = b with b > 0. So square root of a could be taken to be 1 b. Thus if we wish to allow square roots of any number we need only make up a symbol for 1. Engineers generally use j and mathematicians use i (for imaginary no doubt). Since there is no real number with our desired property we must throw in our new number i. Now we have a more complex type of number that looks like c = a + bi. We would like to maintain the rules of arithmetic so we are forced into (a + bi) + (c + di) = (a + c) + (b + d)i and (a + bi)(c + di) = (a + bi)c + (a + bi)di = ac + bci + adi + bdi2 = (ac bd) + (bc + ad)i. In other words with only our symbol i thrown in we can with apparent consistency dene an addition and a multiplication. If we assume (as we must) that 94
there is no relation of the form a + bi = 0 with a or b non-zero then we have a system that is consistent with arithmetic. We also note that if a or b is not 0 then (a + bi)(a bi) = a2 (bi)2 = a2 (i)2 (b2 ) = a2 + b2 > 0. Thus if we set a + bi = a bi. Then if c = a + bi, cc = a2 + b2 thus c c . = 2 cc a + b2 So c c c c= 2 =1 a2 + b2 a + b2
c This tells us that if c 6= 0 then 1 exists and is given by a2 +b2 . We now have c a number system that contains the square root of every real number. Let us call (as does everyone else) these numbers complex numbers. We assert that every complex number has a square root. In fact, the Fundamental Theorem of Algebra (mentioned earlier) asserts that every non-constant polynomial with complex coecients has at least one root. The proofs of this theorem involve a deeper understanding of numbers than we have as yet and we will defer this to the next chapter where we will come to grips with the problem of rigorously explaining numbers. We will content ourselves, in this chapter, to showing that every complex number has n-th roots for all n = 2, 3, ... For this we need trigonometry.
3.5.2
Exercises.
1. Show that (1 + i)2 = 2i. 2. If a and b 6= 0 are given real numbers and if c = show that (c + di)2 = a + ib. 3. Use the formula in 2. to calculate i. q
a+ a2 +b2 2
and d =
b 2c
then
3.5.3
Trigonometry.
We have seen in our discussion of Euclids Elements that the Greeks were very interested in the properties of circles. They also had a notion of angle and studied methods of bisecting and trisecting angles. Our trigonometry is based on the understanding that angles can be represented by points on the unit circle. This is motivated by the following calculation. Consider (x + iy)(u + iv) = (xu yv) + (xv + yu)i = t + is. We calculate t2 +s2 = (xu yv)2 + (xv +yu)2 = x2 u2 2xyuv + y 2 v 2 +x2 v 2 +2xyuv +y 2 u2 = 95
x2 u2 + y 2 v 2 + x2 v 2 + y 2 u2 = (x2 + y 2 )(u2 + v 2 ). The conclusion we have been aiming at is that if x2 + y 2 = 1 and u2 + v 2 = 1 then the point that corresponds to (x + iy)(u + iv) also has this property. If we dene the unit circle to be the set of all complex numbers z = x + iy such that x2 + y 2 = 1. Then we conclude that the product of two elements of the unit circle is on the unit circle. We also note that y 0 and x = iy is on the unit if circle then y = 1 x2 . If y < 0 then y = 1 x2 . Thus up to the sign we have parametrized the unit circle in terms of the value of the x coordinate and a sign. We are looking for a better parametrization in terms of a parameter . We wish to have z() = x() + iy() with z(1 )z(2 ) = z(1 + 2 ). If we multiply out we have x(1 )x(2 ) y(1 )y(2 ) = x(1 + 2 ) and x(1 )y(2 ) + x(2 )y(1 ) = y(1 + 2 ). Also we have assumed that x()2 + y()2 = 1. There is an amazing fact that we will prove in our discussion of calculus. It says that if we have two functions of a real parameter satisfying the above three conditions and one more that asserts that if we make a small change in the value of then this induces a small change in the value of each of x() and y() then there is a xed real number c such that x() = cos(c) and y() = sin(c). This is why the rst two equations look so familiar. For the moment we will assume that we are all experts in trigonometry. We have therefore observed that the points of the unit circle can be described as z() = cos() + i sin(). We also know that there is a number with the property that if we consider the values z() for 0 < 2 then every point of the circle has been parametrized with a unique parameter. We also note that if have set up our parameter so that we traverse the circle counter-clockwise then we most have z(0) = z(2). Then using the property z(a + b) = z(a)z(b) we must have z(0)2 = z(0). Now this implies (z(0) 1)z(0) = 0. Since z(0) is not zero (its on the unit circle). We must have z(0) = 1. We can now make an observation due to Abraham De Moivre (1667-1754) (and perhaps to Jean dAlembert p (1717-1783)). If z is a complex number then we can write z = rz() with r = x2 + y 2 and z() = z/r. We presume that we can take arbitrary roots of non-negative real numbers. So if we want an n-th 1 root of z we can take r n z( n ). This says that a complex number has at least one n-th root for each n. We have seen that a positive real number has two square roots x the square-root symbol always stands for the non-negative square root. The point here is the square roots of 1 are 1. The same sort of
96
thing happens in general. If an = bn = z then (a/b)n = 1. Thus the ambiguity in taking roots is contained in the n-th roots of unity. Here is the observation: z( 2k n ) = z(2k) = z(2)k = z(0)k = 1k = 1. n
This is explained in terms of the following picture (here we have plotted 8 equally spaced points on the circle 2k with k = 0, 1, 2, 3, 4, 5, 6, 7) 8
We can see that we are just putting n (in this place 4) equally spaced points on the circle with the rst one 1. Multiplication by the second one clockwise cycles the points clockwise around the circle. Interpretation of Cardanos formula. We now see that there is a real problem with Cardanos formula (and Ferraris for that matter). In Cardano there are two cube roots and two square roots thus there is a possible thirtysixfold ambiguity in the formula. The only way to make it a formula again is to give a rule for how to choose the roots. Lets look at the equation x3 = ax + b again. The formula says v v s s u u 2 2 ub a 3 u b a 3 3 3 b b t t + + 2 2 2 2 2 2
b 2 3 is a root. We note that the formula involves both square roots of 2 a 2 symmetrically. So the ambiguity involves only how we choose cube roots. We b 2 3 will therefore use the same square root, v,of 2 a in both parts of the 3 b formula. We write u = 2 . Then we must choose a cube root of u+v and a cube 97
root of u v so that x = + is a solution to the equation. Let us calculate. With , arbitrary choices of cube roots. Then x3 = 3 + 32 + 3 2 + 3 . We therefore have x3 = u + v + 32 + 3 2 + u v = b + 32 + 3 2 .
2 Now 32 + 3 = 3( + ). We note that 3 3 = (u + v)(u v) = b 2 b 2 3 a 3 u2 v 2 = 2 2 a = 3 . Thus to help resolve the ambiguity of 3 cube roots we choose and so that = a . (Notice that we can do this by 3 multiplying one of or by a complex number whose cube is 1.) We now note that + = x (by our denition) and 3 = 3 a = a. Thus 32 + 3 2 = ax. 3 With these choices and x = + then the equation x3 = ax + b is satised. We also note that since the choice of forces that of and vice-versa there is now only a threefold ambiguity. Which is what we would have if a = 0. The general polynomial of degree three has 3 roots.
3.5.4
Exercises.
1. What are the 4 eighth roots of 1? 2. Find a fourth root of 1 and show that its powers give up to reader the 8 equally spaced points around the circle in the picture above. 3. Resolve the ambiguity in Cardanos formula for a solution of x3 + ax = b. 4. Write out the three roots of the equation x3 + 2x = 4. 3.5.5 Polynomials of degree 5 or higher.
We will begin this section with a special case of Gausss fundamental theorem of algebra. A fuller explanation of the argument in the next subsection will be given in the next chapter. Also the following theorem will be an ingredient in our development of the full theorem. We will see that the result is based on a deeper understanding of the concept of a real number. We will be studying the full fundamental theorem of algebra in the next chapter. Here we will give an argument for polynomials with real coecients of odd degree that uses methods of analysis (the subject of the next chapter). The proof involves a deep property of real numbers which we will assume. The reader who has not had any introduction to the manipulation of inequalities might nd that the proof below is gibberish. Try reading it anyway. The mysteries will be expanded on in the next chapter.
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Polynomials of odd degree. The purpose of this subsection is to discuss the following Let f (x) = a0 + a1 x + ... + an xn be a polynomial with an 6= 0, a0 , ..., an real numbers and n odd. Then there exists a real number c such that f (c) = 0. Notice this assertion is not about arbitrary polynomials but only ones of odd degree and having real coecients. In particular, if the coecients are rational then there is a real root. As we observed above this result is a consequence of a deep property of real numbers which will be delved into more deeply in the next chapter. We rst note that we may assume that an = 1 since we can divide through by an . Thus we are looking at f (x) = xn + (a0 + a1 x + ... + an1 xn1 ). We assume that C > |ai | for all i = 0, ..., n1. Then |a0 +a1 x+...+an1 xn1 | |a0 | + |a1 ||x| + ... + |an1 ||x|n1 . Then |a0 + a1 x + ... + an1 xn1 | C + C|x| + ... + C|x|n1 . Thus if |x| > 1 then we have |a0 + a1 x + ... + an1 xn1 | nC|x|n1 . We now note that if x is real then f (x) xn + nC|x|n1 . Now suppose that x < 0 and |x| > 2nC. Then since n is negative xn = |x||x|n1 . Thus f (x) |x||x|n1 + nC|x|n1 < 2nC|x|n1 + nC|x|n1 = nC|x|n1 < 0. We conclude that if x < 0 and |x| > 2nC then f (x) < 0. We note that if a and b are real numbers then a + b a |b|. Indeed if b 0 then this is an equality and if b > 0 then b > |b|. Thus if x > 2nC and x > 1 then f (x) xn |a0 + a1 x + ... + an1 xn1 | xn nC|x|n1 = |x||x|n1 nC|x|n1 2nC|x|n1 nC|x|n1 = nC|x|n1 > 0. We have thus shown that if x < nC and x < 1then f (x) < 0 and if x > 2nC and x > 1 then f (x) > 0. Fix u < 1 and u < nC. Fix v > 1 and v > 2nC. Then f (u) < 0 and f (v) > 0. The deep property that we need is that if g is a polynomial and if we have two real numbers a < b then for every real number, c, between g(a) and g(b) there exits a real number, y, with a y b such then g(y) = c. We apply this to f . Then since f (u) < 0 and f (v) > 0 the number 0 is between f (u) and f (v) and thus there exists y with u y v and f (y) = 0. The property we have used is called the intermediate value property and it applies in much greater generality than polynomials (as we shall see in the next chapter).
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Why havent we found a paradox? In the previous subsection we have observed that if we have a polynomial of odd degree with real coecients then it has a real root. The existence of the root is demonstrated using a deep property of the real numbers not by giving a formula. This leads to the question: If we have a polynomial of degree 5 with rational coecients then what is the nature of the real numbers that are its real roots? The point is that for degrees 2, 3 and 4 the roots were found it the collection of complex numbers that can be found by the following operations on rational numbers: 1. Arithmetic (addition, subtraction, multiplication and division). 2. Extraction of roots ( 2 x, 3 x, ...) which we now understand how to do using trigonometry. For example in Cardanos formula we must extract a square root of an expression involving the coecients of the polynomial, do arithmetic with that not necessarily rational number combined with a further coecient and then extract cube roots and then subtract these numbers. The amazing outcome of the work of Runi, Abel and Galois is that these operations are not enough to nd all roots of polynomials with rational coecients of degree at least 5. This goes far beyond showing that we cannot nd an explicit formula using only operations of type 1. or 2. It shows that roots of polynomials with rational coecients form an algebraic object that is much more subtle than was imagined. In the next chapters we will endeavor to explain the analysis that is involved in Gausss fundamental theorem of algebra. This analysis comes from the foundations of the dierential and integral calculus which had been developed for totally dierent purposes (the determination of velocities, accelerations and tangents). Also the study of roots of equations led to what is now called abstract algebra. The abstraction of addition, multiplication and division. The whole is a startling edice that will be one of the main subjects of the remaining chapters. We will content ourself with a discussion of why the Abel, Runi, Galois theory was a surprise to the mathematicians of the eighteenth and nineteenth centuries (as the theory developed). This involves the question of why mathematicians with only the knowledge that there is a formula for a solution of an equation of degree two with (say) rational coecients seemed to expect that there would be analogous formulas in higher degrees? Since it apparently took mankind about 4000 years from the realization that there was a quadratic formula to the time of Cardano and Ferrari when formulae for degrees 3 and 4 were discovered, why were mathematicians not looking for reasons why this couldnt be done? The expectation that it could, in fact, be done was correct for degrees 3 and 4 but denitely incorrect for higher degree. This time it took approximately 200 years to come to the realization that one could not do for degree 5 what was done for 2,3 and 4. This is not unlike the prevailing feeling of mathematicians before the nineteenth century that the parallel postulate was a consequence of the other axioms of Euclids geometry. The point is that 100
the mathematicians had decided that they believed the validity of a statement that they could not prove. Since there was no justication for this belief one should perhaps call it a prejudice. In the latter part of the twentieth century and the beginning of this (the twenty rst) century there has been a new debate on the question of truth without proof Brilliant expositors of mathematics justify their views of computability and articial intelligence with just such an idea. Indeed, the argument is that if a human being can discern the truth of an assertion without a proof then he can nd true statements that could never be found by a computer. Thus human beings must be more than biological computers. We will not enter this fray which is poised at a higher level than we have scaled as yet. Rather, the history of the search for formulae for the solution of polynomial equations and the prejudice that this could be done is perhaps related to a problem with the exibility of the human mind. That so many believe that assertions must be true even though we cant prove them may be related to the prejudices that were at the root core of the horrible events of the twentieth century.
3.5.6
Exercises.
1. Does the intermediate value property: If a polynomial, f (x) with real coecients, takes two values f (a) > 0 and f (b) < 0 then there exists a real number, c, between a and b such that f (c) = 0. Seem obvious? Is it it true if we replace the word real by rational? 2. Let f (x) = x5 + 2x2 + x + 1 show that there is a real root between 1.23 and 1.22 by calculating the two values and seeing that one is negative and the other is positive. If you have access to a computer algebra package you could use it to check that the intermediate root does indeed exist.
4
The dawning of the age of analysis.
Archimedes (287-212 BC) proved that the area of a circle is equal to the area, A, of a right triangle with one side equal in length to the radius and the hypotenuse equal to the circumference (see in discussion in Chapter 2). His method was to observe that the area of the triangle in question is either equal to, strictly less than or strictly greater then the area of the circle. He then inscribes regular polygons with the number of sides increasing indenitely and shows that they are eventually bigger in area than any number strictly less than A he then circumscribes regular polygons of increasingly many sides and shows that eventually they have area less than any number strictly larger than A. He then concludes that the only possibility left is that A is the area. This type of argument replacing direct calculation with upper and lower bounds is the method of modern analysis. The main impetus for the development of a rigorous branch 101
of mathematics which we now call analysis was the need for a consistent underpinning for (what we now call) Calculus (Isaac Newton(1642-1727), Gottfried Leibniz(1646-1716). The term calculus is a generic term that roughly means a method of calculation. It was a revolutionary idea that led to simple methods of calculating areas and tangents in geometry and velocities, accelerations and trajectories in mechanics. In particular, the clever method of Archimedes becomes unnecessary within the framework of Calculus. Unfortunately, the early methods were completely formal implicitly assuming that one can deal with quantities that were so small that their squares could be treated as 0 (uents in the terminology of Newton, innitesimals to others). Although there was no rigorous notion of innitesimal in the seventeenth and eighteenth century the idea led to such amazing simplictions of dicult problems that the theory led to a revolution in mathematics. As we shall see in later chapters a more rigorous approach to solving the same problems was developed in the nineteenth century and was based on the ideas of modern analysis. We wiill see in this chapter that Fermat (1601-1665) had developed methods consistant with modern analysis to compute certain important areas and tangents. But he had no general calculus based on modern analysis. In the twentieth century a more rigorous version of the innitesimal calculus was developed by Abraham.Robinson(19181974) based on a deep understanding of logic which made the formal methods of the seventeenth and eighteenth centuries more acceptable in the twentieth centuries. All attempts at understanding a rm basis of calculus are in the end based on attempts to understand the real number system. This was part of our goal in the previous chapter. There, we showed how Euclid and Descartes had developed numbers out of geometry. Descartes went much further and showed that the algebraic manipulation of numbers could replace the clever methods of geometry. However, Descartes numbers did not have any existance beyond geometry. We also saw that the basic question of whether ther exist roots of polynomial equations and whether or not we can calculate them also devolves on the question: What is a real number and thereby what is a complex number. We will be studying these points in this chapter. The modern formulation of real and complex numbers will have to wait for the next chapter.
4.1
4.1.1
Early aspects of analysis.
Zenos paradox.
We will begin this chapter with a standard puzzle usually attributed to Zeno (490-425 B.C.). Suppose there were a tortoise and a hare (sometimes it is Achilles) such that the hare moves twice as fast as the tortoise. To simplify things we assume that the tortoise can move 1 unit in a second and the hare can move 2 units in a second. Suppose that the tortoise starts moving rst along a straight line and travels a distance d before the hare begins moving along the same line. We then have the following situation in the rst d seconds: the hare 2 is d units from the starting point and the tortoise is 3d . In the next d seconds 2 4 102
the hare has moved to 3d units and the tortoise is at 7d , that is the tortoise is 2 4 still ahead by d units. After the next d seconds the tortoise will be ahead by d 4 8 8 units, etc. Thus the hare will never catch the tortoise! We know that there is something wrong here since it is obvious that the hare will eventually pass the tortoise. Aristotle (384-322 B.C.) used this paradox as evidence for the premise that innity is meaningless. This is, certainly a practical point of view. We cannot do an innite number of operationseach of which take at least a xed amount of time to accomplish. But this is not what is happening in our discussion of the tortoise and the hare. If we redid the steps and did the measurement in xed units of time, say one second. Then after n seconds the tortoise would be at d + n units from the start and the hare would be at 2n units. Thus after (say) d + 1 seconds the tortoise would be at 2d + 1 units from the start and 2 2 the hare would be 2d + 1 units. That is they will pass each other before d + 1 2 seconds elapse. Lets look at our original analysis let us make the problem more 100 concrete by taking d ot be 100. Then after 10 steps the tortoise is 1024 units 100 ahead of the hare. After 20 interations of this procedure it is 1048576 ahead. If say the units were meters then this is less than .0001 meters and the amount of time to travel that far for the tortoise would be that many seconds. This is absurd. There is no way we can measure that small an interval in time (let alone what we would have a few iterations further along). The time intervals are becoming so small as to be meaningless. However this is not a solution to the paradox. For example, it is possible for the toroise to move as far as he wishes even if he moves in certain incriments of time that become arbitrarily small. Here we look at just the tortoise and look at where he is from the start after 1 seccond, a 1/2 half second later, a 1/3 second later,... Then after 2 such time intervals he would have gone 3 = 1.5 units, after 4 he would have gone 2 25 761 2436559 12 2.08, after 8 it would be 280 2.72, after 16 it would be 720720 3.3,after 200 it would be about 5.88. after 10000 it would have gone 9.79 units. We will show that the numbers dened in this way increase without bound (see the section immediately below on the harmonic series). Thus just saying that the time incriments are becoming too small to measure does not resolve the puzzle. Many look upon this puzzle as indicating a need for a better understanding of innity. We will take a dierent approach and explain how the techniques of modern analysis explain that the puzzle is merely a missunderstanding of the nite. 4.1.2 The harmonic series.
In this section we will use the method of Nicole dOresme (1323-1382) to show 1 that the numbers 1 + 1 + 1 + ... + n increase without bound with n. The idea 2 3 of Oresme can be seen as follows: 1+ 1 1 =1+ , 2 2
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1 1 1 1 1 1 1 1 + + >1+ + + =1+ + 2 3 4 2 4 4 2 2 (here we have observed that 1 > 1 ) 3 4 1+ 1 1 1 1 1 1 1 + + + + + + = 2 3 4 5 6 7 8 1 1 1 1 1 1 1 1+ +( + )+( + + + )> 2 3 4 5 6 7 8 1 1 1 1 1 1 1 1 1 1+ + +( + + + )=1+ + + . 2 2 8 8 8 8 2 2 2 The pattern is now clear if we add up 1+ 1 1 1 + + ... + n+1 2n + 1 2n + 2 2
1 We have 2n terms that are all at least as big as 2n+1 . Thus they add up to a 1 1 n number that is at least 2 2n+1 = 2 . The conclusion is that the sum
1+
If we dene the integral logarithm in base 2 by I log2 (N ) = n if 2n1 < N 2n . Then we have 1 1 1 1 + + + ... + 1 + I log2 (n). 2 3 n This beautiful argument actually gives a very good idea of how this series of numbers grows with n. One can show that there exists a constant (Eulers constant) that is usually denoted and another constant we will call which (we shall see just the natural logarithm of 2) such that if we substitute increasingly larger values of n in the expression 1 1 1 + + ... + n n 2 3 2 it becomes smaller than any preassigned (small number). We will discuss this in more detaiol when we talk about logarithms. This constant occurs in many contexts in mathematics and has been calculated to high precision. However, it is not known if it is a rational number. 1+ Exercises. 1. Suppose you have blocks each 1 unit thick 4 units wide and 12 units long (the unit could be inches or cent1meters the dimensions are not terribly important) made of a uniform material. Suppose you were to pile the bocks one on top of each other so that the second overhangs the rst the third overhangs the second, etc. How big an overhang could we achieve? 2. Use a computer algebra package or calculater with high precision and natural logarithms (ln )to calculate 1 1 1 + + ... + ln(n) 2 3 n for large n. What is the value of that your calculation predicts. 1+ 104
1 1 n 1 1 + + ... + n + n 1+ . 2 3 2 1 2 2
4.1.3
Another look at the methods of Archimedes.
As we saw Archimedes developed a method of proving formulas for areas of geometric gures. His method was to have a target value, A, for the area in mind and to show that for any B > A there exists a geometric gure strictly containing the one in question with area that we know how to compute and which is less than B. He then showed that if C < A then there was a gure stricly inside the one in question with area bigger than C. He than concludes that the area is bigger than any number strictly bigger than A and is less than any number strictly bigger than A. He then asserts that this implies that the area must be A. The argument is ingenious but once understood seems self evident. However, there are several (reasonable) assumptions that have been made and there is one problem with the method. We will rst look at the assumptions. The rst is about numbers (which eventually leads to the notion of a Dedekind cut) the then there are two about areas. The assumption about numbers is: 1. If A and D are numbers then A = D if the following two conditions are satised a) Every B satisfying B > A also satsies B > D. b) Every C satsifying C < A also satsies C < D . The rst assumption about areas is: 2. If F and G are subsets of the plane with areas A and B and if every point in F is also in G then A B. It is hard to disagree with these two assumptions. The next is not clearly an assumption at all will eventually become part of the denition of a set with area. 3. Let F be a subset of the plane. Suppose that A is a number such that whenever G is a set that has an area B that contains F then B A and whenever L is a set that is contained in F and has an area C then C A. Then F has area A. The rst assumption is part of the order properties of the real numbers. The second is a property that must be satised if we are to have a reasonable notion of area. The third has to do with the fact that in the contexts that are least weird to mathematicians not every set can be allowed to have an area with condition 2. (and a few more equally obvious conditions) satised. We also indicated that there was a problem with Archimedes method. The problem is that it is a method that proves that a value asserted for an area is the correct one. It gives no method of nding what the value should be. We know that Archimedes was aware that the area of the circle of radius r is r2 . He also clearly knew that the circumference is 2r. The right triangle with
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sides of length r and and the other of length the circumference has area 1 r (2r) = r2 . 2 Thus he has proved the formula that we believe. (Actually what he has done is reduced the problem of calculating the area to the problem of calculating the circumference or vice-versa.) A general method for calculation was one of the main aims in the development of the innitesimal and integral calculus. Archimedes was the rst to calculate the area of a segement of a parabola. We will not go into his derivation but say that his basic axioms of area were also used and the area was given in terms of the area of a triangle that one can only feel was an outgrowth of an amazing insight. After we discuss calculus we will explain Archimedes remarkable formula. Recall that a parabola is a conic section that is determined by a point, P , and a line, L. The curve is the locus of points whose distance to P is equal to its distance to the line L. For instance
The point P is called the focus of the parabola and the line perpendicular to L through P is called the axis of the parabola. The Greeks understood that for a conic section a line could intersect it in two, one and no points. A line that intersected the curve at one point would be called the tangent line to that point (it was known to be unique). Archimedes in ??? set about to calculate the area of what he called a section of the parabola, that is,the set is cut out by a line intersecting with the parabola at two points A and B.
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He assumes that the point A and B are on dierent sides of the axis and A is closer. He then considers the triangle ABC formed by the tangent line through B the segment AB and the line parallel to the axis of the parabola through A.
The theorem of Archimedes is: The area of the parabolic segment is one third the area of the triangle ABC. This theorem is one of the high points of Greek geometry. Archimedes approach (as we have pointed out) involved a guess of the area and then trhough brilliant upper and lower estimates proving that his asserted area is correct. In fact, he had a method of deciding what the appropriate value should be that involved what he called The Method. This method was based on what is now called statics in physics involving the theory of levers and pulleys. So in addition to being one of the greatest mathematicians who ever lived he was also a great physicist. The story of how The Method was rediscovered after seeming to be lost for over a thousand years is also very interesting. We refer to more standard texts in the history of mathematics for this story (e.g. C. Boyer et. al. A History of Mathematics).
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Exercises. 1. Derive an equation for the parabola using the plane coordinates (x, y) if we take the line L to be given by y = 1 and the focus to be the point (0, 1 ). 4 4 2. For the parabola in problem 1. show that if the line AB is parallel to the axis then the endpoints are given with A having x-coordinate a and B having x-coordiane a and the area of the indicated triangle is 4a3 so the area of the 3 sector is 4a . 3 3. Show that the theorem of Archimedes shows that the area of the parabolic segment AB depends only on the sum of the distances of A and B to the axis.
4.2
Precursors to calculus.
As mentioned above, Nicole dOresme had developed methods for studying the growth of innite sequences of numbers. He also understood fractional powers of positive numbers and most astonishingly used graphical methods to plot data (using a horizontal axis for the independent variable and the verticle axis for dependent variable. However, very little progress was made in the years between Archimedes and Oresme in the calculation of areas bounded by curves. One major drawback was that the mathematical notation was still quite cumbersome and the methods of Oresme to visualize were not widely used. In the last chapter we mentioned the work of Vite which explained how to deal with unknown quantities and thereby led to the concept of function. However, his work did not separate the notion of number from geometry. Thus, positive numbers were lengths of intervals, products of positive numbers were areas of rectangles, triple poducts were volumes, etc. He also used cumbersome notation for powers writing something like xcube for what we would write as x3 . Thus he would have x xsquare = xcube. This did not aord a useful formalism for doing algebraic manipulation of polynomials. In our notation, a polynomal: x3 + 2x2 + 5x + 1 would be understood to represent a volume so the 2 would would be in units of length, the 5 in units of area and the 1 would be a volume. You could then visualize a cube of side x a rectangular box with base of side x and height 2, a rectangular box with base of area 5 and height x and a three dimensional gure of volume 1 all attached to each other in some way. This all changed with the French mathematicians of the rst half of the seventeenth century. We have already written about Descartes explanation of how to interpret products of positive numbers as intervals and thereby freed polynomials of units. Of the great mathematicians of this time the one who arguably came the closest to calculus was Fermat. He, in fact, was more in the tradition of Archimedes than the later formal methodology of calculus. That is, more in line with the modern notion of analysis.
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4.2.1
The Pascal triangle and the Leibniz harmonic triangle.
We recall that (x + y)2 = x2 + 2xy + y 2 . Multiplying out we see that (x + y)3 = x3 + 3x2 y + 3xy 2 + y 3 . We can continue to multiply indenitely and we see that (x + y)n has n + 1 terms the ith a multiple of xni y i . If we lay out the coecients we have for n = 0, 1, 2, 3, 4, 5. 1 1 1 1 1 1 5 4 10 3 6 10 2 3 4 5 1 1 1 1 1
Blaise Pascal (1623-1662) observed the pattern that one had a triangle with the two legs all ones and the interior values gotten by adding together the two adjacent values one row up for the interior points. Thus for the fth power we would get 1, 5, 10, 10, 5, 1 and, say, 10 = 4 + 6. The standard method of writing these coecients is n so 5 = 10. With the conventions that n = 0 if i < 0 i 2 i n or i > n. It is convenient to write n = 1 and n = 1 (these account for the 0 outer legs of the triangle). We have n n1 n n2 2 n (x + y)n = xn + x y+ x y + ... + xy n1 + y n . 1 2 n1 If we multiply this identity by (x + y) then we have n n n (x+y)xn1 y+ (x+y)xn2 y 2 +...+ (x+y)xy n1 +y n . (x+y)xn + 1 2 n1 Now n (x + y)xni y i = n xn+1i y i + n xn+1i1 y i+1 . This says that in the i i i product the coecient of xn+1i y i is n n .+ i1 i Which is the pattern Pascal observed. We will call this the generating identity for the binomial coecients. We note that if 0 i n then n 6= 0. i Some years after Pascals discovery Christiaan Huygens (1629-1695) asked Leibniz to sum the series: 2 2 2 + + ... + + ... 1(1 + 1) 2(2 + 1) n(n + 1) that is the sum of the reciprocals
1 , n = 1, 2, 3, .... The rigorous theory of (n+1) 2 summing such series had not as yet been developed. However, Leibniz came up with as solution that incontrovertibly summed the series to 2. Here is what he did. He observed that 1 1 1 = . n(n + 1) n n+1
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This says that if we sum the rst, say, 5, terms we have 1 1 1 1 1 1 1 1 2 1 +2 +2 +2 +2 =2 . 2 1 2 2 3 3 4 4 5 5 6 6
2 If we sum the rst n terms we get 2 n+1 . Thus if we sum a million terms the sum is 2 to 5 signicant gures. The more terms we add the closer the value is to 2. This is essentially the modern version of sumation of innite series. We should, however, point out that one must be careful about formal manipulation of innite series. For example, suppose we want to sum
1 1 + 1 1 + 1 + ... If we sum the rst 2n terms we get (1 1) + (1 1) + ... + (1 1) = 0. If we sum the rst 2n + 1 terms we get (1 1) + (1 1) + ... + (1 1) + 1 = 1. Leibniz felt that a reasonable value for the sum of this series should be 1 . 2 Returning to the reciprocals of the binomial coecients n+1 Leibniz made 2 a beautiful discovery that allowed him to compute many more innite series using exactly the same trick.. He rst observed Pascals triangle could be written somewhat dierently as 1 1 1 1 1 1 1 . . . 1 2 3 4 5 6 7 . . . 1 3 6 10 15 21 28 . . . 1 1 1 1 4 5 6 7 10 15 21 28 20 35 56 84 35 70 126 210 56 126 252 462 84 210 462 964 . . . . .. . . . . . . . . .
One observes that except for the rst row and column if we look at an entry in this double array then it is the dierence between the entry directly below and the entry one below and one to the left. Thus 6 in the third row has 10 directly below and four one down and one to the left We have 6 = 10 4. To this see property is true we note that the entries in the rst row are 0 , 1 , 2 , 3 .... 0 0 0 0 Those in the second row are 1 , 2 , 3 , 4 ... in the third 2 , 3 , 4 , 5 ,..., 1 1 1 1 2 2 2 2 j+i2 etc. Thus the entry in the i, j position is i1 . In other words the element in the 3, 5 positions is 82 = 15 that in the 5, 4 position is 7 = 35. The 2 3 j+i2 j+i1 j+i2 moving the negative term to assertion above is just i1 = i i the right hand side we see that this is the generating identity for the binomial coecients. Now if we use the method of Leibniz above we nd that if we consider an entry not on the rst row or column and add up all the entries in the column directly to its left that are either on the same row or a higher row then we get the original entry . For example: 462 = 210 + 126 + 70 + 35 + 15 + 5 + 1. 110
The harmonic triangle of Leibniz is given by 1
1 2 1 3 1 4 1 5 1 6 1 7 1 2 1 6 1 12 1 20 1 30 1 42 1 56 1 3 1 12 1 30 1 60 1 105 1 168 1 252 1 4 1 20 1 60 1 140 1 280 1 504 1 840 1 5 1 30 1 105 1 280 1 630 1 1260 1 2310 1 6 1 42 1 168 1 504 1 1260 1 2772 1 5544 1 7 1 56 1 252 1 840 1 2310 1 5544 1 12012
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. . .
. .. .
1 Here the rst row consists of the numbers 1, 1 , 3, , 1 , ... that is the terms in 2 4 1 the harmonic series. The second row consists of the numbers n(n+1) that is 1 1 1 1 1 1 1 1 1 2 , 6 , 12 , 20 , ... The third row consists of the numbers 3(n+2) that is 3 , 12 , 30 , 60 , ... 3 1 The k-th row has entries k n+k1 these entries can be read o from Pascals ( k ) triangle. Leibniz observation was that the sum of the entries in the k-th row from the n-th poistion on is given by the number in the k 1-st row in the n-th position. Thus the sum
1 1 1 1 1 + + + + ... = . 60 105 168 252 20 That is we are summing the entries in the third row starting with the fourth entry the sum is the fourth entry in the second row. In particular the answer to Huygens question is twice the sum of the entries in the second row which is twice the rst entry in the rst row which is 2. Although the Leibniz method was ingeneous it was a severely limited ,method of summing innite series. A series that looked very similar to the series 1+ is 1 2 + ... + + ... 3 n(n + 1)
1 1 + ... + 2 + ... 4 n This series baed Leibniz who was asked to sum it in 1673 by Henry Oldenburg (1615-1677) and, in fact, all mathematicians until Euler determined its sum in about 1736. We will come back to this series later in this chapter. 1+ Exercises. 1. Use the harmonic triangle to sum the series 1 1 1 + + ... + + ... 6 24 (n + 2)(n + 1)n 2. Use the method that Leibniz used in answering Huygens to show that the entire harmonic triangle works as advertised. Hint: we need to show that 1 1 1 n+k1 n+k = . k k k (k + 1) n+k k k+1 111
3. You may wonder why we called Leibniz array a triangle. If you rotate the rectangular version of Pascals triangle 45 degrees to the right (clockwise) then it it is a triangle. Do the same with the harmonic version. Explain the geometyr of summing series in terms of the version that is given as a triangle. 4.2.2 Fermats calculation of areas.
Fermat considered the problem of calculating the area of a gure bounded by a line pL1 , a line L2 perpenducular to L1 and a curve which we would write as y = x q with p, q > 0 relatively prime integers. He also allowed p to be negative but his method failed for p = 1. We will discuss this case later, although it q was done chronologically earlier. In fact, the case of q = 1 had been handled by several authors who came before Fermat. Here is a picture of Fermats area 5 corresponding to the curve y = x 3 with the base of length 2.
Let the length be denoted m. His idea was as follows consider a number 0 < E < 1 then one has the points E k m for k = 0, 1, 2, .... Which start with m and decrease to indenitely becoming arbitrarily close to 0. He then drew the corresponding rectangles corresponding to the vertical lines through these points he would then have two collections of rectangles one inside and one outside the area in question. In our example above with E = 1 this looks like:v 2
He then sums the areas of the corresponding rectangles: The inner being (Em) q (m Em) + (E 2 m) q (Em E 2 m) + ... + (E k m) q (E k1 m E k m) + ...
p p p
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and the outer being (m) q (m Em) + (Em) q (Em E 2 m) + ... + (E k1 m) q (E k1 m E k m) + .... The idea of Fermat is to add up the rst k terms of these sums we rst look at the outer sum and write it out m q +1 Em q +1 + E q +1 m q +1 E q +2 m q +1 + ... + E q +k m q +1 E p p p p p = m q +1 (1 E) + m q +1 (1 E)E q +1 + ... + m q +1 (1 E)E k q +k + .. = m q +1 (1 E)(1 + E q +1 + ... + E k q +k + ...).
p p p p p p p p p p p kp p kp q +k+1 p p p
m q +1
p
If we set F = E q +1 then the outer sum is given as m q +1 (1 E)(1 + F + F 2 + ... + F k ) with F = E q +1 . Fermat writes this as F = E close this expression 1 + F + F 2 + ... + F k = He now has the expression m q +1
p p p+q q
. We now recall that we can
1 F k+1 . 1F
(1 E)(1 F k+1 ) . 1F We look at G = E q then F = Gp+q and
1
Now comes the brilliant trick. E = Gq . Thus we have
1 Gq 1 G 1 Gq 1E = . = p+q 1F 1G 1 G 1 Gp+q
p 1 Gq 1G (1 E)(1 F k+1 ) = m q +1 (1 F k+1 ) 1F 1 G 1 Gp+q which is equal to
So
m q +1
p
m q +1 (1 F k+1 )
p
1 + G + ... + Gq1 . 1 + G + ... + Gp+q1
Now the total sum over all values (i.e. not stopping at k) is larger than the indicated area. But the only part of the above expression that depends on k is the term 1 F k+1 . Which is always less than one. We conclude that for all values of E with 0 < E < 1 the number m q +1
p
1 + G + ... + Gq1 1 + G + ... + Gp+q1
p
q is an upper bound for the area. If we evaluate this for E = 1 we get m q +1 p+q as an upper bound for the area. If one looks at the expression for the sum of
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the inner rectangles it is just E q times the expression for the outer ones. We therefore nd that p p 1 + G + ... + Gq1 E q m q +1 1 + G + ... + Gp+q1 is a lower bound for the area. We can evaluate this at E = 1 to see that the p p q q area is at least m q +1 p+q and at most m q +1 p+q . Hence it must be equal to
q m q +1 p+q . If we write r =
p
p
p q
then we have the familiar expression (for those who know
r+1
some calculus) that the area is m . r+1 Fermat used a similar method for negative powers, r = p . Here one should q look at the curve over the half line of all numbers x > m. The method involves taking E > 1 and looking at the points m < Em < E 2 m < .... This time the upper sum for the points m, Em, ..., E k m is: m q (Em m) + (Em) q (E 2 m Em) + (E 2 m) q (E 3 m E 2 m) +
p p p p p
... + (E k m) q (E k+1 m E k m).
p p p 2p
This time we can factor out m q +1 and have m q +1 (E1+E q +2 E q +1 +E q m
p +1 q +3
E q
2p
+2
+...+E
kp q +k+1
E
kp q +k
)=
(E 1)(1 + E
p +1 q
p
+E
2 p +2 q
+E
3 p +3 q
+ ... + E
k p +k q
)=
m q +1 (E 1) with F = E 1 q = E
p pq q
. This time we write G = E q and we have
p
1 F k+1 1F
1
m q +1 E
(1 Gq )(1 G(k+1)(pq) ) . (1 Gpq )
Now if p > q then as k is evaluated at increasinly large values the only term involving k is closer and closer to 1. As in the earlier case we have m q +1 E(1 G(k+1)(pq) )
p
1 + G + ... + Gq1 . 1 + G + ... + Gpq1
We see that the upper sum is always at most m q +1 E
p
1 + G + ... + Gq1 . 1 + G + ... + Gpq1
Now eveluating at E = 1 (thus G = 1) we have as an upper bound on the area m
p +1 q
q m q +1 = p . pq q 1
p
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If 0 < p < 1 one can see that this formula is also true In the case of positive q powers it is clear that if we wish an area for 0 < a < x < m then one can subtract the area between 0 and a and get ar+1 mr+1 r+1 r+1 p for r = q . We can see that the formula for the area over the same interval for r = p < 0 but not 1 is q mr+1 ar+1 . r+1 r+1 Exercises. 5 1. For the indicated case of y = x 3 , and m = 2 calculate (using a high precision calculator or math software package) the upper sums for E = 1 , 1 , 1 2 3 5 and say 100 terms. Compare with the answer. 2. Why didnt the method above work for r = 1? 3. Complete the argument for the inner sum in the rst part of the discussion. 4. Complete the argument for r < 1 by analyzing the lower sum. 5. What do you think Fermat did for rational numbers r with 0 > r > 1? 4.2.3 Fermats derivation of tangents.
Notice that this method only works for p > 1. If we need the area over a q nite interval 0 < m < M then we can just subtract the area above M from the area above m and get p p m q +1 M q +1 p . p q 1 q 1
In addition to his calculation of the area under the curves y = xr with r rational but not 1. Fermat also calculated the tangent lines. Here he also used methods that were clear precursors to what we call calculus. He observed that if one has a curve given as y = xn then the slope of the line through the points (x, xn ) and (x + E, (x + E)n ) is (x + E)n xn (x + E)n xn = . (x + E) x E
The gure below is y = x3 and A and B are two such points.
115
He observed that if E is chosen progressively smaller the connecting line would rotate to a tangent line (for the moment we will take this to mean that any line through A gotten by slightly rotating the tangent line intersects the curve at a nearby point, we will come back to the idea of a tangent line). Fermat (and probably many others) observed that if n is an integer then (x + E)n xn = nExn1 + n(n 1) 2 n2 + ... + E n . E x 2
Thus every term is divisible by E. We therefore have n(n 1) n2 (x + E)n xn + ... + E n1 . = nxn1 + Ex E 2 He could then put E = 0 and nds the slope of the tangent line at (a, an ) to be nan1 . However, Fermat did more, he in fact calculated the slope of the tangent if n is only rational. Here we write n = p and assume that p, q > 0. q We are looking at (x + E) q x q ((x + E) q )p (x q )p = . E E We note that if E > 0 then (x + E) q = x q + F with F > 0. Thus taking q-th poweres of both sides of this equation we have x + E = x + qF x1 q +
1 1 1 p p 1 1
q(q 1) 2 1 2 F x q + ... + F q . 2
Thus subtracting x form both sides of this equation and dividing by F we have
1 2 q(q 1) 1 q E + ... + F q1 . = qx1 q + Fx F 2
This means that we can substitute E equals zero in this equation since if E = 0 1 then F = 0. We therefore have if E = 0 then we can evaluate E and get qx1 q . F We now have p p 1 1 (x + E) q x q (x q + F )p (x q )p F = E F E in both we can substitute E = 0 and get p(x q )p1
1
x q 1 = nxn1 . q
1
This certainly shows that Fermat knew a great deal of what we usually think of as basic calculus. However, he did not invent calculus. The point here is that by its very name calculus is a method of computation. Fermat relies on brilliant relationships between rational and integral powers. He is not in the tradition of Archimedes either since he does not use true limits but rather uses a more algebraic formalism that allows substitution. We will discuss these distinctions more carefully when we get to our discussion of calculus. 116
4.2.4
Further precursors to calculus.
Mathematics ourished in the seventeenth century, Mathematicians nally had a notational system that had enough exibility that they could study very general mathematical relationships. Also numbers had nally been divorced from units. Thus numbers could be manipulated algebraically without recourse (unless so desired) to geometric constructs. In Europe mathematicians were analysing areas, volumes, and tangents as they had never been before. As an example, we will take a look at the work of Isaac Barrow (1630-1663) who held the Lucasian Chair at Cambridge before Newton. He was more a geometer than an algebraist and had a low regard for abstract manipulation. His approach to the tangents studied by Fermat would be substantially as in the following discussion (we will, however, replace his geometric arguments with more algebraic ones). He would consider two positive relatively prime integers p and q yielding the curve that is the locus of points (x, y) with y q xp = 0. To calculate the tangent to this curve at the point (a, b), xed and on the courve, he would substitute x = a + u, y = b + v. Thus he would have y q xp = (b + v)q (a + u)p = with E(u, v) a sum of terms involving ur or v s with r, s 2. The term ap bq = 0 by assumption. Thus if u, v had been chosen very small and such that (a + u, b + v) is on the curve then the quantity qbq1 v pap1 u must be very close to 0 (since if u is smaller than 1 then u2 is smaller than u). This indicates that if we had a particle moving along the curve then at the point (a, b) it would be moving in the direction of the line qbq1 v pap1 u = 0. That is along the line y=
p1 p
bq ap + qbq1 v pap1 u + E(u, v)
pap1 x. qbq1
since aq1 = a q 1 . This agrees with Fermats solution. Barrows approach is b now called implicit dierentiation. Exercises. 1. Complete the calculation that Barrows method gives the same answer as Fermats. 2. Use Barrows method to calculate the tangent to the ellipse x2 + y 2 = 1.
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4.3
Calculus.
As we have seen, the rst half of the seventeenth century was brimming with activity on calculations of areas and tangents. A substantial part of what we call calculus had already been discovered before either Newton or Leibniz had begun their work. However, it was these exceptional mathematicians who actually established the calculus. The term calculus means a method of calculation. This is precisely what they developed. Their method unied what had been done before and established rules which if followed would lead to solutions to problems which heretofore were solved using ingeneious methods. As we have seen, one reason for the explosion of activity was the development of a notational system and an abstract formalism that simplied the task of communicating mathematics. In most aspects of the rivalry between Newton and Leibniz (actually the rivalry was between their adherents and desciples) the history gives the edge to Newton. However, when it comes to the notation that would be used in communicating and working with the calculus Leibniz wins hands down. Their independent work was published in several places. Leibniz published A new method for maxima minima as well as tangents in Acta Eruditorum, 1684. A year later Newton published De methis Fluxionen and claimed that the paper was written in 1671. Newtons masterpiece Principia Mathematica was published in 1687. In the introduction to the rst edition he said: In letters that passed between me and that most excellent geometer G.W.Leibniz 10 years ago, when I signied that I knew a method of determining maxima and minima, of drawing tangents and the like, and when I concealed it in transposed letters... the most distinguished man wrote back that he had also fallen on a method of the same kind, and communicated his method which hardly diered from mine except in his forms of symbols. The rst calculus text was published in 1696 by the Marquis de LHospital called Analyse des innement petits which was a compendium of lessons by his private tutor John Bernouli. 4.3.1 Newtons method of uxions.
In this subsection we will describe Newtons approach to dierential calculus. Since Leibniz approach is essentially the same, we will emphasize the notational dierences in the next subsection. Suppose we x the independent variable to be x and y varies with x according to a predetermined rule. We think of the symbol o as indicating a very small change in x this symbol is a uxion and at rst we will take it to be an independently varying very small value. Then we think of x + o as a very small change in x. Now when x has moved to x + o the value of y changes to a new value y + z (not Newtons notation). This z is an arbitrarily small change in y and so to Newton it should be proportional to o. That is z = yo and this proportionality should be a new function of x. The term yo is called a uent and y is the derivative. We will now look at an example. y = xn and n a positive integer. Then
118
using the binomial formula we have (x + o)n = xn + nxn1 o + higher powers of o. Since o is to be thought of as arbitrarily small the terms beyond the rst power cannot contribute to the uent. Thus the uent if nxn1 o and y = nxn1 . This isnt much dierent from what Fermat might do. We nor look at the case when p p n = a .Then y = x q so y q = xp (looks a bit like Barrows start). This says that (y + yo)q = (x + o)p Now (x + o)p = xp + pxp1 o and (y + yo)q = y q + qy q1 yo thus equating coecients of o we have qy q1 yo = pxp1 o. This implies that yo =
p p p p p 1q p1 y x o = x(1q) q xp1 o = x q 1 . q q q
so
Notice that we have neglected the higher powers of o. This is a consistant part of the method. The point here is that it is a method and not just a clever trick. We look at one more example (which is a special case of the chain rule). 1 Consider y = 1x . Then (1 x)y = 1 (1 x o)(y + yo) = 1. (1 x)y oy + (1 x)yo = 1. (1 x)yo = oy yo = y 1 o o= 1x (1 x)2 y= 1 . (1 x)2
Expanding we have Using the relation we have so
and we conclude
Exercises. 1. Calculate y for y = x3 + 3x + 1 using the method of uxions. 2. Suppose that you know y use the method of uxions to calculate z if 1 z = y. 3. Compare for the case of y = xn with n rational compare the method in this subsection with that of Fermat and that of Barrow. 119
4.3.2
The Leibniz notation.
The notation of Leibniz is now the standard method of expression in calculus. He wrote dx for Newtons xo and when y = f (x) then what we denoted by y dy (this is not Newtons notation) he wrote dx . In his notation one sees whar us called the chain rule in modern calculus immediately if y = f (z) and z = g(x) then dy dy dz = . dx dz dx Leibniz is also credited with the product rule (also called the Leibniz rule). Suppose that f (x) = g(x)h(x). That is y = uv witt u = g(x) and v = h(x) then du dv dy = v+u . dx dx dx In fact y + dy = (u + du)(v + dv) = uv +vdu +udv + dudv = y + vdu + udv. Now subtract y from both sides of the equation and divide by dx. This condition has come to be called Leibnizs rule. Exercise. 1. Do problem 2. of the previous section using the chain rule. 4.3.3 Newtons binomial formula
Newton thought of o as small to rst order, that is o2 is negligable and he understood that one could equally well introduce objects small tosecond order say u with u, u2 not negligable but. One would have 1 f (x + u) = f (x) + f(x)u + g(x)u2 2 with g(x) to be dertermined. He looked at f (x) = x m then f 0 (x) = We now expand out m m (x)u + 1 g(x)u2 f (x + u) = f (x) + f = 2 m m1 (x)u + 1 g(x)u2 f (x) + mf (x) f 2 1 m(m 1) + f (x)m2 (f(x)u + g(x)u2 )2 . 2 2 Expanding in powers of u we have x + u = f (x + u)m = x + mx 1 1m 1 m(m 1) m2 x m u + g(x)u2 ) + x m m 2 2 m m1 m 1 1 x+u+ u2 . x m g(x) + x 2 2m
m1 m 1 1 1 1 m . mx
(
1 1m x m m
2
u2 =
120
m 1 1 m m1 x m g(x) + x =0 2 2m 1 we can solve the equation and get g(x) = m1 x m 2 . Observe that m2 So the third term is m1 1 1 = ( 1). m2 m m
1 1 m(m
Thus
1) 1 2 xm . 2 Proceding in this way Newton derived his formula that the k + 1 term is
1 1) ( m k + 1) 1 k xm k1 . Newton intruduced a notation analogous to ours for binomial coecients to denote this expression. In modern notation we write a a(a 1) (a k + 1) = . k k! 1 1 m(m
From this derivation he asserted that 1 1 1 1 1 1 1 1 m x m 1 t + m x m 2 t2 + ... + m x m k tk + ... (x + t) m = x m + 1 2 k This formally derived formula he checked by taking the m-th power. One can then do the same for rational powers and get a a1 a a2 2 a ak k x t+ x t + ... + x t + ... (x + t)a = xa + 1 2 k
This is Newtons binomial series. Exersizes. 1. Consider t or be the independent variable and calculate the derivative of y = (x + t)a . Then dierentiate the individual terms in Newtons series. Check that the two series for the same thing agree. 4.3.4 The fundamental theorem of calculus.
We consider rhe following picture
121
If we think of h as the innitesimal o then we have the area Ao under the curve above the interval is between yo and (y + yo)o. But we can ignore the o2 terms so Ao = yo. This says the A = y. In other words the area under the curve y = f (x) from a to x thought of as a function of x has derivative f (x) at x. Thus the area under the curve between a and b with a < b is F (b)F (a) for any function such that F (x) = f (x). This is the fundamental theorem of calculus which was rst enunciated by Leibniz. Of course, this argument is not in any way complete but it gives a method and that method yields the correct answer in all cases where another technique 1 could be used. For example if f (x) = xm with m 6= 1 then F (x) = m+1 xm+1 and so we have the same outcome as Fermat (which was completely justied). Exercises. 1. Use the fundamental theorem of calculus to derive the special case of Archimedes theorem in exercise 2 of subsection 1.3 of this chapter. 2. This problem is dicult and can be considered a research project. Use the fundamental theorem of calculus to derive the theorem of Archimedes on the area of a sector of a parabola. 4.3.5 Logarithms
As we saw in the last subsection the Newton-Leibniz method has no problem calculation areas once a function is found with the appropriate derivative. The 1 appropriate function for xr is r+1 xr+1 except, of course, for r = 1. We will use the notation y 0 for what we wrote as y and f 0 (x) for f (x). The question then 1 remains what about y 0 = x ? This is serious since it is necessity if we wish to calculate areas related to the hyperbola uv = 1. As it turns out the appropraite function had been discovered before calculus and for dierent reasons. We will digress from our main line and study the history of the missing function. We rst consider x as a function of y. Then y(x(y)) = y. Thus the chain rule says 1 that y 0 (x(y))x0 (y) = 1. But y 0 (x(y)) = x(y) . So x0 (y) = x(y).
1 In other words if we nd a function such that f 0 (x) = x then we would also nd 0 a function such that g (x) = g(x). Such a function with value g(0) = 1 has the remarkable property that g(a + b) = g(a)g(b). That is, it changes addition into multiplication or vice-versa. We are ahead of our story. John Napier (1550-1617) had an interest in making mechanical devices that would allow one to do complicated calculations precisely and easily. Before him there were several methods found of converting multiplication and division to addition and subtraction. One based on trigonometry that was perfected by the Arab mathematicians called prosthapaeresis. We will not go into this method here but suce to say, it helped Tycho Brahe do his intricate calculations and was based on tables involving for sets of trigonometric identities involving multiplication addition and subtraction. Others, notably Michael Stifel (1487?-1567)
122
had observed that if we x a number a then ax ay = ax+y , ax = axy . ay
This certainly changes multiplication and division into addition and subtraction. However, for Stifel there were only caculations of such powers for a, x, y integers and for rational numbers one would need very accurate methods of extracting roots. Napier decided to just use integral powers of a number that had the property that the successive powers were close enough together that if one drew a straight line between the successive values (interpolated) the value would still be within a desired tolerance. This would allow one to use only integral powers in the tables or on the device to be constructed. Napier chose the number N = 0.9999999. He then considered N L to have logarithm L. In order to avoid small decimals he in fact considered 10000000N L as having logarithm L.Now to multiply 10000000N L by 10000000N K all you need do is add K + L look at the table to nd the number with logarithm K + L (or interpolate to get it) and then shift by 7 decimal positions. This was implemented in a slide rule type mechanism. Note that if L = 1 corresponds to the number 9999999 and L = 0 to 10000000. Now if we calculate he value 10000000N L for L = 10000000 we get 3678794 to seven digit accuracy. It therefore gave an ecient method of doing 7 digit multiplication and division. But except for turniing multiplication into division what does it have to do with 1 the problem of nding a function whose derivative is x ? A hint can befound in the following observation. If f (x) satisfyies f 0 (x) = f (x) and f (0) = 1 then to seven signicatnt gures f (1) = 2.718281. The reciprical of this number is 0.3678794 to 7 decimal places. This cannot be an accident. 1 The upshot is that a function whose derivative is x is very dierent then n a function whose derivative is x for any integer other than 1. The function that has this derivative and value 0 at 1 is usually denoted ln(x) and is called the natural logarithm. It is also denoted simply as log(x) when logarithms to base 10 are note being used. Convarting a base involves the simple maneuver of ln(x) multiplying by the logarithm of the inverse. Thus log(x) = ln(10) . The number that yields a natural logarithm of 1 is usually denoted by e. This number is not rational and as observed above it is 2.718281 to seven decimal places. Exercise. Suppose Napier had used 100000000 = 108 so he would have been looking 8 at powers of N = .99999999. What would the Napier logarithm of 108 N 10 be? 4.3.6 The trigonometric functions.
We have seen that the ancient Greeks had an extensive knowledge of trigonometry. We have given an interpretation of trigonometry in section 3.5.3. In particular, the two basic trigonometric functions cos(x) and sin(x). Notice that we are using x for the variable rather than a more traditional Greek letter and 123
thinking of the functions as being attributes of angles. We have seen that these functions have the following properties: 1. 2. 3. 4. cos(0) = 1, sin(0) = 0. cos(x)2 + sin(x)2 = 1. sin(x + y) = sin(x) cos(y) + sin(y) cos(x). cos(x + y) = cos(x) cos(y) sin(x) sin(y).
We will calculate the derivatives of these functions using techniques of analysis. We will use the prime rather than the dot notation. Consider the picture below of a circle of radius 1.
The lengh of AB is the sin() where is the angle AOB. The length of OA is sin() cos(). The length of the arc CB is and the length of CD is tan() = cos() . We note that the area of the triangle COB is sin() . The area of the triangle 2 COD is tan() (see exercise 1below) and the area of the part of the interior of 2 the circle COB is (at least for positive and a rational multiple of that is 2 2 at most ). To see this last assertion note that the area of a quadrant is 2 4 (since the area of the interior of the circle is ). To get a quadrant we take = . If = then we would get half the area which is = . If = 2k 2 4 8 2 1 with k a positive integer than we would have area k times that of the area of the quadrant. That is . If we multiply by a positive integer m and is very 2 small then we get the area of m equal pieces corresponding to . Thus the area is m . We therefore see that if is a positive rational multiple of less than 2 2 or equal to then the area of the piece of the circle is . We now observe that 2 2 since the three areas are nested we have sin() sin() < < 2 2 2 cos() in the range 0 < < . We therefore see that sin() < 1 and sin() > cos() = 2 p p 1 sin()2 . Using 0 < sin() < we see that sin() > 1 2 . From this we see that as we choose positive and progressively smaller the value of sin() is 124
being crushed to 1. This says that the slope of the tangent line to y = sin(x) at x = 0 is 1.That is sin0 (0) = 1. It is easier to see that cos0 (0) = 0. Indeed, since cos(x)2 + sin(x)2 = 1 we can use Leibniz rule to see that 2 cos0 (x) cos(x) + 2 sin0 (x) sin(x) = 0. Subsituting x = 0 gives 2 cos0 (0) cos(0) + 2 sin0 (0) sin(0) = 0. We have cos(0) = 1 and sin(0) = 0. So cos0 (0) is indeed 0. To calculate all derivatives we note sin(x + o) = sin(x) cos(o) + cos(x) sin(o). Now cos(o) = cos0 (0)o = 0 and sin(o) = sin0 (0)o = o. So sin(x + o) = cos(x)o. Similarly, cos(x + o) = cos(x) cos(o) sin(x) sin(o) = sin(x)o. This yields 4. cos0 (x) = sin(x) and sin0 (x) = cos(x). In p above derivation we used the fact that by choosing is small we can the make 1 2 as close to 1 as we wish. This is true and you might think that it is obvious but it does need proof in modern mathematics. We will discuss this point in the exercises. Exersizes, 1. Use the theory of similar triangles to deduce that in the gure above CD = tan , (Hint: CD = OC . AB OA p 2. Here we will sketch that assertion that if 1 1 then 1 1 2 < 2 . First check that p p 2 = (1 1 2 )(1 + 1 2 ). p Next observe that in the range indicated 1 + 1 2 1. Conclude 1 4.3.7 p 1 2 = 2 p 2 . 2 (1 + 1 )
The exponential function.
In this section we will discuss Eulers unication of logarithms and trigonometry. This was essential done section 3.5.3. We will rst take another look at logarithms. We saw that in Napiers work on logarithms the number u = (1 1 n ) n
with n = 10000000 = 107 played an important role. Also, we pointed out that if f (x) were a function with f 0 (x) = f (x) and f (0) = 1 then to seven signicant 1 gures f (1) = u . Euler established the standard notation e = f (1). Now, f (x) satises f (x + y) = f (x)f (y). This is reminicent of the known formula ax+y = 125
ax ay for x, y rational. Further, a0 = 1 is the standard interpretation of the 0-th power and we have by denition a1 = a. This led to the notation f (x) = ex . The distinction is that this function is dened for all real numbers. If we have a = eL then we say that L = ln(a). This denes the natural logarithm that is up to a sign and a shift essentially Napiers logarithm (that is to 7 signicant 1 gures). We have seen that ln0 (x) = x giving the missing derivative. We note that this new function allows us to dene ax for a > 0 and any real number, x, by ax = eln(a)x . In section 3.5.3 we also saw that in the realm of complex numbers if we set z() = cos() + i sin() then the trigonometric identities of the previous section can be written z( + ) = z()z(). We also note that z(0) = 1. This led Euler to dene eix = cos(x) + i sin(x). This allowed the exponential function to be dened for all complex numbers as ex+iy = ex (cos(y) + i sin(y)) . The basic properties are still satised in this context. 1.e0 = 1. 2.ez+w = ez ew . Euler was especially intrigued with the formula ei + 1 = 0 which he called the relationship between the 5 most important constants of mathematics. At this point we are ahead of our story. We need to learn a few things from Eulers teachers. Exercises. 1. What are the 5 constants in Eulers formula? 2. What value would you assign to (ei )i ? How would you interpret the value?
4.3.8
Power series expansions.
We have already encountered Newtons bininomial formula which is an innite series. This formula showed how one might express a function as an innite 1 series. In this case it is the function (1x)a with a rational and 1 < x < 1. One notes that if we dierentiate this k times one gets a(a + 1) (a + k 1) . (1 x)k+a 126
Thus if we call the function f (x) then Newtons binomial formula becomes f (x) = f (0) + f 0 (0)x + f 00 (0) 2 f (k) (0) k x + ... + x + .... 2 k!
Here f (k) is gotten by dierentiating f repeatedly k times. This result was generalized by many authors but most notably Brook Taylor (1685-1773) and later Colin Maclaurin(1698-1746) who are interchangeably named for the generalization. It says that a function can be expanded in the form f (c + x) = f (c) + f 0 (c)(x c) + f 00 (c) f (k) (c) (x c)2 + ... + (x c)k + .... 2 k!
We will call this series the Taylor series of f (x) at c. For example of f (x) = ex . Then f 0 (x) = f (x) and f (0) = 1. Thus f 00 (x) = 0 0 (f ) (x) = f 0 (x) = f (x). So f (k) (0) = 1 for all k. This says that the Taylor series of ex is x2 x3 xk 1+x+ + + ... + + ..... 2 6 k! Similarly we have cos0 (x) = sin(x) and sin0 (x) = cos(x). This gives cos00 (x) = sin0 (x) = cos(x). and sin00 (x) = cos0 (x) = sin(x). This says that even repeated derivatives are given as follows cos(2k) (x) = (1)k cos(x) and sin2k (x) = (1)k sin(x). The odd repeated derivatives are given as cos2k+1 (x) = (1)k cos0 (x) = (1)k+1 sin(x) and sin(2k+1) (x) = (1)k sin0 (x) = (1)k cos(x). Since cos(0) = 1 and sin(0) = 0 we have the Taylor series cos(x) = 1 and sin(x) = x x2 x4 x2k + ... + (1)k + ... 2 4! (2k)!
x2k+1 x3 x5 + ... + (1)k + .... 6 5! (2k + 1)!
If we add together cos(x) + i sin(x) then we have 1 + ix x4 x5 x2k x2k+1 x2 ix3 + + i + ... + (1)k + (1)k + .... 2 3! 4! 5! (2k)! (2k + 1)! 127
if we write z = ix then z 2k = (i2k )x2k = (i2 )k x2k = (1)k x2k and z 2k+1 = ix(z 2k ) = ix(1)k x2k = (1)k ix2k+1 . Thus in terms of z we have 1+z+ z2 z3 zk + + ... + + .... 2 3! k!
This says that Eulers interpretation of eix as cos(x) + i sin(x) is completely consistant with Taylor series. 4.3.9 Eulers summation of a series.
As we mentioned the value of the series 1 1 1 1 + 2 + 2 + ... + 2 + ... 2 3 n was a mystery to some of the greatest minds of the seventeenth and early eighteenth centuries. We have discussed the method of Leibniz that summed the series 1 2 2 1+ + + ... + + ... = 2. 3 34 n(n + 1) One notes that each term of the rst series is less than the corresponding one in the second. This implies that the series sums to a number that is infact less than 2. Before we give Eulers ingeneous deduction of the sum there is another sum that the previous section allows us to calculate. 1+ 1 1 1 + + ... + + ... = e. 2 6 n!
The terms in the sum are fairly simple but the number e is not a simple rational number. One doesnt guess such a value and in fact it had no name until Euler named it. It is therefore not a reasonable idea to just guess an answer. He rst observes that if we have a polynomial of the form 1 a1 x + a2 x2 + ... + an xn and if this polynomial has roots r1 , ..., rn counting multiplicity then if this roots are non-zero we have 1 1 1 + + ... + . a1 = r1 r2 rn To see this we observe that the polynomial with value 1 at 0 and roots r1 , ..., rn is x x x 1 1 1 . r1 r2 rn Now compare the coecient of x. We will come back to this in the exercises. Eulers leap was to apply this observation to an innite series (as in the last section). He considers sin(x) = x x3 x5 x2n+1 + ... + (1)n + ... 3! 5! (2n + 1)! 128
Thus
x2 x4 x2n sin(x) =1 + ... + (1)n + ... x 3! 5! (2n + 1)!
as series with only even powers. Assuming that we can expand it as a polynomial the roots being n wint n = 1, 2, 3, ... So we could expect that this is given by x x x x x x (1 + ) 1 (1 + ) 1 (1 + ) 1 2 n n 2 x2 x2 x2 1 2 1 2 2 = 1 2 4 n which also has even powers. Thus if you consider the series 1 x x2 xn + ... + (1)n + ... 3! 5! (2n + 1)!
then it is reasonable to think that it is given by x x x 1 2 1 2 2 1 2 4 n 1 1 1 1 = 2 + 2 + ... + 2 2 + ... 3! 4 n This yield s
Thus if we apply the observation (valid for polynomials) we have
1 1 1 2 = 1 + + + ... + 2 + ... 6 4 9 n Although no one doubted this as the sum of the series after they saw the marvelous argument the reader should be cautioned that this is not a proof of the formula (as Leibniz derivation is of his value for his series). As it turns out this argument can be made rigorous using a theory of innite products. Indeed, the above innite products can be be proved to converge in a well dened sense to the desired function and the suggested formal manipulation actually gives the Taylor series. Exercise. 1. If f is a polynomial of degree n with roots r1 , ..., rn (allowing for repititions) then f is a multiple of (x r1 )...(x rn ). Assuming that f (0) = 1 show that x x x f (x) = 1 1 1 . r1 r2 rn Hint: The multiple is
(1)n r1 r2 rn .
129
4.3.10
The question of rigor.
There were two controversies that arose in the development of the Calculus. The rst was the question of priority between Newton and Leibniz. It can be said that Newton came out ahead on that issue (although few doubt the independence of Leibniz contribution). However, Newtons apparent victory was one of the causes of the eclipse of English mathematics during the eighteenth century. There are many explanations of this but one the strongest (to our mind) is just that the Leibniz notation was superior. The second contraversy had to do with the very roots of the Calculus. The scientic community knew that caculus gave them an entirely new arsonal of tools to study problems in simple mechanical ways that had been only handled in special cases by methods that were extremely clever and complicated. However, the method of both Newton and Leibniz involved the multiplication and division of objects that were not exactly numbers. Newtons symbol o was an object that one should consider to be such that o2 can be neglected in expressions where it occurs.. The ratio f (x+o)f (x) = f 0 (x) is the derivative. o Leibniz approach was similar he had dx and one should think of (dx)2 = 0 but dy dx was the derivative. Many scientists, philosophers,etc. felt that there was a dangerous lack of foundation for these methods. However, the methods always gave correct answers to the problems to which they were applied. However, the application of the methodology was becoming more and more of a specialty. For example, in the derivation that we gave in the previous section we saw an argument that as it turns out gives the correct answer but is based on a premise that has not been checked. One starts with (at least the hope that) something like the following statement is true. 1. f (0) = 1. 2. f 0 (0) = a. Then a is the sum of the reciprocals of the roots of f (the numbers c such that f (c) = 0. This is true for polynomials if one includes complex roots. However it is denitely false for even very nice functions. For example, ex has the properties 1. and 2. with a = 1. But it is never 0 (this includes using the extended denition ex+iy = ex (cos y + i sin y) of Euler. This says that the argument of Euler is only rigorous if he shows that the function that he dened has the property that the sum of the reciprocals of its roots is the negative of its derivative at 0.This can be done, as we indicated in the previous section by giving a rigorous meaning to the product formula for sin x . x Even before Euler at the very beginning of the development of Calculus there were skeptics about the foundations (not the applications). One of the most serious attacks was made by Bishop George Berkeley (1685-1753). In his pamphlet The Analyst in 1734 he expressed doubts about the foundations of Calculus in particular of Newtons uxions. His point was you cannot have something that behaves like a very small but non-zero number and still has the property that its square is 0. It is clear that you must exercise great care when yo divide by something whose square is o. He labled such objects innitesimals
130
and argued that they cannot have an independent reality. Here is a quote from The Analyst in his discussion of uxions: ...they are neither nite quantities nor quantities innitely small, not yet nothing. In fact, one can develop a rigorous theory with highly restricted classes of functions. For example, if we only consider polynomials then we well see in the next chapter that we can have a completely consistant theory with polynomials in two variables with one variable having the property that it is not 0 but its square is 0. This theory would also wallow for power series and explain why the Newton-Leibniz method always gave the right answer for functions given by power series. A more radical consistant theory which allowed for objects like the ones that Berkeley disparaged was developed by Abraham Robinson in his theory of non-standard analysis. Roughly speaking, he hypthesized the existance of non-standard numbers that were allowed to t between actual numbers. To describe these numbers we must understand our usual number system in a more rigorous manner than we have so far. There were other problems with the foundations of calculus that were less apparent in the seventeenth and eighteenth centuries. This had to do with how careful one must be in the choice of functions that are analyzable using the methods at hand. For example if we considered the function |x| (x if x 0 and x if x < 0) Then if x > 0 we have |x + o| |x| x+ox = =1 o o and if x < 0 we have (x + o) (x) o |x + o| |x| = = = 1. o o o However if we consider x = 0 then we are dealing with |o| . o In other words we must gure out a meaning for |o|. much worse phenomena are possible and can actually occur in useful applications of mathematics. Calculus had to be given a rm footing,
131

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