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Unformatted text preview: SECTION 3.5 Primes and Greatest Common Divisors The prime numbers are the building blocks for the natural numbers in terms of multiplication, just as the
elements (like carbon, oxygen, or uranium) are the building blocks of all matter. Just as we can put two
hydrogen atoms and one oxygen atom together to form water, every composite natural number is uniquely
constructed by multiplying together prime numbers. Analyzing numbers in terms of their prime factoriza tions
allows us to solve many problems, such as ﬁnding greatest common divisors. Prime numbers have fascinated
peOple for millennia, and many easy—tdstate questions about them remain unanswered. Students interested
in pursuing these topics more should deﬁnitely consider taking a course in number theory. . In each case we can just use trial division up to the square root of the number being tested. 3) Since 21 : 3  7, we know that 21 is not prime. b) Since 2 X29, 3 X29, and 5 3’29, we know that 29 is prime. We needed to check for prime divisors only up
to m, which is less than 6. c) Since 2 [(71, 3 X71, 5 [71, and 7X71, we know that 71 is prime. (1) Since 2/,‘97, 3 X97, 5X97, and 7,l’97, we know that 97 is prime. e) Since 111 = 3  37, we know that 111 is not prime. f) Since 143 : 11  13, we know that 143 is not prime. . In each case we can use trial division, starting with the smallest prime and increasing to the next prime once we ﬁnd that a given prime no longer is a divisor of what is left. A calculator comes in handy. Alternatively,
one could use a factor tree. a) \Ve note that 2 is a factor of 88, and the quotient upon division by 2 is 44. \Ve divide by 2 again, and then
again, leaving a quotient of 11. Since 11 is prime, we are done, and we have found the prime factorization:
as = 23  11. b) 126:263=2321=233T=2327 e) 7'29=3243=3381233327:33339:33333323G (l) 10012714327 1113 e) 1111 = 11 101 (we know that 101 is prime because we have already tried all prime factors less than 92 11. 13. 15. 17. 19. 21. Chapter 3 The thdamentals: Algorithms, the Integers, and Matrices f) 909090 = 2451545 = 23151515 = 23350505 = 233316835 : 233353367 = 233357481=
2333571337=233571337 101:234'5678‘910:2v3225‘(23)72332(25)=233‘1527 . We give a proof by contradiction. Suppose that in fact log2 3 is the rational number p/q, where p and q are integers. Since log2 3 > 0, we can assume that p and q are positive. Translating the equation log23 : p/q
into its exponential equivalent, we obtain 3 = 21’”. Raising both sides to the qth power yields 3‘? = 2”. Now
this is a violation of the Fundamental Theorem of Arithmetic, since it gives two different prime factorizations
of the same number. Hence our assumption (that log2 3 is rational) must be wrong, and we conclude that log2 3 is irrational. . This is simply an existence statement. To prove that it is true, we need only exhibit. the primes. Indeed, 3, 5, and 7 satisfy the conditions. (Actually, this is the only example, and a harder problem is to prove that there are no others.) The prime factors of 30 are 2, 3, and 5. Thus we are looking for positive integers less than 30 that have none
of these as prime factors. Since the smallest prime number other than these is 7, and T2 is already greater
than 30, in fact only primes (and the number 1) will satisfy this condition. Therefore the answer is 1, 7, ll ,
13, 17, 19, 23, and 2.9. a) Since gcd(11,15) = 1, gcd(11, 19) = 1, and gcd(15, 19) = 1, these three numbers are pairwise relatively
prime. b) Since gcd(15, 21) : 3 > 1, these three numbers are not pairwise relatively prime. c) Since gcd(12,17) : 1, gcd(12,31) = 1, ged(12,37) = l, gcd(17,31) = l, gcd(17,37) = 1, and gcd(31,
37) = l , these four numbers are pairwise relatively prime. (Indeed, the last three are primes, and the prime
factors of the ﬁrst are 2 and 3.) (1) Again, since no two of 7, 8, 9, and 11 have a common factor greater than 1, this set is pairwise relatively prime. The identity shown in the hint is valid, as can be readily seen by multiplying out the righthand side (all the
terms caneel—felescope—exeept for 2‘15 and ~1). We will prove the assertion by proving its contrapositive.
Suppose that n is not prime. Then by deﬁnition n = ab for some integers a and b each greater than 1. Since
a > 1, 2“ e l, the ﬁrst factor in the suggested identity, is greater than 1. Clearly the second factor is greater
than 1. Thus 2" — l = 2‘“ — 1 is the product of two integers each greater than 1, so it is not prime. We compute ¢(n) here by enumerating the set of positive integers less than n that are relatively prime to n.
8) M4) = {1,3} = 2 b) M10) = {1,3,7,9} =4
c) d)(13)=l{1,2,3,4,5.6,7,8,9,10,11,123}! =12 All the positive integers less than or equal to pk (and there are clearlyr pl“ of them) are less than 1)“ and
relatively prime to pk unless they are a multiple of p. Since the fraction l/p of them are multiples of p, we have apt) = p“ (1  Up) :1)" — in“. To find the greatest common divisor of two numbers whose prime factorizations are given, we just need to
take the smaller exponent for each prime. a) The first number has no prime factors of 2, so the gcd has no 2's. Since the ﬁrst number has seven factors
of 3, but the second number has only ﬁve, the gcd has five factors of 3. Similarly the gcd has a factor of 53.
So the gcd is 35  53. Section 3.5 23. 27. 29. . First we ﬁnd the prime factorizations: Primes and Greatest Common Divisors 93 b) These numbers have no common prime factors, so the gcd is 1. c) 2317 d) 41 43  53 e) These numbers have no common prime factors, so the god is 1. f') The god of any positive integer and 0 is that integer, so the answer is 1111. To ﬁnd the least common multiple of two numbers whose prime factorizations are given, we just need to take
the larger exponent for each prime. a) The ﬁrst number has no prime factors of 2 but the second number has 11 of them, so the 1cm has 11
factors of 2. Since the ﬁrst number has seven factors of 3 and the second number has ﬁve, the 1cm has seven
factors of 3. Similarly the 1cm has a factor of 59 and a factor of 73. So the 1cm is 211  37  59  73. b) These numbers have no common prime factors, so the 1cm is their product, 29  37  5'3  73  ll  13‘ 17. c) 2331 d) 414363 e) 212 4313617721, as in part (b) f) It makes no sense to ask for a positive multiple of 0, so this question has no answer. Least common multiples
are deﬁned only for positive integers. 9292s 2 2S  3 112 and 123552 : 25  33  11  13. Therefore
gcd(92928,123552) = 25  3 11 z 1056 and lem{92928,123552) : 2*3 . 33  11*3  13 2 10872576. The re
quested products are (25  3 i 11)  (28  33  l12  l3) and (28  3 v 112)  (25 v 33 ~ 11  13), both of which are
21334113 13 :11,481,440,256. The important observation to make here is that the smaller of any two numbers plus the larger of the two
numbers is always equal to the sum of the two numbers. Since the exponent of the prime p in gcd(a, b) is the
smaller of the exponents of p in a and in b, and since the exponent of the prime p in lcm(a, b) is the larger
of the exponents of p in e and in b, the exponent of p in gcd(o,b)lcm(o,b) is the sum of the smaller and
the larger of these two values. Therefore by the observation, it equals the sum of the two values themselves,
which is clearly equal to the exponent of p in ab. Since this is true for every prime p, we conclude that
ged(o., b)lcm(o, b) and ab have the same prime factorizations and are therefore equal. Obviously there are no deﬁnitive answers to these problems, but we present below a reasonable and satisfying
rule for forming the sequence in each case. a) There are 1’s in the prime loeations and 0’s elsewhere. In other words, the 71““ term of the sequence is 1
if n. is a prime number and 0 otherwise. b) The suspicious 2’s occurring every other term and the appearance of the 11 and 13 lead us to discover
that the 71‘" term is the smallest prime factor of n. (and is 1 when n = l). c) The nth term is the number of positive divisors of n. For example, the twelfth term is 6, since 12 has the
positive divisors 1, 2, 3, 4, 6, and 12. A tip011' to get us going in the right direction is that there are 2’s in
the prime locations. (1) Perhaps the composer of the problem had something else in mind, but one rule here is that the nth term is
0 if and only if n. has a repeated prime factor; the 1’s occur at locations for which n is “squarefree” (has no
factor, other than 1, that is a perfect square). For example, 12 has the square ‘22, so the twelfth term is 0.
e) We note that all the terms (after the ﬁrst one) are primes. This leads us to guess that the 71”] term is the
largest prime less than or equal to n (and is 1 when n = 1). 1') Each term comes from the one before it by multiplying by a certain number. The multipliers are 2, 3,
5, 7, 11, 13, 17, 19, and 23*the primes. So the rule seems to be that we obtain the next term from the
1111‘ term by multiplying by the 71”] prime number (and we start at 1). In other words, the nth term is the product of the smallest. n. — 1 prime numbers. 94 31. 33. 35. 37. Chapter 3 The Fundamentals: Algorithms, the Integers, and Matrices
Consider the product n.(n + l)(n. + 2) for some integer n. Since every second integer is even (divisible by this product is divisible by 2. Since every third integer is divisible by 3, this product is divisible by 3.
Therefore this product has both 2 and 3 in its prime factorization and is therefore divisible by 2  3 2 6. it is hard to know how to get started on this problem. To some extent, mathematics is an experii'nental
science, so it. would probably be a good idea to compute n12 — 7917 + 1601 for several positive integer values
of n, to get a feel for what is happening. Using a computer, or at least a calculator, would be helpful. if
we plug in n : l,2,3,4,and 5, then we get the values 1523, 1447, 1373, 1301, and l231, all of which are
prime. This may lead us to believe that the proposition is true, but it gives us no clue as to how to prove it.
Indeed, it seems as if it would be very hard to prove that this expression always produces a prime number,
since being prime means the absence of nontrivial factors, and nothing in the expression seems to be very
helpful in proving such a negative assertion. (The fact that we cannot factor it algebraically is irrelevantfin
fact, if it factored algebraically, then it would essentially never be prime.) Perhaps we should try some more
integers. if we do so, we find a lot more prime numbers, but. we are still skeptical. Well, perhaps there is some
way to arrange that this expression will have a factor. How about 1601'? Well, yes! if we let a : 160l, then
all three terms will have 1601 as a common factor, so that 1601 is a factor of the entire expression. In fact,
16012 — 79  l601 + 1601 = 1601  1523. So we have found a counterexample after all, and the proposition is
false. Note that this was not. a problem in which we could proceed in a calm, calculated way from problem
to solution. Mathematics is often like thatﬁlots of false leads and approaches that get us nowhere, and then
suddenly a burst of insight that solves the problem. (The smallest in for which this expression is not prime is
n : 80; this gives the value 1681 : 41 '41.) Recall that the proof that there are inﬁnitely many primes starts by assuming that. there are only finitely
many primes p1 , p2, ..., pm and forming the number 331392    pr, + 1. This number is either prime or has
a prime factor different from each of the primes p1, p2, ..., pm; this shows that there are inﬁnitely many
primes. So, let us suppose that there are only ﬁnitely many primes of the form at: +3, namely ql, qg, .. . , q,,, where q1 = 3, qz = 7, and so on. \V hat number can we form that is not divisible by any of these primes, but that must be divisible by a
prime of the form 4k + 3'? Vile might consider the number 4(1qu    qn + 3. Unfortunately, this number is not
prime, as it is is divisible by 3 (because q1 = 3). Instead we consider the number Q = 4q1q2  q,, ~ 1. Note
that Q is of the form 41: + 3 (where k : qlqg   q,, e 1). 11' Q is prime, then we have found a prime of the
desired form different from all those listed. If Q is not prime, then Q has at least one prime factor not in
the list q1, q2, . . . , qn, because the remainder when Q is divided by qj is qj — 1, and (13 i 1 0. Therefore
qj )l’Q for j : 1, 2, . . . , n. Because all odd primes are either of the form 4k + l or of the form all: + 3, and the
product of primes of the form 41: + 1 is also of this form (because (41‘: + 1)(4m + l) = 4(4km + k + m) + 1),
there must be a factor of Q of the form 4}: + 3 different from the primes we listed. This complete the proof. we need to show that this function is oneto—one and onto. In other words, if we are given a positive integer
(r, we must show that there is exactly one positive rational number m/n (written in lowest terms) such that
K(m/n,) : x. To do this, we factor x into its prime factorization and then read off the m and n such that
K (m/n)
of m, with the exponents being half the corresponding exponents in 1:; and the primes that occur to odd = 3;. The primes that occur to even powers are the primes that occur in the prime factorization
powers are the primes that occur in the prime factorization of n , with the exponents being half of one more
than the exponents in 3:. Since this uniquely determines m. and n, there is one and only one fraction, in lowest terms, that maps to a; under K. ...
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This note was uploaded on 02/18/2012 for the course MATH 640:244 taught by Professor Ming during the Spring '09 term at Rutgers.
 Spring '09
 Ming

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