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Unformatted text preview: 34 Chapter l The Foundations: Logic and Proofs SECTION 1.6 Introduction to Proofs This introduction applies jointly to this section and the next (1.? . Learning to construct good mathematical proofs takes years. There is no algorithm for constructing the
proof of a true prOposition ( there is actually a deep theorem in mathematical logic that says this). Instead, the
construction of a valid proof is an art, honed after much practice. There are two problems for the beginning
studentifiguring out the key ideas in a problem ( what is it that really makes the proposition true?) and
writing down the proof in acceptable mathematical language. Here are some general things to keep in mind in constructing proofs. First, of course, you need to find out
exactly what is going on—why the proposition is true. This can take anywhere from ten seconds ( for a really
simple proposition) to a lifetime (some mathematicians have spent their entire careers trying to prove certain
conjectures). For a typical student at this level, tackling a typical problem, the median might be somewhere
around l5 minutes. This time should be spent looking at examples, making tentative assumptions, breaking
the problem down into cases, perhaps looking at analogous but simpler problems, and in general bringing all
of your mathematical intuition and training to bear. It is often easiest to give a proof by contradiction, since you get to assume the most (all the hypotheses
as well the negation of the conclusion), and all you have to do is to derive a contradiction. Another thing
to try early in attacking a problem is to separate the proposition into several cases; proof by cases is a valid
technique, if you make sure to include all the possibilities. In proving propositions, all the rules of inference
are at your disposal, as well as axioms and previously proved results. Ask yourself what deﬁnitions, axioms,
or other theorems might be relevant to the problem at hand. The importance of constantly returning to the
deﬁnitions cannot be overstated! Once you think you see what is involved, you need to write down the proof. l n doing so, pay attention both
to content (does each statement follow logically? are you making any fallacious arguments? are you leaving out
any cases or using hidden assumptions?) and to style. There are certain conventions in mathematical proofs,
and you need to follow them. For example, you must use complete sentences and say exactly what you mean.
(Au equation is a complete sentence, with “equals” as the verb; however, a good proof will usually have more
English words than mathematical symbols in it.) The point of a proof is to convince the reader that your line
of argument is sound, and that therefore the proposition under discussion is true; put yourself in the reader’s
shoes, and ask yourself whether you are convinced. Most of the proofs called for in this exercise set are not extremely difficult. Nevertheless, expect to have
a fairly rough time constructing proofs that look like those presented in this solutions manual, the textbook,
or other mathematics textbooks. The more proofs you write, utilizing the different methods discussed in this
section, the better you will become at it. As a bonus, your ability to construct and respond to nonmathematical
arguments (politics, religion, or whatever) will be enhanced. Good luck.’ We must show that whenever we have two odd integers, their sum is even. Suppose that a. and b are two
odd integers. Then there exist integers s and t such that a : 25 + 1 and b 2 2t + 1. Adding, we obtain
a + b = (2s + l) + (2t + l) = 2(s + t + 1). Since this represents a + b as 2 times the integer s + t + l, we
conclude that a. + h is even, as desired. We need to prove the following assertion for an arbitrary integer u: “If n. is even, then 11.2 is even.” Suppose
that n. is even. Then n. = 2k for some integer it. Therefore 77.2 = (2k)2 = 4h2 2 2(2k2). Since we have 2 written n as 2 times an integer, we conclude that. n2 is even. We can give a direct proof. Suppose that m. + n is even. Then m + n = 25 for some integer 3. Suppose that
n. + p is even. Then 71. + p = 2t for some integer t. If we add these [this step is inspired by the [act that we
want to look at m+ p], we get m +P+ 2n. = 2s + 2t. Subtracting 2n from both sides and factoring, we have Section 1.6 11. 13. 15. 17. 19. Introduction to Proofs 35 m + p = 2s + 225 — 2n. = 2(s + t — 71). Since we have written m. + p as 2 times an integer, we conclude that m —— p is even, as desired. . The difference of two squares can be factored: (1.2 — b2 = (a + b)(a — b). If we can arrange for our given odd integer to equal a + b and for a — b to equal 1, then we will be done. But we ean do this by letting a and
b be the integers that straddle n/2. For example, if n = 11, then we take a. 2 G and b 2 5. Specifically, if
n. = 2k+ 1, then we let a : 16+] and b = is. Here, then, is our proof. Since n is odd, we can write a : 2k+ 1
for some integer k. Then (.1; + 1)2 + k2 = k2 + 2k + 1 — k2 : 21:: + 1 : n. This expresses n, as the difference
of two squares. The proposition to be proved here is as follows: If r is a rational number and i is an irrational number, then
s : r +7' is an irrational number. So suppose that r is rational, ’t is irrational, and s is rational. Then by
Example 7 the sum of the rational numbers 3 and +1" must be rational. (Indeed, if s : 0/1) and r : c/d,
where (1., b, c, and d are integers, with b 79 0 and d 713 0, then by algebra we see that s+(+r) : (ad—bc)/(bd),
so that patently s + (—r) is a. rational number.) But 3 + (+7") = r + i + r = 71, forcing us to the conclusion
that i is rational. This contradicts our hypothesis that 1' is irrational. Therefore the assumption that s was
rational was incorrect, and we conclude, as desired, that s is irrational. To disprove this proposition it is enough to find a counterexample, since the proposition has an implied unis
versal quantification. We know from Example 10 that \/2 is irrational. If we take the product of the irrational
number \/2 and the irrational number ﬂ, then we obtain the rational number 2. This counterexample refutes the proposition. We give an proof by contraposition. The contrapositive of this statement is “If 1 is rational, then a: is
rational” so we give a direct proof of this contrapositive. Note that since l/x exists, we know that (c 7'9 0. If
1/32 is rational, then by definition 1/3: = p/q for some integers p and q with q gﬁ 0. Since 1/3; cannot be 0
(if it were, then we’d have the contradiction 1 = a:  0 by multiplying both sides by .17), we know that p 75 0.
Now .7: : 1/(1/rr) : 1/(p/q) = q/p by the usual rules of algebra and arithmetic. Hence 1' can be written as
the quotient of two integers with the denominator nonzero. Thus by definition, :0 is rational. We will prove the eontrapOsitive (that if it is not true that e 2 1 or y 2 1, then it is not true that $+ y 2 2),
using a direct argument. Assume that it is not true that a: 2 1 or y 2 1. Then (by De Morgan’s law) .1: < 1
and y < 1. Adding these two inequalities, we obtain a: + y < 2. This is the negation of :L' + y 2 2, and our proof is complete. a) We must prove the eontrapositive: If n is odd, then H3 + 5 is even. Assume that n is odd. Then we can
write 71 = 2k+1 for some integer 1:. Then n3+5 : (213+ 1)3 +5 = 81:3 +12k2+6k+6 = 2(4k3 +6k2+3k+3).
Thus 713 + 5 is two times some integer, so it is even. b) Suppose that n3 + 5 is odd and that n. is odd. Since a is odd, and the product of odd numbers is odd,
in two steps we see that n3 is odd. But then subtracting we conclude that 5, being the difference of the two
odd numbers n3 + 5 and n3, is even. This is not true. Therefore our supposition was wrong, and the proof by contradiction is complete. The proposition we are trying to prove is “If 0 is a positive integer greater than 1, then 02 > 0.” Our
proof is a vacuons one, exactly as in Example 5. Since the hypothesis is false, the conditional statement is automatically true. 36 21. 23. 25. 27. 29. 31. Chapter 1 The Foundations: Logic and Proofs
The proposition we are trying to prove is “If a and b are positive real numbers, then (a. + 1))1 2 (1‘ +111.”
Our proof is a direct one. By the deﬁnition of exponentiation, any real number to the power 1 is itself. Hence
(a + b)1 2 a, + b 2 (1.1 + ()1. Finally, by the addition rule, we can conclude from (n + 1.5)1 2 a1 + b1 that
(a. + b)L 2 a1 + b" (the latter being the disjunction of (a. + b)1 2 01 +151 and (a. + b)‘ > (21 + bl One might
also say that this is a trivial proof, since we did not use the hypothesis that a and b are positive (although of
course we used the hypothesis that they are numbers). We give a proof by contradiction. If there were nine or fewer days on each day of the week, this would account
for at most 9 ‘ 7 2 63 days. But we chose 64 days. This contradiction shows that at least ten of the days
must be on the same day of the week. One way to prove this is to use the rational root test from high school algebra: Every rational number that
satisﬁes a polynomial with integer coefficients is of the form p/q, where p is a factor of the constant term of
the polynomial, and q is a factor of the coefﬁcient of the leading term. In this case, both the constant and the
leading coefﬁcient are 1, so the only possible values for p and q are :tl. Therefore the only possible rational
roots are 21:1/ (i1), which means that 1 and —1 are the only possible rational roots. Clearly neither of them
is a root, so there are no rational roots. Alternatively, we can follow the hint. Suppose by way of contradiction that a/b is a rational root, where
a and b are integers and this fraction is in lowest terms (that is, a. and b have no common divisor greater
than 1). Plug this proposed root into the equation to obtain (13/!)3 + a/b+ 1 2 0. Multiply through by b3 to
obtain a3 + ab2 + b3 2 0. If a and b are both odd, then the left~hand side is the sum of three odd numbers
and therefore must be odd. If a. is odd and b is even, then the leftehand side is odd + even + even, which is
again odd. Similarly, if a is even and b is odd, then the left—hand side is even + even + odd, which is again
odd. Because the fraction a/b is in simplest terms, it cannot happen that both a and b are even. Thus in
all cases, the left—hand side is odd, and therefore cannot equal 0. This contradiction shows that no such root exists. We must prove two conditional statements. First, we assume that n is odd and show that 511+ 6 is odd (this
is a direct proof). By assumption, in 2 2k: +1 for some integer ls. Then 511+ 6 2 5(2k + 1) + 6 2 10k + 11 2
2(5)}: + 5) + 1. Since we have written 571+ 6 as 2 times an integer plus 1, we have showed that 5n + 6 is
odd, as desired. Now we give an proof by contraposition of the converse. Suppose that n is not odd——iu other
words, that n. is even. Then n. 2 2!: for some integer k. Then 5n + 6 2 10k + 6 2 2(5)}: + 3). Since we
have written 571 + 6 as 2 times an integer, we have showed that 571. + 6 is even. This completes the proof by
contraposition of this conditional statement. This proposition is true. We give a proof by contradiction. Suppose that m is neither 1 nor —1. Then 17m
has a factor (namely (ml) larger than 1. On the other hand, mu. 2 1, and 1 clearly has no such factor.
Therefore we conclude that m 2 1 or m 2 —1. It is then immediate that n 2 1 in the first case and n. 2 —1 in the second case, since mu. 2 1 implies that n 2 1/171. Perhaps the best way to do this is to prove that all of them are equivalent to at being even, which one can
discover easily enough by trying a few small values of If a: is even, then a: 2 2k for some integer 1:.
Therefore 3ft + 2 2 3 ' 2k + 2 2 6k + 2 2 2(3k + l), which is even, since it has been written in the form 2t,
where t 2 3k+1. Similarly, rc+5 2 2k+5 2 2k+4+12 2(k+'2)+l , so n+5 is odd; and 3:2 2 (2.1s)"2 2 2(2k'2),
so :52 is even. For the converses, we will use a proof by contraposition. So assume that :c is not even; thus ct
is odd and we can write 3; 2 2k+1 for some integer is. Then 39: +2 2 3(2k+ 1) +2 2 6!: + 5 2 2(3k+2)+1,
which is Odd (i.e., not even), since it has been written in the form 215+ 1, where t 2 3!: + 2. Similarly,
$+5 2 2k+1+5 2 2(k+3), so n+5 is even (i.e., not odd). That (52 is odd was already proved in Example 1.
This completes the proof. Section 1.6 33. 35. 37. 39. 41. Introduction to Proofs 37 It is easiest to give proofs by contraposition of —> (ii), a (i), —: and (iii) —> For the
ﬁrst of these, suppose that 3m + 2 is rational, namely equal to p/q for some integers p and q with q # 0.
Then we can write :1: = ((p/q) e 2)/3 : (p — 2q)/(3q), where 3g 71 0. This shows that a: is rational. For
the sec0nd conditional statement, suppose that a; is rational, namely equal to p/q for some iutegers p and q
with q 335 0. Then we can write 3:1: + 2 = (330 + 2q)/q, where q 75 0. This shows that 33: + 2 is rational. For
the third conditional statement, suppose that 2 is rational, namely equal to p/q for some integers p and q
with q 0. Then we can write a: = 2p/q, where q 0. This shows that a: is rational. And for the fourth
conditional statement, suppose that :1: is rational, namely equal to p/q for some integers p and q with q 0.
Then we can write :r/2 2 p/(2q), where 2g at 0. This shows that 113/2 is rational. The steps are valid for obtaining possible solutions to the equations. If the given equation is true, then we
can conclude that :L' = 1 or :r = 6, since the truth of each equation implies the truth of the next equation.
However, the steps are not all reversible; in particular, the squaring step is not reversible. Therefore the
possible answers must be checked in the original equation. “Te know that no other solutions are possible, but
we do not know that these two numbers are in fact solutions. If we plug in .1: : 1 we get the true statement
2 = 2; but if we plug in :r = 6 we get the false statement 3 = —3. Therefore a: = l is the one and only
solution of : 3 — 9:. Suppose that we have proved 391 e) 304 6 pg 6 135 6 p3 —> p1. Imagine these conditional statements arranged
around a circle. Then to prove that each one of these propositions (say 13,) implies each of the others (say p} ),
we just have to follow the circle, starting at p,, until we come to 1),, using hypothetical syllogism repeatedly. \Ve can give a very satisfying proof by contradiction here. Suppose instead that all of the numbers a1, 0.2,
.., on are less than their average, which we can denote by A. In symbols, we have a, < A for all i. If we
add these T; inequalities, we see that
a1+a2++an<nA.
By deﬁnition,
0.1 +0.2 ++en_
—n .
The two displayed formulae clearly contradict each other, however: they imply that 71A < nA. Thus our assumption must have been incorrect, and we conclude that at least one of the numbers a, is greater than or A: equal to their average. We can prove that these four statements are equivalent in a circular way: —> a a (to) a For the ﬁrst, we want to show that if n. is even, then H + 1 is odd. Assume that n is even. Then 72. = 2A:
for some integer it. Thus it + 1 = 21.“. + 1, so by deﬁnition 71 + l is odd. This completes the ﬁrst proof.
Next we give a direct proof of a Suppose that n. + 1 is odd, say 7:. + 1 = 2!: + 1. Then
3n+1 = 271+ (31+ 1) : 212+ 2k+1 = 2(n+k) + 1. Since this shows that 3n+1 is 2 times an integer plus 1,
we conclude that 3n+1 is odd, as desired. For the next proof, suppose that 3n+l is odd, say 3n+l : 2tt+ 1.
Then 3n = (371+ l) — 1 = (2!: + 1) — l = 2k. Therefore by deﬁnition 3n is even. Finally, we must prove that
if 311 is even, then 71 is even. We will do this using a proof by contraposition. Suppose that n. is not even.
Then 71 is odd, so we can write 17. : 2k: +1 for some integer k. Then 3n = 3{2k+ l) = bit+3 = 2(3k + 1) + 1.
This exhibits 371. as 2 times an integer plus 1, so 377. is odd, completing the proof by contraposition. ...
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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
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