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Unformatted text preview: 26 Chapter 1 The Foundations: Logic and Proofs SECTION 1.6 Introduction to Proofs 2. 10. 12. 14. 16. 18. 20. We must show that whenever we have two even integers, their sum is even. Suppose that a and 1') are
two even integers. Then there exist integers s and t such that a : 2s and b : 2t. Adding, we obtain
a + I) : 2s + 26 = 2(s + t). Since this represents [1+ b as 2 times the integer s + t, we conclude that a +1) is
even, as desired. I I We must show that whenever we have an even integer, its negative is even. Suppose that e is an even integer.
Then there exists an integer s such that a. 2 23. Its additive inverse is —2s, which by rules of arithmetic and
algebra (see Appendix 1) equals 2(—s). Since this is 2 times the integer —s, it is even, as desired. An odd number is one of the form 211 + 1, where n. is an integer. “re are given two odd numbers, say 2n + 1
and 2b — 1. Their product is (2o +1)(2b + 1) : deb + 20 + 21) +1 = 2(201) + a + b) + 1. This last expression
shows that the product is odd, since it is of the form 2n + l, with n : 2n!) + a + 1'). Let n = 1712. If m. = 0, then 11+ 2 : 2, which is not a perfect square, so we can assume that m. 2 1. The
smallest perfect. square greater than n is (m + 1)2, and we have (m + 1)2 = 1712 + 2m + 1 : n + 2m + l >
a, + 2  1 + l > 71+ 2. Therefore 71 + 2 cannot be a perfect square. A rational number is a number that can be written in the form :c/y where cc and y are integers and y 3/: 0.
Suppose that we have two rational numbers, say (1/6 and c/d. Then their product is, by the usual rules for
multiplication of fractions, (ac) / (lid). Note that both the numerator and the denominator are integers, and
that fJCf 74 0 since [2 and cf were both nonzero. Therefore the product is, by definition, a rational number. This is true. Suppose that a/b is a nonzero rational number and that :1: is an irrational number. We must
prove that the product ﬂea/b is also irrational. We give a proof by contradiction. Suppose that rte/b were
rational. Since (1/!) r/: 0, we know that a. aé 0, so We is also a rational number. Let us multiply this rational
number b/a by the assumed rational number cro/h. By Exercise 26, the product is rational. But the product is
(b/a)(.1:a /b) = :5, which is irrational by hypothesis. This is a contradiction, so in fact 320/1) must be irrational,
as desired. If a: is rational and not zero, then by deﬁnition we can write a: = p/q, where p and q are nonzero integers.
Since 1/.c is then q/p and p yé 0, we can conclude that 1/1: is rational. \Ve give a proof by contraposition. If it is not true than 77:! is even or u. is even, then m and n are both odd.
By Exercise 6, this tells us that mu is odd, and our proof is complete. a) We must prove the contrapositive: If n. is odd, then 3n + 2 is odd. Assume that n is odd. Then we can
write it I 2!: + 1 for some integer k. Then 3n. + 2 : 3(2k + 1) + 2 : 6!; + 5 : 2(31: + 2) + 1. Thus 371+ 2 is
two times some integer plus 1, so it is odd. b) Suppose that 3n + 2 is even and that. n. is odd. Since 3n. + 2 is even, so is 311.. If we add subtract an odd
number from an even number, we get an odd number, so 311 a n. = 211 is odd. But this is obviously not true.
Therefore our supposition was wrong, and the proof by contradiction is complete. We need to prove the proposition “If 1 is a positive integer, then 12 _>_ 1.” The conclusion is the true statement
1 2 1 .
showed that. the conclusion was true. Therefore the conditional statement is true. This is an example of a trivial proof, since \ve merely Section 1.6 22. 24. 26. 28. 30. 32. 34. 36. 38. Introduction to Proofs 27 We give a proof by contradiction. Suppose that we don't get a pair of blue socks or a pair of black socks.
Then we drew at. most one of each color. This accounts for only,r two socks. But we are drawing three socks.
Therefore our supposition that we did not get a pair of blue socks or a pair of black socks is incorrect, and
our proof is complete. We give a proof by contradiction. If there were at nrost two days falling in the same month, then we could
have at most 2  12 2 24 days, since there are 12 months. Since we have chosen 25 days, at least three of
them must fall in the same month. We need to prove two things, since this is an “if and onl;r if” statement. First let us prove directly that
Then
771+ 4 2 14k + 4 2 2(71'r + 2). This is 2 tinres an integer, so it is even, as desired. Next we give a proof by if n is even then 77? + 4 is even. Since 72. is even, it can be written as 2!: for some integer Ir. contrapositiorr that if 7n + 4 is even then n is even. So suppose that :n is not even, i.e., that n is odd. Then a can be written as 2k + 1 for some integer It. Thus "in + 4 2 14k + 11 2 2(7k + 5) + 1. This is 1 more than 2 times an integer, so it is odd. That completes the proof by contrapoaition. There are two things to prove. For the “if” part, there are two cases. If m 2 n, then of course mg 2 n2;
2 2 (2n) = everything on the left and factoring, we have (m + n)(m v n.) = 0. Now the only way that. a product of two . r)
if m 2 n, then m‘ 2 n2. 2 (—1)2n."3 2 11.2. For the “only if” part, we suppose that m Puttirg
numbers can be zero is if one of them is zero. Therefore we conclude that either in + n 2 0 (in which case in 2 —n ), or else in. k n 2 0 (in which case in 2 n), and our proof is complete. We write these in symbols: 0 < b, (a. + b)/2 > a, and (a + b)/2 < D. The latter two are equivalent to
n + b > 2a and (1+ b < 211, respectively, and these are in turn equivalent to b > a. and a. < b. respectively. It
is now clear that all three statements are equivalent. We give direct proofs that implies (it), that (it) implies (iii), and that (iii) implies That will suffice.
For the first, suppose that. :1: 2 p/q where p and q are integers with q 2 0. Then a:/2 2 p/(2q), and this is
rational, since p and 2e are integers with 2G 0. For the second, suppose that {17/2 2 p/q where p and q
are integers with q 79 0. Then 1’ 2 (2p}/q, so 3.1: — 1 2 (6p)/q — 1 2 (6p — q)/q and this is rational, since
Up 7 q and q are integers with q 9’5 0. For the last, suppose that 3.1: e 1 2 p/q where p and q are integers
with q 2 0. Then :1: 2 (p/q + 1)/3 2 (p + q)/(3q), and this is rational, since p + q and So are integers with
3r; 74 0. No. This line of reasoning shows that if V2.12 — l 2 re, then we must have :17 2 1 or :5 2 —l. These are
therefore the only possible solutions, but we have no guarantee that the;r are solutions, since. not all. of our
steps were reversible (in particular, squaring both sides). Therefore we must substitute these values back into
the original equation to determine whether they do indeed satisfy it. Tire onlrr conditional statements not shown directly are p1 4) pg, 1);; H [34, and 103 H 1)... But these each
follow with one or more intermediate steps: p1 <—> pg. since p1 H p3 and p; H 192.: p2 <—i 194, since 132 H 371
(just established) and 191 (—r 13.1; and 193 <——» p.) ,sincc p3 4—) p1 and 191 H pr. We must find a number that cannot be written as the sum of the squares of three integers. We claim that 7
is such a number (in fact, it is the smallest such number). Tire only squares that can be used to contribute
to the sum are 0. 1 , and 4. “7e cannot use two 4’s, because their sum exceeds 7. Therefore we can use at
most one 4, which means that we n'rust get 3 using just 0’s and 1’s. Clearly three 1’s are required for this,
bringing the total number of squares used to four. Thus 7 cannot be written as the sum of three squares. 28 40. 42. Chapter 1 The Foundations: Logic and Proofs
Suppose that we look at the ten groups of integers in three consecutive locations around the circle (ﬁrst—
second—third, second—thirdfourth, ..., eightheninth—tenth, ninth—tenth—ﬁrst, and tenthﬁrstsecond). Since
each number from 1 to 10 gets used three times in these groups, the sum of the sums of the ten groups must
equal three times the sum of the numbers from 1 to 10, namely 3 v 55 : 165. Therefore the average sum is
165/10 2 165. By Exercise 39, at least one of the sums must be greater than or equal to 16.5, and since the
sums are whole numbers, this means that at least one of the sums must be greater than or equal to 17. We show that each of these is equivalent to the statement (1)) n. is odd, say 31 = 2k+ 1. Example 1 showed that
(2)) implies (i), and Example 8 showed that (6) implies (v). For (1;) —> we see that 1 — n = 1— (2k+ 1) :
2(—.i:) is even. Conversely, if n were even, say It : 2m, then we would have 1 — n = 1 e 2772 : 2('H_l) + 1, so
1  u would be odd, and this completes the proof by coritraposition that (it) —i (v). For (a) —> (iii), we see
that 77.3 : (2L'+1)3 : 8k3+12k2+6k+1 = 2(4k3+6k2+3k)+1 is odd. Conversely, if n were even, say a = 2m,
then we would have n3 = 2(4m13), so 113 would be even, and this completes the proof by contraposition that ——i Finally, for (in) —> (iv), we see that 71.2 +1 : (2!: + 1)2 + 1 = 4&2 + 4!: + 2 = 2019 + 2k +1) is
even. Conversely, if n were even, say n : 2171, then we would have n2 + 1 = 2(27112) + 1 , so :12 + 1 would be
odd, and this completes the proof by contraposition that (is) —i (1:). ...
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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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