{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

1.6 (evens)

# 1.6 (evens) - 26 Chapter 1 The Foundations Logic and Proofs...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 a. + 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 .9 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 (2e +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 271 + l, with n. : 20.!) + 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 > n + 2 - 1 + l > u. + 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/!) and CM. 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 lid 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 RIG/f) is also irrational. We give a proof by contradiction. Suppose that rte/b were rational. Since (1/!) .—/; 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 eta/b 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/.e 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 311+ 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. = 2n 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. Therefore the conditional statement is true. This is an example of a trivial proof, since we merely showed that. the conclusion was true. Section 1.6 Introduction to Proofs 27 22. 24. 26. 28. 30. 32. 34. 36. 38. 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 = 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 if n is even then 77? + 4 is even. Since a. is even, it can be written as 2!: for some integer I:. Then 771+ 4 : 141: + 4 : 2(71’: + 2). This is 2 tinres an integer, so it is even, as desired. Next we give a proof by contraposition that if 7n + 4 is even then u 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 = 14!; + 11 : 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 = n, then of course mg : n.2, 2 2 n2. Puttir‘g if m. : -n, then m2 2 (so)2 = (—1)2u"3 = 11.2. For the “only if” part, we suppose that in everything on the left and factoring, we have (m + n)(m v n.) = 0. Now the only way that. a product of two numbers can be zero is if one of them is zero. Therefore we conclude that. either in + n : 0 (in which case in : —-n ), or else in. k n : 0 (in which case in = n), and our proof is complete. We write these in symbols: 0 < b, (a. + b)/2 > a, and (a + 130/2 < D. The latter two are equivalent to o + 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 (i) implies (it), that (it) implies (iii), and that (iii) implies (i). That will suffice. For the ﬁrst, suppose that. :1: : p/q where p and q are integers with q 5:9 0. Then 59/2 : p/(2q), and this is rational, since p and 2e are integers with 2q 3/: 0. For the second, suppose that {if/2 : p/q where p and q are integers with q 79 0. Then 1’ 2 (2p}/q, so 3.1: — 1 : (6p)/q — 1 = (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: ! 1 : p/q where p and q are integers with q # 0. Then :1: : (p/q + 1)/3 = (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 : :5, then we must have :17 2 1 or :5 : ~1. 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 most substitute these values back into the original equation to determine whether they do indeed satisfy it. The onlrr conditional statements not shown directly are p1 4—) pg, ,1}; H [34, and p3 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 <——» 79,] ,sincc p3 (—r p1 and p1 t—r pi. 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). The 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 must 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—third-fourth, ..., eightheninth—tenth, ninth—tenth—ﬁrst, and tenth-ﬁrst-second). 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 . 5-5 : 165. Therefore the average sum is 165/10 : 16.5. 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;) —> (ii) 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 - 72 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 it = 2m, then we would have n3 = 2(4m13), so 113 would be even, and this completes the proof by contraposition that (iii) ——i {1}). Finally, for (a) —> (iv), we see that 71.2 +1 : (2!: + 1)2 + 1 = 4&2 + 4!: + 2 = 2(21c2 + 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 (a). ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern