c Let x be a rational number for which x 2 is an integer Suppose for the sake

C let x be a rational number for which x 2 is an

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(c) Let x be a rational number for which x 2 is an integer. Suppose, for the sake of contra- diction, that x is not an integer. (d) Let p and q be primes for which p + q is also prime. Suppose, for the sake of contradic- tion, neither p nor q is equal to 2. (e) Let be a line and let T be a triangle. Suppose, for the sake of contradiction, that intersects all three sides of T . (f) Let C 1 and C 2 be distinct circles. Suppose, for the sake of contradiction, there are three (or more) distinct points where C 1 and C 2 intersect. (g) Suppose, for the sake of contradiction, there were only finitely many primes. 20.5 Let x and x + 1 be consecutive integers. Suppose, for the sake of contradiction x and x + 1 were both even. Since x is even, there is an integer a with x = 2 a . Since x + 1 is even, there is an integer b with x + 1 = 2 b . Adding 1 to both sides of x = 2 a we have x + 1 = 2 a + 1, and since x + 1 = 2 b , we have 2 b = 2 a + 1, which we can rewrite as 2 ( b - a ) = 1 and so b - a = 1 2 . Since b and a are integers, so is 1 2 . But 1 2 is not an integer. ⇒⇐ Therefore, consecutive integers cannot both be even. 20.6 We could write a proof nearly identical to the one from the previous problem (20.5), but here is another approach. Suppose x and x + 1 were consecutive odd integers. Then x + 1 and x + 2 are consecutive even integers (easy to prove), contradicting Exercise 20.5. ⇒⇐ Therefore, consecutive integers cannot both be odd. 20.7 Let p and q be primes for which p + q is also prime. Suppose neither p nor q is 2. It follows that p and q are not divisible by 2, and so are odd. Furthermore, p , q > 2, so p + q > 4. Since p and q are odd, p + q is even, and so divisible by 2. Therefore p + q is not prime. ⇒⇐ Therefore one of p or q must be 2. 20.8 Suppose, for the sake of contradiction, there is a real number x whose square is negative. By the trichotomy property, one of the following holds: x > 0, x = 0, or x < 0. (a) If x > 0, multiplying both sides by x (a positive quantity) gives x 2 > x · 0 = 0, so x 2 is positive. ⇒⇐ (b) If x = 0, then x 2 = 0. ⇒⇐ (c) If x < 0, then x 2 = ( - x )( - x ) and from case (a) we again have x 2 > 0. ⇒⇐
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78 Mathematics: A Discrete Introduction In all cases we reach a contradiction, and so x 2 is not negative. 20.9 Let a and b be real numbers with ab = 0. Suppose, for the sake of contradiction, that neither a nor b is 0. Since b 6 = 0, it has a reciprocal b - 1 . Thus ab = 0 ab · b - 1 = 0 · b - 1 a = 0 which contradicts a 6 = 0. ⇒⇐ Therefore a = 0 or b = 0. 20.10 We must show that 1 < a < a ; we prove each part of this inequality separately. Suppose, for the sake of contradiction, that a 6 > 1. This means that a 1. Since a 1, we have ( a ) 2 1 2 which gives a 1, but a > 1. ⇒⇐ Therefore a > 1. Suppose, for the sake of contradiction, that a 6 < a . This means that a a . Squaring both sides gives a a 2 and then dividing both sides by a gives 1 a , but a > 1. ⇒⇐ Therefore a < a . 20.11 Suppose 4 | n and, for the sake of contradiction, 4 | ( n + 2 ) . Then there are integers a and b such that n = 4 a and n + 2 = 4 b . Subtracting the former from the latter we have 2 = 4 b - 4 a which leads to b - a = 1 2 which is impossible because a and b are integers (and 1 2 is not). ⇒⇐ Therefore n + 2 is not divisible by 4.
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