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Transcript
Discreet Mathematics Chapter 1: The Foundations: Logic and Proofs Section 1.7: Introduction to Proofs Q.2. Use a direct proof to show that the sum of two even integers is even. Solution: Assume n, m even numbers. So we can conclude there are a, b where: n=(2×a) 1 m=(2×b) 2 from 1 and 2 n+m=2×a + 2×b=2(a+b) which is even number. Q.18. Prove that if n is an integer and 3n + 2 is even, then n is even using a) a proof by contraposition. b) a proof by contradiction. Solution: 1. Proof by contraposition: We need to prove that if n is odd, then 3n + 2 is odd. Assume that n is odd. Then we can write n = 2k + 1 for some integer k. Then 3n + 2 = 3(2k + 1) + 2 = 6k + 5 = 2(3k + 2) + 1. – odd number Thus 3n + 2 is odd. 2. proof by contradiction. We need to show that if 3n+2 is even and n is odd, then there is a contradiction. Let n be an arbitrary integer. Assume by way of contradiction that 3n + 2 is even and n is odd. Then n = (2k +1) for some integer k and so 3n+2 = 3(2k +1)+2 = 6k +5 = 2(3k + 2) + 1 So 3n + 2 is odd. So 3n + 2 is odd and even. Contradiction! Q.8. Prove that if n is a perfect square, then n + 2 is not a perfect square. Solution: By contradiction. Assume that both n and n + 2 are perfect square. This means there exist nonnegative integers a, b such that: n = a2 n+2 = b2. Then 2 = (n+2)-n = b2 -a2 = (b-a)(b+a): Therefore : b+a = 2 and b-a = 1, so 2b = 3 and b = 3/2; contradiction with the fact that n is an integer.