Assignment 3 Solutions

Assignment 3 Solutions - Regular session Assignment#3 Due...

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Regular session: Assignment #3 Due Tuesday, 2/3 Complete the following exercises from chapter 1.6: �   5, 6,  9, 10,  13, 14, 16, 20, 24, 26 Exercise 5: Prove that if m + n and n + p are even integers, where m, n, and p are integers, then m + p is even. What kind of proof did you use? m + n is even Hypothesis m + n = 2s by definition of even integer n + p is even Hypothesis n + p = 2t by definition of even integer m + n + n + p = 2s + 2t Algebra m + p + 2n = 2s + 2t Algebra m + p = 2s + 2t – 2n Algebra m + p = 2(s + t – n) Algebra m + p is even by definition of even integer This is a direct proof. Exercise 6: Use a direct proof to show that the product of two odd numbers is odd. Let m and n be two odd numbers. m = 2a + 1 by definition of odd integer n = 2b + 1 by definition of odd integer m × n = (2a + 1) × (2b + 1) Algebra m × n = 4ab + 2a + 2b + 1 Algebra m × n = 2(2ab + a + b) + 1 Algebra m × n is odd by definition of odd integer Exercise 8: Prove that if n is a perfect square, then n + 2 is not a perfect square.

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Let n = m 2 . If m = 0, then n + 2 =2, which is not a perfect square, so we can assume that m 1. The smallest perfect square greater than n is (m + 1) 2 , and we have (m + 1) 2 = m 2 + 2m + 1 = n + 2m +1> n+2 1+1>n +2. Therefore n +2 cannot be a perfect square. Exercise 9: Use a proof by contradiction to prove that the sum of an irrational number and a rational number is irrational. Let p be the proposition “ i is an irrational number and r is a rational number” and q be the proposition “sum s of i and r is irrational”. Prove by contradiction means we are assuming q is false and that leads to contradiction to the proposition p . Assuming q is false or “sum s of i and r is rational”. s is rational Assumption s = a / b by definition of rational with a and b being integers and b 0 r = c / d by definition of rational with c and d being integers and d 0 i r s + = Algebra i d c b a + = Algebra i d c b a = - Algebra i bd bc ad = - Algebra i
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