f2006Solns4 - ECE 190 Fall 2006 Assignment 4 Solutions...

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ECE 190 — Fall, 2006 Assignment 4 Solutions Section 3.1 26 Suppose that m is any odd integer and n is any even integer. By definition of odd and even there exist integers p and q such that m = 2 p +1 and n = 2 q . Then m - n = 2 p +1 - 2 q = 2( p - q )+1. Since p and q are both integers, r = p - q is also an integer and m - n = 2 r + 1 where r is an integer. Thus m - n is odd. 46 Suppose that m and n are any two integers with n - m even. By definition of an even integer, we can let n - m = 2 p where p is an integer. Since n 3 - m 3 = ( n - m )( n 2 + nm + m 2 ), then n 3 - m 3 = 2 p ( n 2 + nm + m 2 ). Since sums and products of integers are integers, then q = p ( n 2 + nm + m 2 ) is an integer and n 3 - m 3 = 2 q where q is an integer. Thus n 3 - m 3 is even. Section 3.6 11 Suppose that a + br is rational with b = 0. By definition of rational numbers, we can write a = i j , j = 0 , b = k l , l = 0 , a + br = m n , n = 0. Then r = m n - a ÷ b = m n - i j ÷ k l = mj - ni nj × l k = ( mj - ni ) l jkn . Now j = 0 , n = 0 and, further, k = 0 because b = 0. By properties of integers, the numerator and denominator are both integers and, since the denominator is the product of non-zero integers, it is not zero. Thus r is rational. We are, however, given that r is irrational and we have a contradiction. Our supposition must be wrong; a + br must be irrational. 18 The contrapositive of the given statement is: “If it is not the case that at least one of two numbers is less than 25, then it is not the case that their sum is less than 50.” Simplifying this, we get “If each of two numbers is at least 25, then their sum is at least 50.” Let
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