Question

An integer n is called even if n = 2m for some integer m, and odd if n + 1 is even. Prove the following statements:

(a) An integer cannot be both even and odd.

(b) Every integer is either even or odd.

(c) The sum or product of even integers is an even integer. What can you say about the sum or product of odd integers?

Answer 1

(a) If an integer is both even and odd, there exist integers m and n such that 2m = 2n + 1.

This implies that 1 = 2(m−n), which means that 1 is even but this is a contradiction since 1(= 0 + 1) is odd. Hence every integer is either ever or odd.

(b) Suppose on the contrary that the set

S = {k ∈ Z| k is neither even nor odd}

is non-empty. Now, we know that T = S^c (complement of S in Z) is non-empty. Choose n ∈ S such

that t := n − 1 ∈ T. Such an n exists because Z is an inductive set. If t is even, this means that n is

odd; if t is odd then n is even. In either case, we get a contradiction since we assumed that n ∈ S.

(c) We have (2n)(2m) = 4nm and 2n + 2m = 2(n + m), so product and sum of even integers is even. Also,

(2m+1)(2n+1) = 2(2mn+m+n)+1, so the product of odd integers is odd. Finally, (2n+1)+(2m+1) = 2(n + m + 1), so the sum of odd integers is even.

(a) If an integer is both even and odd, there exist integers m and n such that 2m = 2n + 1. This implies that 1 = 2(m−n), which means that 1 is even but this is a contradiction since 1(= 0 + 1) is odd. Hence every integer is either ever or odd.

(b) Suppose on the contrary that the set S = {k ∈ Z| k is neither even nor odd} is non-empty. Now, we know that T = S^c (complement of S in Z) is non-empty. Choose n ∈ S such that t := n − 1 ∈ T. Such an n exists because Z is an inductive set. If t is even, this means that n is odd; if t is odd then n is even. In either case, we get a contradiction since we assumed that n ∈ S.

(c) We have (2n)(2m) = 4nm and 2n + 2m = 2(n + m), so product and sum of even integers is even. Also, (2m+1)(2n+1) = 2(2mn+m+n)+1, so the product of odd integers is odd. Finally, (2n+1)+(2m+1) = 2(n + m + 1), so the sum of odd integers is even