Prove that if n is an integer and 3n+2 is even then n is even

Given

If 3n + 2 is even then n is even ∀ n ∈ Z

Proof By Contradiction

We will prove the given statement by contradiction

Suppose 3n+2 is even. Assume n is odd.

If n is odd, then we can write n as

n = 2k + 1, k ∈ Z

Substitute n in 3n+2

3(2k + 1) + 2

= 6k + 5

The term 6k is always even and 5 is odd. Accordingly, 6k + 5 is odd. This implies that 3n + 2 is odd. But this is contradictory. Our assumption that “n is odd” is incorrect. Hence “n is odd” is false. This implies that n must be even. This proves that if 3n + 2 is even then n is even ∀ n ∈ Z.

Proof By Contrapositive

The given statement is “If 3n+2 is even, then n is even”.

The contrapositive of the given statement is “If n is odd, then 3n+2 is odd”

Suppose n is odd. If n is odd, then we can write n as

n = 2k + 1, k ∈ Z

3n+2 = 3(2k + 1) + 2

= 6k + 5

The term 6k is always even and 5 is odd. Accordingly, 6k + 5 is odd. Thus, 3n + 2 is odd.

This implies that the statement “if n is odd then 3n+2 is odd”. Therefore, the contrapositive “If 3n+2 is even, then n is even” is also true. Hence proved.

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