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

Given

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

Proof By Contradiction

We will prove the given statement by contradiction

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

If n is even, then we can write n as

n = 2k ∀ k ∈ Z

Substitute n in 3n+2

3(2k) + 2

= 6k + 2

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

Proof By Contrapositive

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

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

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

n = 2k ∀ k ∈ Z

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

= 6k + 2

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

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

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