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” is true. Therefore, the contrapositive “If 3n+2 is even, then n is even” is also true. Hence proved.