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.
