Let 𝑛 be an integer. Prove that if 3𝑛 + 2 is even, then 𝑛 is even usinga) a proof by contraposition.b) a proof by contradictio

a) Proof by contraposition:

The contrapositive of the statement "if 3n + 2 is even, then n is even" is "if n is not even (i.e., n is odd), then 3n + 2 is not even (i.e., 3n + 2 is odd)".

Let's assume n is odd. Then, n can be written as 2k + 1 for some integer k. Substituting n into the expression 3n + 2 gives 3(2k + 1) + 2 = 6k + 3 + 2 = 6k + 5. Since 6k is even (as it's a multiple of 2), and the sum of an even number and an odd number (5 in this case) is odd, we can conclude that 3n + 2 is odd. Therefore, the contrapositive is true, which means the original statement is true.

b) Proof by contradiction:

Assume the opposite of the statement, i.e., "3n + 2 is even and n is not even (i.e., n is odd)".

If n is odd, then n can be written as 2k + 1 for some integer k. Substituting n into the expression 3n + 2 gives 3(2k + 1) + 2 = 6k + 3 + 2 = 6k + 5. Since 6k is even (as it's a multiple of 2), and the sum of an even number and an odd number (5 in this case) is odd, we can conclude that 3n + 2 is odd.

But this contradicts our assumption that 3n + 2 is even. Therefore, our assumption that n is odd must be false, which means n is even. Hence, the original statement "if 3n + 2 is even, then n is even" is proven true by contradiction.