Prove by contraposition For any integer n if n2 is a multipl

Prove by contraposition. For any integer n, if n² is a multiple of 3, then n is a multiple of 3. There must be two cases involving subproofs.

Solution

This is a proof by contrapositive.

The contrapositive of the statement is,

let  “If n is a multiple of 3, then n² is multiple of 3.”

If n is a multiple of 3, then n = 3k for some integer k.

Then n² = (3k)² = 9k²

So n² is divisible by 3

hence proved