Proof by contraposition.

Question

Use proof by contraposition to prove that the following statement is true for all positive integers:

If $n^2$ is a multiple of $3$, then $n$ is a multiple of $3$. Hint: every integer $n$ can be expressed as $n=3k$, $n=3k+1$ or $n=3k+2$, for some integer k.

So far I have; If $n$ is not a multiple of $3$ then $n^2$ is not a multiple of $3$.

Let $n$ be a multiple of $3$. Then $n=3k$ for some integer $k$ and so $$n^2=(3k)^2=9k^2=3(3k^2)$$

Thus $n^2$ is a multiple of $3$, since $9k^2 is an integer$. So the contrapositive is true and hence the original statement is true.

Is this all I need to do? Doesn't seem enough for a 6 mark question, or should I have done something differently?

Answer 1

This is correct, there is essentially nothing else to show. Have you been given any other information about the problem if you don't think it is enough for a "6 mark question" ?

EDIT :

I must have misread the question, my apologies. You must prove it in the other direction as well, as it currently stands the proof is incomplete.

Answer 2

It looks to me like you've proven that if $n$ is a multiple of $3$, then $n^2$ is a multiple of $3$. The contrapositive of that is that if $n^2$ is not a multiple of $3$, then $n$ is not a multiple of $3$. You need to start by proving that if $n$ is not a multiple of $3$, then $n^2$ is also not. Then the original statement follows by contraposition.

I say this because you start by assuming n is a multiple of three rather than assuming it isn't.

Answer 3

I think you must prove that if $n = 3k + 1$ or $n = 3k + 2$, $n^2$ cannot be a multiple of $3$. you have proved $q\implies p$ but not the contrapositive of the statement.