Use proof by contraposition to prove that the following statement is true for all positive integers 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?
3 Answers
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.
It looks to me like you've proven that if n is a multiple of three, then n^2 is a multiple of three. The contrapositive of that is that if n^2 is not a multiple of three, then n is not a multiple of three. You need to start by proving that if n is not a multiple of three, 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.
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→p) not the contrapositive of the statement.
