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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?

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3 Answers

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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.

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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.

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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.

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