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Assume that $n^2$ is even

Therefore $n^2 = 2k$ for some integer $k$.

How do I finish this proof?

user3924310
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6 Answers6

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Suppose instead that $n$ is odd. Then, $n = 2k+1$, so $n^2 = (2k+1)^2 = 4k^2+4k+1$. But $4k^2+4k$ must be even, so...

Emily
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    Shouldn't that be "contraposition" rather than "contradiction"? – beep-boop Oct 06 '14 at 18:20
  • Yes, but I'm not intending the use of the word to refer to the specific proof technique, rather just a descriptive word. I could say "suppose instead that $n$ is odd," if it helps clear anything up. – Emily Oct 06 '14 at 18:24
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    There's no contradiction, though. You're proving that $$n \quad \text{is odd} \ \implies n^2 \quad \text{is odd.}$$ – beep-boop Oct 06 '14 at 18:30
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    I know the difference between contrapositive and contradiction. And I also know that in common language, which I am trying to use, "contradiction" is often used to mean "the opposite." And when I use the word "opposite" in this comment, I don't mean "the thing multiplied by -1". I've changed my wording to be more reflective of plain language and less reflective of precise terminology, the distinctions of which are likely presently irrelevant to someone who could not, on their own, come up with this proof. – Emily Oct 06 '14 at 18:33
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    I'm not disputing whether you know it, but it's imprecise to talk about contradictions when there are none. – beep-boop Oct 06 '14 at 18:38
  • Your assumption was that I was aiming for precision -- and that precision makes any difference in the current context. – Emily Oct 06 '14 at 18:49
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$2k = n^2 = \begin{cases} (2m +1)^2 = 4m^2 +4m +1 \\ (2m)^2=4m^2 \end{cases}$

Which case do you think is rather possible?

Daniel Valenzuela
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I think it's best to use a proof by cases here. .

Case 1: $n$ is even. Both $n^2$ and $n$ are even, so the statement "if n^2 is even then n is even" holds.

Case 2: $n$ is odd. $n^2$ is odd, so the statement "if n^2 is even then n is even" is inconsequential and still holds.

NovaDenizen
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First note that the statement $n^2$ is even $\Rightarrow$ $n$ is even is logically equivalent to its contrapositive. Namely, $\neg$($n$ is even) $\Rightarrow$ $\neg$($n^2$ is even), or in other words $n$ is odd $\Rightarrow$ $n^2$ is odd. Hence, it will be enough to prove the conditional statement $n$ is odd $\Rightarrow$ $n^2$ is odd.

Proof: Assume n is odd. That is $n=2k+1$ for some integer $k$. Then, $$n^2 = (2k+1)^2$$ $$\Rightarrow$$ $$n^2=4k^2+4k +1=2(2(k^2+k))+1$$

Note that $2(k^2+k)=z$ for some integer $z$, because integers are closed under addition and multuplication. So, $n^2 = 2z+1$.

Therefore, by the definition of odd, $n^2$ is odd.

Valentino
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What constitutes an acceptable proof depends, in part, on what you are allowed to assume as already established. Here's a proof based on Euclid's Proposition VII.30, namely that that if $p$ is a prime number and $p$ divides a product $ab$, then $p$ divides either $a$ or $b$ (or both).

Note that $2$ is a prime number. If $2$ divides $n^2$, then, by Euclid's theorem, $2$ divides either $n$ or $n$ -- which is to say, $2$ divides $n$.

Barry Cipra
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Suppose that $n$ is odd, then the prime decomposition of $n$ does not contain a 2, so the prime decomposition of $n^2$ does not contain a 2 too, hence must be odd.

Marm
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