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I'll quickly go over my understanding of it:

If a number $n^2$ is even, then $n$ is even. The contrapositive is that if $n$ is not even, then $n^2$ is not even.

We represent n as $n=2p+1$. $n^2=4p^2 + 4p + 1 = 2(2p^2+2p) + 1$. We see that $n^2$ is odd. Therefore, the original statement must be true.

Now here's my question, set $n^2=2$. $n^2$ is even but $n=\sqrt2$ isn't. Or is it? I'm confused

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    $\begingroup$ The statement that you're looking for is: "if $n$ is an integer, then $n$ is even if and only if $n^2$ is." $\endgroup$
    – Marcus M
    Jul 19, 2015 at 18:21
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    $\begingroup$ To be more precise one should say "if the square of an integer is even, then that integer is even". When writing $n^2$ it is implied $n$ is an integer, so $2$ is not the square in that sense. $\endgroup$
    – Joel Cohen
    Jul 19, 2015 at 18:24
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    $\begingroup$ only integers can be even or odd $\endgroup$ Jul 20, 2015 at 6:19

5 Answers 5

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Both $n^2$ and $n$ must be integers for this theorem to hold.

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Everything you wrote is correct up until the last line. If $n^2$ is even, then $$ n^2 = 2m \quad\text{ for some } m. $$

When you wrote $n^2 = 2$, you were (falsely) assuming that $m=1$.

By the way the even/odd language is only used when you're talking about integers (whole numbers), so there is no square root of $2$.

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    $\begingroup$ $m$ is not a particular number, it's defined as being equal to $\frac{n^2}2$ (assuming $n^2$ is even). What's wrong with setting $m = 1$? $\endgroup$
    – user253751
    Jul 19, 2015 at 22:53
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The original statement about "the square of a number" is true in a context where "number" means "integer." If that context is not clear, you had better say explicitly "the square of an integer". The square of any even integer can be written $2m$ where $m$ is an integer, but not every number that can be written as $2m$ where $m$ is an integer is a square of an integer of any kind.

Examples of squares of even integers written in the form $2m$:

$$\begin{align} 2^2 &= 4 = 2m & \text{where }\ & m=2. \\ 4^2 &= 16 = 2m & \text{where }\ & m=8. \\ 6^2 &= 36 = 2m & \text{where }\ & m=18. \\ 8^2 &= 64 = 2m & \text{where }\ & m=32. \\ \end{align}$$

Examples of $2m$ where $m$ is an integer, but $2m$ is not the square of any integer:

$$\begin{align} 2 = 2m & & \text{where }\ & m=1. & & \text{Not the square of any integer: }\ 1^2 < 2 < 2^2.\\ 6 = 2m & & \text{where }\ & m=3. & & \text{Not the square of any integer: }\ 2^2 < 6 < 3^2.\\ 8 = 2m & & \text{where }\ & m=4. & & \text{Not the square of any integer: }\ 2^2 < 8 < 3^2.\\ 10 = 2m & & \text{where }\ & m=5. & & \text{Not the square of any integer: }\ 3^2 < 10 < 4^2.\\ \end{align}$$

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if n is not even (odd)

Not being even is the same as being odd only if n is an integer. Concretely, in your case, sqrt(2) is certainly not even. But that doesn't mean it's odd. :-)

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[This answer was sparked by a question $4$ years later about the above question].

Almost surely the intended context is the ring of integers, so $\,n\in \Bbb Z,\,$ which excludes $\,n = \sqrt 2$.

But it is instructive to examine what occurs in the more general number ring that you consider.

In fact it remains true if we adjoin $\,\sqrt 2\,$ to $\,\Bbb Z\,$ to obtain $\,\Bbb Z[\sqrt 2] = \{ j + k \sqrt 2\ :\ j,k\in\Bbb Z\}$

As I explain here this ring has a sense of parity: $\,\alpha = j+k\sqrt 2\,$ is even $\iff \sqrt 2\mid \alpha \iff 2\mid j$

which immediately yields $\ \alpha^2\,$ even $\iff \alpha$ even. As explained in the linked post, integer parity results immediately generalize to any ring which has $\,\Bbb Z/2 = $ integers $\!\bmod 2\,$ as an image.

As above, many results from elementary number theory generalize to algebraic numbers. These topics are covered in any course on algebraic number theory.

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