# If the square of a number is even, then the number is even. Is that true for 2?

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

• The statement that you're looking for is: "if $n$ is an integer, then $n$ is even if and only if $n^2$ is." Jul 19, 2015 at 18:21
• 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. Jul 19, 2015 at 18:24
• only integers can be even or odd Jul 20, 2015 at 6:19

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

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 $2$.

• $m$ 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 $m = 1$? Jul 19, 2015 at 22:53

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}

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. :-)

[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.