9
$\begingroup$

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

$\endgroup$
3
  • 5
    $\begingroup$ 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 '15 at 18:21
  • 3
    $\begingroup$ 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 '15 at 18:24
  • 2
    $\begingroup$ only integers can be even or odd $\endgroup$ – Jake Sellers Jul 20 '15 at 6:19
20
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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 $2$.

$\endgroup$
1
  • 1
    $\begingroup$ $m$ 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 '15 at 22:53
1
$\begingroup$

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:

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

$\endgroup$
0
$\begingroup$

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. :-)

$\endgroup$
0
$\begingroup$

[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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.