$2d^2=n^2$ implies that $n$ is multiple of 2

Question

I'm reading a proof of the irrationality of $\sqrt 2$. In a step it states that $2d^2=n^2$ implies that $n$ is multiple of $2$. How?

If $n$ is odd then its square is also odd. But the left-hand side is even. — njguliyev, Sep 02 '13 at 12:18
related to my earlier question (why this only works for prime numbers): http://math.stackexchange.com/q/162119/29313 — KutuluMike, Sep 02 '13 at 14:23

score 8 · Answer 1 · answered Sep 02 '13 at 12:19

8

$\;2\;$ is a prime and divides the left side in $\,2d^2=n^2\;$ , so by the fundamental theorem of arithmetic it also divides the right hand side...

answered Sep 02 '13 at 12:19

DonAntonio

3

The FTA seems like overkill to me... – Benjamin Dickman Sep 02 '13 at 12:24
Perhaps so, @BenjaminDickman, yet do you know any other, inequivalent and formal way to prove the statement? – DonAntonio Sep 02 '13 at 12:25
1

Yes; I have already provided one. – Benjamin Dickman Sep 02 '13 at 12:26
That doesn't look so formal to me, @BenjaminDickman: how can you prove any integer is either odd or even, as you implicitly assumed? You may want to go to Peano axioms and/or induction, which would imply more or less the same work as the FTA. After all, someone asking such a basic question may want to axiomatically base things... – DonAntonio Sep 02 '13 at 12:32
I would accept also your answer now that I've understood, but actually my question was basic just because they didn't teach me very well (at all?) how to make demostrations at university or college. – bigstones Sep 02 '13 at 13:13

score 7 · Accepted Answer · answered Sep 02 '13 at 12:21

If $n^2 = 2d^2$, then $n^2$ is a multiple of $2$ hence even.

We now prove: $n^2$ even implies $n$ even.

Proof: We tackle the contrapositive, i.e., $n$ odd implies $n^2$ odd.

Since $n$ is odd, we can write $n = 2k+1$ for some integer $k$.

Then $n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$, which is an odd number.

(It's of the form $2m + 1$ for an integer $m = 2k^2 + 2k$.)

This completes the proof, and the contrapositive is the statement you asked about. QED

score 3 · Answer 3 · answered Sep 02 '13 at 12:19

3

Since $d$ is an integer, $2d^2$ must be even. For any odd integer $n$, $n^2$ must also be odd. Therefore, $n$ must be even, and thus a multiple of 2.

answered Sep 02 '13 at 12:19

Glen O

score 3 · Answer 4 · answered Sep 02 '13 at 12:20

3

It works for any prime $p$: $$p|n^2 \Rightarrow p|n$$ This is because of the uniqueness of prime factor decomposition.

answered Sep 02 '13 at 12:20

AlexR

Ömer · Answer 5 · 2013-09-02T12:28:45.557

-2

square of even numbers are even, so $n$ is even

edited Sep 02 '13 at 12:28

answered Sep 02 '13 at 12:18

Ömer

3

This is a logical fallacy. "If $n$ is even, then $n^2$ is even," does not mean that "If $n^2$ is even, then $n$ is even." – Thomas Andrews Sep 02 '13 at 12:30
@ThomasAndrews: yes you are right my answer was so quick and it wasn't well-considered. – Ömer Sep 02 '13 at 13:10

5 Answers5