2n^2=m^2

Why is 2 a factor of m?

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- Jun 7th 2010, 05:47 AMStuck Manfactors
2n^2=m^2

Why is 2 a factor of m? - Jun 7th 2010, 05:52 AMUnknown008
Put this into this form:

$\displaystyle m^2 = 2n^2 = 2 \times n \times n$

It is like factorising a number.

You can divide m^2 by 2 without getting a remainder, but a quotient equal to n^2.

I hope it helps. - Jun 7th 2010, 05:53 AMundefined
So, m and n are integers.

If m^2 is twice n^2 that means m^2 is even. (Which is the same as saying 2 is a factor of m^2.) But if m^2 is even then m must also be even.

As it turns out, we know something stronger: that 4 is a factor of m.

Are you looking at a proof that the square root of 2 is irrational, by chance? - Jun 7th 2010, 06:03 AMStuck Man
Yes good guess.

- Jun 7th 2010, 07:17 AMKrahl
if m is even say 2k for some integer k, then $\displaystyle m^2=(2k)^2=4k^2$ and so $\displaystyle m^2$ is even.

if m is odd say 2k+1 for some integer k, then $\displaystyle m^2=(2k+1)^2=4k^2+4k+1$ is odd.

so if 2 divides $\displaystyle m^2$, then m^2 is even and by above m is even and so 2 divides m. - Jun 7th 2010, 07:20 AMStuck Man
This has been more complicated than I first thought. Wikipedia has the full proof which was useful.