1. Jan 10, 2012

f24u7

I saw a proof on a textbook, but I don't understand why m and n should not both be even?

Prove p^2 =2 then p is irrational

(1) p^2 = 2
is not satisfied by any rational p. If there were such a p, we could write p = m/n
where m and n are integers that are not both even.

Then (1) implies

(2) m^2=2n^2

This shows that m^2 is even.

Hence m is even (if m were odd, m 2 would be odd),and so m^2 is divisible by 4.
It follows that the right side of (2) is divisible by 4,
so that n  is even, which implies that n is even.

The assumption that (1) holds thus leads to the conclusion that both m
and n are even, contrary to our choice of m and n. Hence (1) is impossible for
rational p.

Last edited: Jan 10, 2012
2. Jan 10, 2012

micromass

Staff Emeritus
We can choose m and n not both even.

Indeed, let's say that $p=\frac{12}{8}$ then both the numerator as the denominator are even. But we can simplify it as $p=\frac{3}{2}$, not not both are even.

So we can always write p=m/n with not both m and n even. If m and n were both even, then we can just simplify the fraction.

3. Jan 10, 2012

f24u7

thanks for the reply, but I still don't understand why is m and n both are even a conclusive evidence that p is not a rational?

4. Jan 10, 2012

micromass

Staff Emeritus
We assumed p to be rational. We could write p=m/n with m and n not both even. But then we reasoned further and we found that m and n were both even. This is a contradiction, so the hypothesis that p be rational was false. Thus p is irrational.