# In proof of SQRT(2) is irrational, why can't a,b both be even

1. Sep 14, 2014

### Thecla

In the classic proof of irrationality of SQRT(2) we assume that it can be represented by a rational number,a/b where a, b are integers. This assumption after a few mathematical steps leads to a contradiction, namely that both a, b are even numbers.
Well you can say that the rational fraction has to be in its lowest terms;therefore either a or b or both must be odd.
However, that wasn't in the assumption(lowest terms). The assumption was just two integers a,b.
Why can't they both be even?

2. Sep 14, 2014

### gopher_p

In the correct version of the classic proof, it is assumed that a and b are not both even.

3. Sep 14, 2014

### Simon Bridge

And the reason you want to make the assumption that both are even right at the start is...

4. Sep 15, 2014

### bhillyard

The proof I use starts by assuming a and b are integers with no common factors i.e. a/b is in its lowest terms. Then you end up with a loop in which a and b always have a common factor of 2 - but this is impossible as the numerator and denominator would get smaller without limit. An impossibility for positive integers.

5. Sep 15, 2014

### pwsnafu

If a and b are both even, then let $a=2c$ and $b=2d$. Repeat the argument on c and d, so we conclude they too are even. Let $c=2e$ and $d=2f$....which are also even. And so on. This argument should keep going forever.

But here's the thing. The natural numbers have a minimum (namely 1). So we can't keep halving forever. At some point this process must terminate. This contradicts our original argument.

6. Sep 15, 2014

### Thecla

Thanks for the help

7. Sep 17, 2014

### CKH

Another proof:

In $2 = (a/b)^2$, assume that $a$ and $b$ have been reduced to have no common factors.
Then $2b^2=a^2$, so $a$ must be even (contain a factor 2).
Thus $a=2c$ for some $c$
Thus $2b^2=4c^2$
Thus $b^2=2c^2$, so now $b$ must be even also.

This contradicts the assumption that $a$ and $b$ have no common factor, so no such $a$ and $b$ can exist.

8. Sep 21, 2014

### lurflurf

Just do this

Suppose $$\sqrt{2}=a/b$$
where
a=2^m c
b=2^n d
c and d are odd integers
then
$$\sqrt{2}=a/b \\ \sqrt{2}b=a \\ 2b^2=a^2 \\ 2(2^n d)^2=(2^m c)^2 \\ 2^{2n+1} d^2=2^{2m} c^2 \\$$
thus
2n+1=2m
2n+1 is odd while 2m is even
an even number cannot equal an odd number