Question

why is sqrt 2 irrational?

Answer 1

it cannot be written as a ratio.

Answer 2

Any number in Q can be wrote as "m/n" where m and n can't be simplified between themselves. You suppose that

\[\sqrt2 = \frac{m}{n}\]
So.
\[2 = \frac{m^2}{n^2}\]
So....
\[2n^2 = m^2\]

The number at the first member must be even, because any number multiplied by 2 is even.
But we have an identity here, so the number at the second member must be even as well. But we just begun by saying that these two numbers couldn't be simplified. So here's an absurd,
sqrt can't be in Q, it must be somewhere else. Then you'll have to understand how's that defined, but that's another story.

Answer 3

Floored. I dont like it this way though. I'd be happy if someone walked me through the steps .

Answer 4

Fine. Well let's start from the beginning then, what's a rational number?

Answer 5

a number that can be expressed as a fraction

Answer 6

That's right. Now let's recall that in order to be rational, they need to be either a fixed number, or a periodic one.
Means that if you do 1/2 = 0.5 this is okay, and 4/3 is periodic so is fine as well.

Now, step two is two suppose, using an absurd hypotesis, that there are two numbers, which we don't know yet, so that:

\[\frac{a}{b} = \sqrt2\]

Because we want to proove that sqrt2, is/isn't a rational number, since we know that in order to

\[\sqrt 2 \in Q \rightarrow \exists m,n : \frac{m}{n} = \sqrt2\]

So all we do know is doing some math here:

\[\frac{a}{b} = \sqrt2\]

Answer 7

ok what is the definition of a fixed number and a periodic one?

Answer 8

By fixed number I meant that it has no periodicity nor it is irrational what's so ever. So it's not a repeating decimal. It has a finite number of "characters" (didn't want to use numbers again :P).

Look at this for a better idea: http://en.wikipedia.org/wiki/Repeating_decimal

P.s: in order to find an absurdum back there we have to suppose the fraction is irriducible.