[tex] \sqrt{2} [/tex] is, well, [tex] \sqrt{2} [/tex]. It's what's called an irrational number, which means that it can't be written as the ratio of two integers (i.e. numbers like 3/5, 8/13, 256/689), though I might as well take this time to go into one of my favorite proofs. I'm going to show you why it's impossible to write [tex] \sqrt{2} [/tex] as a fraction, and we're going to start by assuming that it is possible.
Proof: [tex] \sqrt{2} [/tex] is irrational.
Assume [tex] \sqrt{2} [/tex] is rational. If [tex] \sqrt{2} [/tex] is rational, it must be able to be written as the ratio of two integers - let's call them p and q - [tex] \frac{p}{q} [/tex]. Let's assume that [tex] \frac{p}{q} [/tex] is in simplest form; we can't reduce it any more.
We know that [tex] \sqrt{2}= \frac{p}{q} [/tex], which means that, squaring both sides:
[tex]2 = \big( \frac{p}{q}\big)^2 [/tex]
[tex]2= \frac{p^2}{q^2} [/tex]
Multiplying both sides by [tex]q^2[/tex], our equation becomes:
[tex]2q^2=p^2[/tex]
Since this shows that [tex]p^2[/tex] is a multiple of 2, we know that p is even, since squaring it gives us another even number.
Any even number is just 2 multiplied by some integer - let's call that integer k - so we can say that, since p is even:
[tex]p = 2k[/tex]
Plugging that back in, we get:
[tex]2q^2=(2k)^2[/tex]
Expanding the right side:
[tex]2q^2=4k^2[/tex]
And dividing both sides by 2:
[tex]q^2 = 2k^2[/tex]
So [tex]q^2[/tex] is even, which means that q is even, too. But wait, what did we say at the beginning?
"Let's assume that [tex] \frac{p}{q} [/tex] is in simplest form; we can't reduce it any more."
But, since both p and q are even, we can reduce our fraction by dividing p and q by 2, so we just contradicted ourselves! Since we assumed something was true, and following a completely logical set of steps made it come out false, we know that our assumption was wrong! So,
[tex] \sqrt{2}[/tex] is irrational.
End of story. Or, as mathematicians tend to say it, Q.E.D. (Quod erat demonstrandum - "which was to be demonstrated")