>Assume there are two integers A and B in a ratio such that A/B=√2.
*Sigh!* Sure, fine, two integers in a ratio.
>Then A^2=2B^2.
Yes, yes, I've seen this before. So A^2 is an even number, which means A is an even number, which means we can replace A with 2C where C is an integer.
>Right! Therefore 4C^2=2B^2, but that means 2C^2=B^2, so B^2 is even, and we can replace B with 2D!
And so on ad-infinitum. *Yawn!*
>So that means my premise was wrong! There is no such ratio of integers!
Then what are A and B? If they aren't integers, what are they?
>W-what? No, I choose the pretend the ratio part is the part I've disproven.
Can √2 be multiplied by 10?
>Of course it can!
So then whatever that result is—let's call it √200—dividing it by 10 would equal √2, right?
>Y-yeah.
So that's a ratio, isn't it?
>Um, yeah, I guess.
And we agree 10 is an integer, right?
>Y-yeah.
So then √200 must not be an integer. Your conclusion was that there is no ratio of two integers A and B that equals √2, but I've just shown that the problem isn't with the ratio part, and the denominator is definitely an integer, so the numerator must not be one. So what is it?
>Um, uh, um, there's a decimal point.
I'm assume you agree it can written using a string of numerals. We don't have to resort to seance or something, right?
>Of course it can be written using numerals, AND a decimal point!
And if we keep multiplying by 10, we can keep shifting that decimal point to the right as far as we need to in order to get an integer. Or are you saying that at some length, it ceases to be an integer over a power of 10?
>Exactly!
But you can't seem to tell me exactly when it stops being an integer.
>Only when it goes on forever!
You mean like 0.333... as the decimal expression of one-third? That one goes on forever, yet you seem to consider this a valid radio of two integers. Why doesn't that numerator do the magical not-an-integer thing?
>SHUT UP! YOU'RE UGLY!

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Aren't irrational numbers... well... irrational?