Originally Posted by

**mfetch22** I know to prove the irrationality of $\displaystyle \sqrt{2}$ we simply minipulate the following:

$\displaystyle (\frac{p}{q})^2 = 2$

until we hit the contradiction of not being in reduced terms. I also know that we must take into account that all number can be written in the form $\displaystyle 2n$ or $\displaystyle 2n+1$ for $\displaystyle n=1,2,3...$; and the fact that if $\displaystyle x^2$ is an even number then $\displaystyle x$ is an even number. Heres my question, how do you prove the irrationality of a certain opertation on numbers in general? And for odd square roots?

Two examples of what I mean:

[1] How would you prove that if $\displaystyle n=1, 2, 3...$ that $\displaystyle \sqrt{n}$ is always irrational?

[2] How would you prove $\displaystyle \sqrt{3}$ is irrational? Would you find a way to define numbers in terms of 3's ? Just like with the $\displaystyle \sqrt{2}$? So all numbers would be of the form $\displaystyle 3n$, $\displaystyle 3n+1$, and $\displaystyle 3n+2$; and would you some how use that fact to prove the irrationality of $\displaystyle \sqrt{3}$?

Thanks in advance