For example, $\sqrt2$ isn't a rational number, since there is no rational number whose square equals two. And I see this example of a real number all the time and I'm just curious about how you can find or determine other numbers like so. Or better yet, how was it found that if $\mathbb Q$ is the set of all rational numbers, $\sqrt2\notin\mathbb Q$?
I appologize if the number theory tag isn't appropriate, I'm not really sure what category this question would fall under.
Best Answer
I'm not sure if you're asking about finding a real number, or determining whether a given real number is rational or not. In any case, both problems are (in general) very hard.
Finding a real number
There are lots and lots of real numbers. How many? Well the set of all real numbers which have a finite description as a string in any given countable alphabet is countably infinite, but the set of all reals is uncountably infinite $-$ if we listed all the reals that we could possibly list then we wouldn't have even scratched the surface!
Determining whether or not a given real number is rational
No-one knows with certainty whether $e+\pi$ or $e\pi$ are rational, though we do know that if one is rational then the other isn't. In general, finding out if a real number is rational is very hard. There are quite a few methods that work in special cases, some more sophisticated than others.
An example of a ridiculous method that can be used to show that $\sqrt[n]{2}$ is irrational for $n>2$ is as follows. Suppose $\sqrt[n]{2} = \dfrac{p}{q}$. Then rearranging gives $2q^n=p^n$, i.e. $$q^n + q^n = p^n$$ but since $n>2$ this contradicts Fermat's last theorem. [See here.]
The standard proof of the irrationality of $\sqrt{2}$ is as follows. Suppose $\sqrt{2} = \frac{p}{q}$ with $p,q$ integers and at most one of $p$ and $q$ even. (This can be done if it's rational: just keep cancelling $2$s until one of them is odd.) Then $2q^2=p^2$, and so $2$ divides $p^2$ (and hence $p$); but then $2^2$ divides $2q^2$, and so another $2$ must divide $q^2$, so $2$ divides $q$ too. But this contradicts the assumption that one of $p$ and $q$ is odd.