Firstly, the definition.

A number $x$ (real or complex) is algebraic if and only if there exists a polynomial $P$ with integer coefficients such that $P(x)=0$. A number which is not algebraic is said to be transcendental. If $x$ is algebraic then the minimal polynomial of $x$ is the polynomial $P$ with minimal degree such that $P(x)=0$.

Here are the three theorems.

1. Liouville's Theorem

   If $x$ is algebraic then there exists $c>0$ such that for all integer $p$ and natural $q$ we have:

   $|x-p/q|>c/q^n$

   where $n$ is the degree of $x$. (That is, the degree of the minimal polynomial $P$ such that $P(x)=0$.)

   Equivalently, if the quantity $|x-p/q|q^n$ can be made arbitrarily small then $x$ is transcendental.

   This theorem effectively shows if a number is very "close" to being rational (by which I mean it has very good rational approxiations) then it is transcendental.

   For example, using this theorem, it is a straightforward exercise to show that the series:

   $1/2^{1!}+1/2^{2!}+\dots$

   is transcendental. The reason is because the terms of the series are decreasing in absolute value so rapidly that the series has excellent rational approximations hence Liouville's theorem shows it's transcendental.

   In fact this number was the first number ever to be proved to be transcendental. I'm not sure, but I think that before they'd proved it was transcendental, most people believed all numbers were algebraic. Later Cantor stepped on the scene and showed that älmost all" numbers are transcendental.

2. Lindemann's theorem

   This one shows that if $a_1$, ..., $a_n$ are distinct algebraic numbers and $b_1$, ..., $b_n$ are non-zero and algebraic then $b_1\exp (a_1)+\dots +b_n\exp (a_n)$ is non-zero.

   Three useful facts that follow immediately from this theorem are:

   • $e$, $\pi$ are transcendental.
   • $e^x$, $\cos x$, $\sin x$, $\tan x$ are transcendental if $x$ is algebraic and not 0.
   • $e^{\pi ix}$, $\cos (\pi x)$, $\sin (\pi x)$, $\tan (\pi x)$ are transcendental if $x$ is algebraic and irrational.
3. Gelfond-Schneider

   If $a$, $b$ are algebraic with $a$ not equal to 0,1 and $b$ not rational then $a^b$ is transcendental.

   I will not be proving this one as the proof is incredibly complicated, intricate and uses complex analysis. To give some idea, the proof was first obtained in the 1930s. It was one of the 20 problems Hilbert proposed in 1900.

   It follows pretty quickly from this theorem that:

   • $\sqrt{2}{}^{\sqrt{2}}$ is transcendental (this has been asked at least 3 times before!)
   • $e^{\pi }$ is transcendental (to prove this, use the fact that $e^{\pi i}=-1$).
   It is not however known whether ${\pi }^e$ is transcendental or not (or even whether it's irrational). Nor is it known whether $\pi +e$ or indeed any individual linear combination of $\pi$ and $e$ is transcendental.