A short proof that 2^{\sqrt{2}} is not rational.

Here is an adaptation of Lang’s proof of Gelfond’s theorem to prove that $2^{\sqrt{2}}$ is not rational – needed for http://mathoverflow.net/questions/138247/prove-that-sqrt2-sqrt2-is-an-irrational-number-without-using-a-theorem/138291#138291. Consider the functions $f_1=e^{\sqrt{2}x}, f_2=e^x$. Let $x_i=i\log 2, 1\le i\le m$, where $m$ is big enough (we shall see how large $m$ should be below). Suppose that $2^{\sqrt{2}}$ is rational. Then $f_i(x_j)$ is a rational number for every $i,j$.

Lemma 1. Let us have a homogeneous system of $r$ linear equations with $n$ variables with integer coefficients, the abs. value of each coefficient is at most $A$. Then there exists a non-trivial integer solution with $|x_i|\le 2(nA)^{\frac{r}{n-r}}$ provided $n>r$.

Proof (half a page) see in Lang “Algebra”.

Consider the function $F=\sum_{i,j=1}^r b_{i,j}f_1^if_2^j$ with rational coefficients. Set $n=\frac{r^2}{2m}$ (choose $r$ so that it is an integer). By Lemma 1, we can find integers $b_{i,j}$ such that all derivatives $D^k F(x_i)=0$, $0\le k, moreover absolute values of $b_{i,j}$ are bounded by something like $O(n^{2n})$ for very large $n$ by Lemma 1.  Since $f_1, f_2$ are algebraically independent  (here we use  that $\sqrt{2}$ is irrational), the function $F$ is not identically 0. Let $s\ge n$ be the smallest integer such that all derivatives up to the order $s-1$ of $F$ are 0 at each $x_i$, but the derivative $g=D^s(F(x_i))\ne 0$ for some $i$ (for simplicity let $i=1$, the proof is the same for any $i$). Then $g$ belongs to the field $K=\mathbb{Q}[\sqrt{2}]$, that is $g=a+b\sqrt{2}$ where $a,b$ are rational numbers. Moreover the least common denominator $c$ of $a, b$  is bounded by $C_1^s$ for some constant $C_1$. The conjugate of $cg$ (i.e.,$ca-cb\sqrt{2}$) is bounded by $C_2s^{5s}$ for some constant $C_2$. Then the norm of $cg$ in $K$, i.e. the product of $cg$ and its conjugate,  which is an integer $\ge 1$, is bounded by $C_2s^{5s}|g|$. Now consider the function $H(z)=\frac {F(z)}{\prod_{i=1}^{m}(z-x_i)^{s-1}}$. Then $H(x_1)$ differs from $D^sF(x_1)$ by some factor bounded by $C_3^s s!$ (which is “small”). By the maximum modulus principle, the value of $H(x_1)$ is bounded above by the value of $H(z)$ where $z$ is a complex number on a circle of large enough radius $R$ (we need only that $R$ is bigger than each $x_i$). Taking $R$ large enough, we get that $H(x_1)$ is bounded from above by a number of the form $\frac{s^{3s}C_4^{2rR}}{R^{ms}}$. For $R=s^{C_4s}$ ($C_4$ is an appropriate constant) this gives a contradiction with the previous inequality for $|g|$ (the denominator grows much faster than the numerator, so the fraction cannot be always greater than 1).