definition of $2^{- \sqrt{2}}$ I have to write the definition of $2^{- \sqrt{2}}$.
But I have never seen this kind of exercise.
Can someone help me to understand what I have to do?
 A: You know the definition of $x ^ a$ when $a \in \mathbb{N}$ as $a \times a  \times ...  \times a$.
This exercice is asking you to find a way to extend this definition when $a \in \mathbb{R}$, such that your definition of $x ^ a$ provides the usual nice proprerties that we expect (morphisme between $+$ and $ \times $).
Hint : The idea is to extend it to $a \in \mathbb{Z}$, then $a \in \mathbb{Q}$, then $a \in \mathbb{R}$.
A: Note that $a^{-b}=1/a^b$ and $a^b=\exp(\ln(a)\cdot b)$ for $a,b>0$.
$$2^{-\sqrt{2}}=\frac{1}{2^\sqrt 2}=\frac{1}{\exp(\ln(2)\cdot\sqrt2)},$$
and in the last term, all expressions are well-defined.
A: Using the basic "$a^n$ means a multiplied by itself n times" when n is a positive integer, we can derive two very nice properties:  $(a^n)(a^m)= a^{m+n}$ and $(a^n)^m= a^{mn}$. Now, we can extend $a^x$ in such a way as to keep those true.  For example, we want $a^{n+ 0}= (a^n)(a^0)$.  Of course 0 is the "additive identity": n+ 0= n.  So we want to $(a^n)(a^0)= a^n$.  That means we must define $a^0= 1$ for all a.  For negative powers, -n, we want $a^{n- n}= (a^n)(a^{-n})= a^0= 1$.  So we must define $a^{-n}= \frac{1}{a^n}.  
For fractional powers we use $(a^n)^m= a^{mn}$.  In order for that to be true, even for n= 1/n, we need $(a^n)^{1/n}= a^{n/n}= a^1= a$.  That means that we must define $a^{1/n}= \sqrt[n]{a}$.  Then, of course, $a^{m/n}= (a^{1/n})^m= (\sqrt[n]{a})^m= \sqrt[n]{a^m}$.  
Irrational numbers, like $\sqrt{2}$, cannot be defined "algebraically", they must be defined "analytically" using some "limite" process.  There are a number of ways of doing that, "Dedekind Cuts" or "equivalence classes of Cauchy sequences" or "equivalence classes of increasing sequences bounded above" where the "equivalence relation" in the last two is "$\{a_n\}$ and $\{b_n\}$ are equivalent if and only if the sequence $\{a_n- b_n\}$ converges to 0.  It is in this sense that we can say "$\sqrt{2}= 1.4142...$".  What that means is that the sequence 1, 1.4, 1.41, 1.414, 1.4142, … is a sequence belong to the equivalence class defining $\sqrt{2}$ (we say that this sequence "converges to $\sqrt{2}$".  Given that definition we define $a^{\sqrt{2}}$ to be the equivalence class (real number) that the sequence $a^1, a^{1.4}, a^{1.41}, a^{1.414}, a^{1.4142}, \cdot\cdot\cdot$.  In practice to find, say $5^{\sqrt{2}}$, we would approximate by, say, 1.4142, and approximate $5^{1.4142}$ which is 5 to a rational number.
