Real numbers to real powers Suppose that $a,x,y\in\mathbb R$ with $a\ge1$ and $x,y\ge0$. We define the real power $a^x$ as follows:
$$a^x=\sup\{a^r:0\le r\le x,r\in\mathbb Q\}$$
I've tried some ways to prove that $a^xa^y=a^{x+y}$, for example 
If we let $\alpha=a^x$, $\beta=a^y$ and $\gamma=a^{x+y}$, as a first stage I should prove that $\alpha\beta$ is an upper for the set $\{a^t:0\le t\le x+y,t\in\mathbb Q\}$. So I assume that $0\le t\le x+y$, so $-x\le t-x\le y$ which implies that $a^{t-x}\le\beta$. Since $a^x=\alpha$, this implies that $a^t\le\alpha\beta$. But I think that this proof has a problem, since $t-x$ can be in $\mathbb R\setminus \mathbb Q$.
As a second stage, for $c<\alpha\beta$, I should find a $t\in \mathbb Q,0\le t\le x+y$ such that $a^t>c$. Since $c<\alpha\beta$, so $\frac c\alpha<\beta$, so there is an $s\in\mathbb Q,0\le s\le y$ such that $a^s>\frac c\alpha$. Similarly there is a $r\in\mathbb Q,0\le r\le x$ such that $a^r>\frac c\beta$. Therefore $a^{r+s}>\frac{c^2}{\alpha\beta}$. So from here I think we can't obtain anything.
How can I solve these problems? Thanks.
 A: Right, your argument for $\gamma \leqslant \alpha \beta$ has the problem that you cannot guarantee that $t - x \in \mathbb{Q}$.
What you would need to make it work without a real modification are $r,s \in \mathbb{Q}$ with $r \leqslant x$, $s\leqslant y$ and $t \leqslant r+s$. Then you'd have
$$a^t \leqslant a^{r+s} = a^r\cdot a^s \leqslant \alpha \beta$$
and be done. But you cannot find such $r,s$ if $x,y \in \mathbb{R}\setminus \mathbb{Q}$ and $t = x+y \in \mathbb{Q}$.
This argument would work if the definition used a strict inequality,
$$a^x = \sup \:\{ a^r : r\in \mathbb{Q}, r < x\}\,,\tag{$\ast$}$$
(the restriction to nonnegative exponents is unnecessary) because for every rational $t < x+y$ we can find rational $r < x$ and $s < y$ with $t < r+s$.
Since the definition uses a nonstrict inequality, we need some additional work. Probably the easiest way (one of several equally easy ways) is to show that
$$\sup\: \{ a^r : r\in \mathbb{Q}, r < x\} = \sup\: \{ a^r : r \in \mathbb{Q}, r \leqslant x\}\,,$$
i.e. that $(\ast)$ is equivalent to the given definition.
The direction $\alpha\beta \leqslant \gamma$ doesn't suffer from that problem. If $r,s$ are rational with $r \leqslant x$ and $s\leqslant y$, then $r+s \leqslant x+y$ and hence
$$a^r\cdot a^s = a^{r+s} \leqslant a^{x+y} = \gamma\,.\tag{1}$$
Fixing an arbitrary rational $r \leqslant x$ and taking the supremum of $(1)$ over all rational $s \leqslant y$, we obtain
$$a^r\cdot \beta \leqslant \gamma\,.\tag{2}$$
Now taking the supremum over all rational $r \leqslant x$ in $(2)$ yields
$$\alpha\cdot\beta \leqslant \gamma\,.$$
