exponentiation - Proof that irrational exponents exist, and are unique?

Can someone provide a proof that for any given irrational number, $b$, exponentiation by that number defined as a limit of rational powers always converges, and that if we choose a particular base, $a$, that the no two $b$'s yield the same limit. (I suppose in the case where $a$ is not $0$ or $1$?)

Answer

We can restrict to the case $a>1$ (noting that $0Note that $\frac pq<\frac rs$ implies $ps$x\mapsto a^x$ is strictly increasing when looking only at rational $x$. If we take the convergence for granted, this implies that $x\mapsto a^x$ is injective as a function defined on all of $\mathbb R$: If $b\in\mathbb R$ and $b<\frac pq\in\mathbb Q$, then almost all fractions in a sequence converging to $b$ are $<\frac pq$, hence their powers are $

Now for the convergence: For rational exponents, the usual power laws hold, and as we have just seen order is also respected (for bases $a>1$). Hence whenever $\left|\frac pq-\frac rs\right|<\frac 1N$, we have $$\left|a^{\frac pq}-a^{\frac rs}\right|=a^{\frac rs}\cdot \left|a^{\frac pq-\frac rs}-1\right|
We may assume wlog that $\frac rs<\lceil b\rceil +1=:M$ so that $$\left|a^{\frac pq}-a^{\frac rs}\right|As $\sqrt[N]a\to 1$ and $\sqrt[N]{1/a}\to 1$, we conclude that any sequence $\{\frac{p_k}{q_k}\}_{k\in\mathbb N}$ of fractions converging to $b$ produces a Cauchy sequence $\{a^{\frac{p_k}{q_k}}\}_{k\in\mathbb N}$.