Proof that irrational exponents exist, and are unique?

Question

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 1

We can restrict to the case $a>1$ (noting that $0<a<1$ can then be treated by taking reciprocals). Note that $\frac pq<\frac rs$ implies $ps<rq$, hence $a^{rs}<a^{rq}$ and by taking $qs$th roots, $a^{\frac pq}<a^{\frac rs}$. Therefore the map $x\mapsto a^x$ 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 $<a^{\frac pq}$, consequently the limit (if it exists) is $\le a^{\frac pq}$. Similarly for $\frac pq<b$, we have $a^{\frac pq}\le a^b$. Then for $b_1<b_2$ pick fractions with $b_1<\frac pq<\frac rs<b_2$ and conclude that $a^{b_1}\le a^{\frac pq}<a^{\frac rs}\le a^{b_2}$. This shows the second part.

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 $$\left|a^{\frac pq}-a^{\frac rs}\right|=a^{\frac rs}\cdot \left|a^{\frac pq-\frac rs}-1\right|<a^{\frac rs}\cdot\max\left\{a^{\frac 1N}-1,1-a^{-\frac 1N}\right\}$$ We may assume wlog that $\frac rs<\lceil b\rceil +1=:M$ so that $$\left|a^{\frac pq}-a^{\frac rs}\right|<a^M\cdot\max\left\{\sqrt[N]a-1,1-\sqrt[N]{\tfrac 1a}\right\}.$$ 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 $\{a^{\frac{p_k}{q_k}}\}_{k\in\mathbb N}$.