# $(x^r)^s=x^{rs}$ for the real case

Hi everyone I'd like to know if the following is correct and if someone knows a better way to do it.

Definition Let $x>0$ be a real, and $\alpha$ be a real number. We define the quantity $x^{\alpha}$, by the formula $\text{lim}_{n\rightarrow\infty} x^{q_n}$ where $(q_n)$ is a sequence of rationals which converges to $\alpha$. (I've shown that the definition is well-defined).

Lemma: Let $r,s \in \mathbb{R}$ and $x\in \mathbb{R}^{>0}$. Then $(x^r)^s=x^{rs}$.

Proof: Let $(s_n), (r_m)$ be sequences of rational numbers which converges to $r$ and $s$ respectively.

\begin{align} \lim_{n\rightarrow \infty}( \lim_{m\rightarrow \infty}x^{r_m})^{s_n}=\lim_{n\rightarrow \infty} \lim_{m\rightarrow \infty}((x^{r_m})^{s_n}) \\ = \lim_{n\rightarrow \infty} \lim_{m\rightarrow \infty} x^{r_ms_n} \end{align}

We will show that $\lim_{n\rightarrow \infty} \lim_{m\rightarrow \infty} x^{r_ms_n}= \lim_{n\rightarrow \infty} x^{r_ns_n}=x^{rs}$

Claim 1: $\lim_{m\rightarrow \infty} x^{r_ms_n}=x^{rs_n}$

It will suffice to show that $$\lim_ {m\rightarrow \infty}x^{r_ms_n-rs_n}= 1$$ Since the claim would then follows by the limit laws and the properties of the exponents (since $x^{r_ms_n}=x^{r_ms_n-rs_n}x^{rs_n}$). We know by hypothesis that $r_m \rightarrow r$, thus using the limit laws we can conclude $r_ms_n-rs_n \rightarrow 0$.

Write $t_m = r_ms_n-rs_n$. We have to show that for every $\varepsilon>0$, the sequence $(x^{t_m}) \rightarrow 1$. Let $\varepsilon>0$ be given. We already know that $(x^{1/k}) \rightarrow 1$ and also $(x^{-1/k}) \rightarrow 1$ by the limit laws. So there is some $K$ for which $x^{1/K}$ and $x^{-1/K}$ are simultaneously $\varepsilon$-close to $1$. Let us fix $K$. Now since $t_m$ converges to zero as we have shown, so there is some $M$ such that $|t_m|\le 1/K$ for all $m\ge M$. Thus $$-1/K \le t_m \le 1/K$$

If $x>1$ we have $x^{-1/K}\le x^{t_m}\le x^{1/K}$, and in particular $x^{t_m}$ is $\varepsilon$-close to 1. A similar argument works when $x<1$ just with the inequality in reverse order. Hence $x^{t_m}$ converges to $1$ and the claim follows.

Then

$$\lim_{n\rightarrow \infty} \lim_{m\rightarrow \infty} x^{r_ms_n}=\lim_{n\rightarrow \infty} x^{rs_n}=x^{rs}$$

as desired.

Only I need to justify the step when the limit and the rational exponent commutes $\lim_n {a_n}^q=(\lim _n a_n)^q$ assuming that $a_n$ converges to a positive real number.

Is this a correct argument?

• Your argument seems to be correct. About the commuting of rational exponents with limits, you should be able to show that function $f(x) = x^{q}$ with $q$ rational is continuous for $x > 0$. This is easily proved using algebraic inequalities. – Paramanand Singh Feb 7 '14 at 5:43
• You can get the proof of continuity of $x^{q}$ via inequalities at math.stackexchange.com/a/667040 – Paramanand Singh Feb 8 '14 at 3:00