If $q, r \in \mathbb{R}, x \in \mathbb{R}^+$ then $(x^q)^r=x^{qr}$

Question

I'm stuck on this exercise from Tao's Analysis 1 textbook:

show that if $q, r \in \mathbb{R}, x \in \mathbb{R}^+$ then $(x^q)^r=x^{qr}$.

DEF. (Exponentiation to a real exponent): Let $x>0$ be real, and let $\alpha$ be a real number. We define the quantity $x^\alpha$ by the formula $x^\alpha=\lim_{n\to\infty} x^{q_n}$, where $(q_n)_{n=1}^\infty$ is any sequence of rational numbers converging to $\alpha$.

I've already proved any property of real exponentiation when the exponent is a rational number (for example that the property in question holds when $x \in \mathbb{R}^+$ and $q,r \in \mathbb{Q}$) and that $x^q$ (with $x \in \mathbb{R}^+$ and $ q \in \mathbb{R}$) is a positive real number.

What puzzles me is how to get around the fact that we are considering two limits simultaneously, in fact from the definition above it follows that $(x^q)^r=\lim_{n\to\infty}(\lim_{m\to\infty}x^{q_n})^{r_n}$.

(This question: $(x^r)^s=x^{rs}$ for the real case talks about this exercise, but I don't understand how the author can say that $(x^q)^r=\lim_{n\to\infty}(\lim_{m\to\infty}x^{q_n})^{r_n}=\lim_{n\to\infty}\lim_{m\to\infty}((x^{q_n})^{r_n})=\lim_{n\to\infty}\lim_{m\to\infty} x^{q_nr_n}$.)

So, I would appreciate any hints about how to start/carry out its proof.

Best regards,

lorenzo

Answer 1

By definition: $$(x^q)^r = \lim_n (x^q)^{r_n}$$ Now, $x_q$ is the limit of any sequence $x^{q_n}$ where $q_n$ rationals and $\to q$. For every $n$, choose a $q_n$ so that $x^{q_n}$ is very close to $x^q$. How close? So that $ |(x^{q_n})^{r_n} - (x^q)^{r_n}| < 1/n$ and also $|q_n - q| < 1/n$. Now, since $(x^q)^{r_n} \to (x^q)^r$ and $(x^{q_n})^{r_n} - (x^q)^{r_n} \to 0$ we get $(x^{q_n})^{r_n} \to (x^q)^r$. But $(x^{q_n})^{r_n} = x^{q_n \cdot r_n}$ ( OK for rational exponents) and $q_n \to q$, $r_n \to r$, so $q_n r_n \to q r$. So from the above we have $x^{q_n r_n} \to (x^q)^r$. But $x^{q_n r_n} \to x^{q r}$ by the definition of the power $x^{qr}$. We conclude that $(x^q)^r = x^{q r}$

Obs: The idea was to choose first $r_n\to r$ arbitrary, but then to take $q_n\to q$ that also has some extra properties.

Answer 2

$ \newcommand{\seqlim}[1]{\lim\limits_{#1\to\infty}} $ I find a simple answer for this old question.
Show that $ (x^s)^r = x^{sr} $ for $ s\in\mathbb{Q},r\in\mathbb{R} $ $$(x^s)^r=\seqlim{n}{(x^s)^{r_n}}=\seqlim{n}{x^{sr_n}}=a^{sr},\quad [\seqlim{n}sr_n=sr] $$ Then we have $$ (x^q)^r=\seqlim{n}(x^q)^{r_n}=\seqlim{n}x^{qr_n}=x^{qr},\quad [\seqlim{n}qr_n=qr] $$