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

The following proof builds upon the ideas introduced in the text so far and formalizes the notions in the other answer by Orest Bucicovschi.

Let us assume that $x > 1$ and $r > 0$. First we note that since $q,r\in\mathbb{R}$ there exist convergent sequences of rational numbers $(q_{n})_{n=0}^{\infty}$ and $(r_{n})_{n=0}^{\infty}$ such that $q = \lim_{n\rightarrow\infty}q_{n}$ and $r = \lim_{n\rightarrow\infty}r_{n}$. Since $r > 0$, it can be written as the limit of a sequence whose terms are positively bounded away from $0$ so without loss of generality we may assume that $r_{n}>0$ for all $n\in\mathbb{N}$.

Since $(q_{n})_{n=0}^{\infty}$ and $(r_{n})_{n=0}^{\infty}$ are convergent sequences, by the limit laws the sequence $(q_{n}r_{n})_{n=0}^{\infty}$ is also convergent (and hence bounded) with limit $qr$. Let $M\in\mathbb{Q}$ be a bound of this sequence. Since $\lim_{m\rightarrow\infty}x^{1/m} = \lim_{m\rightarrow\infty}x^{-1/m} = 1$ (as proved earlier in the text) there exists a sufficiently large $m\in\mathbb{N}$ such that $$|x^{\pm 1/m}-1|\leq x^{\mp M}\frac{\varepsilon}{2}$$ Since $(q_{n})_{n=0}^{\infty}$ is convergent there exists $N\in\mathbb{N}$ such that for all $n\geq N$ $$q_{n}-\frac{1}{m}\leq q\leq q_{n}+\frac{1}{m}$$ Let $n\geq N$. First, one must prove the analogue of Lemma $5.6.9\:(e)$ in the text for real exponents (which is not too difficult). Then, since $x > 1$ $$x^{q_{n}-\frac{1}{m}}\leq x^{q}\leq x^{q_{n}+\frac{1}{m}}$$ Since $r_{n}> 0$ by Lemma $5.6.9\:(d)$ (for rational exponents) $$(x^{q_{n}-\frac{1}{m}})^{r_{n}}\leq (x^{q})^{r_{n}}\leq (x^{q_{n}+\frac{1}{m}})^{r_{n}}$$ Using Lemma $5.6.9\:(b)$ (for rational exponents) $$x^{q_{n}r_{n}}(x^{-\frac{1}{m}}-1)\leq (x^{q})^{r_{n}}-x^{q_{n}r_{n}}\leq x^{q_{n}r_{n}}(x^{\frac{1}{m}}-1)$$ Again using Lemma $5.6.9\:(e)$ (for rational exponents) $$-\frac{\varepsilon}{2}\leq x^{-M}(x^{-\frac{1}{m}}-1)\leq (x^{q})^{r_{n}}-x^{q_{n}r_{n}}\leq x^{M}(x^{\frac{1}{m}}-1)\leq \frac{\varepsilon}{2}$$ Therefore, $$\lim_{n\rightarrow\infty}((x^{q})^{r_n}-x^{q_nr_n}) = 0$$ The case $r<0$ can be proved in a similar manner and the case $r = 0$ is trivial. The case $x = 1$ is also trivial and the case $x < 1$ can be dealt with using the limit laws (writing $x = 1/y$ for $y > 1$).

Answer 3

$ \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 $$ (x^q)^r=\seqlim{n}(x^q)^{r_n}=\seqlim{n}x^{qr_n}=x^{qr},\quad [\seqlim{n}qr_n=qr] $$