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,


  • $\begingroup$ Do you already that $x^\alpha$ is continuous (as a function of $x$)? $\endgroup$ – Hagen von Eitzen Dec 16 '15 at 15:54
  • $\begingroup$ @HagenvonEitzen : in the same paragraph where this exercise is stated, there's a lemma called 'Continuity of exponentiation' which says that if $x\in\mathbb{R}^+$, $\alpha\in\mathbb{R}$ and $(q_n)_{n=1}^\infty$ is any sequence of rational numbers converging to $\alpha$, then $(x^{q_n})_{n=1}^\infty$ is a convergent sequence. Are you talking about this? $\endgroup$ – lorenzo Dec 16 '15 at 17:18

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.


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 exists 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 WLOG we may assume that $\forall n\in\mathbb{N}\:r_{n}>0$.

Since $(q_{n})_{n=0}^{\infty}$ is convergent we have $$\forall m\in\mathbb{N}\:\exists N_{m}\in\mathbb{N}\:\text{s.t.}\:\forall n\geq N_{m}\:q_{n}-\frac{1}{m}\leq q\leq q_{n}+\frac{1}{m}$$ Fix $m$ and and let $n\geq N_{m}$. First, one must prove the analogue of Lemma $5.6.9\:(e)$ in the text for the reals (which is not too hard). Then, since $x > 1$ $$x^{q_{n}-\frac{1}{m}}\leq x^{q}\leq x^{q_{n}+\frac{1}{m}}$$ Since $r_{n}> 0$ $$(x^{q_{n}-\frac{1}{m}})^{r_{n}}\leq (x^{q})^{r_{n}}\leq (x^{q_{n}+\frac{1}{m}})^{r_{n}}$$ Using Lemma $5.6.9\:(b)$ for the rationals we have $$x^{q_{n}r_{n}}(x^{-\frac{1}{m}}-1)\leq (x^{q})^{r_{n}}-x^{q_{n}r_{n}}\leq x^{q_{n}r_{n}}(x^{\frac{1}{m}}-1)$$ 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 with limit $qr$. Therefore, this sequence is also Cauchy and hence bounded. Let $M$ be the bound of this sequence (WLOG $M$ may be asusmed to be rational). Again using Lemma $5.6.9\:(e)$ $$x^{-M}(x^{-\frac{1}{m}}-1)\leq (x^{q})^{r_{n}}-x^{q_{n}r_{n}}\leq x^{M}(x^{\frac{1}{m}}-1)$$ Since $\lim_{m\rightarrow\infty}x^{1/m} = \lim_{m\rightarrow\infty}x^{-1/m} = 1$ (as proved earlier in the text) $$\forall\epsilon > 0\:\exists N_{\epsilon}\in\mathbb{N}\:\text{s.t.}\:\forall m\geq N_{\epsilon}\:|x^{\pm 1/m}-1|\leq x^{\mp M}\frac{\epsilon}{2}$$ So $\forall n\geq N_{N_{\epsilon}}$ we have $$-\frac{\epsilon}{2}\leq (x^{q})^{r_{n}}-x^{q_{n}r_{n}}\leq \frac{\epsilon}{2}$$ Also, be definition of real exponentiation we have $(x^{q})^{r}=\lim_{n\rightarrow\infty} (x^{q})^{r_{n}}$ $$\implies\forall\epsilon>0\:\exists K_{\epsilon}\in\mathbb{N}\:\text{s.t.}\:\forall n\geq K_{\epsilon}\:|(x^{q})^{r_{n}}-(x^{q})^{r}|\leq\frac{\epsilon}{2}$$ Finally, we have by the triangle inequality $$\forall n\geq\max\{N_{N_{\epsilon}}, K_{\epsilon}\}\:|(x^{q})^{r}-x^{q_{n}r_{n}}|\leq\epsilon$$ Since this can be done for any $\epsilon>0$ we have $\lim_{n\rightarrow\infty}x^{q_{n}r_{n}} = (x^{q})^{r}$. But by the definition of real exponentiation $\lim_{n\rightarrow\infty}x^{q_{n}r_{n}} = x^{qr}$ so we have proved the result.

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 $y > 1$).


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