Hi everyone I have serious problems with this exercise. I would appreciate any help. Thanks.

(I can use freely any fact regard to real exponentiation when the exponent is a rational number, other tools are not develop in the book yet. Only basic facts about limits and that's all).

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 $\alpha$.

Proposition: Let $x,y$ be a positive real numbers and let $(q_n)_{n=1}^{\infty},(r_n)_{n=1}^{\infty}$ be sequences of rational which converges to the real numbers $q,r$ respectively.

(a) $\,x^q$ is a positive real number.

(b) If $q>0$, $x>y \iff x^q>y^q$

(c) If $x>1$, then $x^q>x^r \iff q>r$. If $x<1$, then $x^q>x^r \iff q<r$.


(a) It will suffice to show that the sequence $(x^{q_n})$ is positively bounded away from zero. Suppose that $x>1$ (notice that the case when $x=1$ it's trivial). We have either $q$ is zero or is different to zero, if $q=0$ the sequence $(x^{q_n})$ eventually converges to $1$. Now, we may assume $q\not=0$. Let $k$ be a natural number such that $1/k \le|q|/2$. By hypothesis, we know $(q_n)\rightarrow q$, so we have for each $n\ge N(1/k)$ [where $N(1/k)\in \mathbb{N}$], $\,d(q_n, q)\le 1/k$. Then $|q_n| \ge |q|-|q_n-q|\ge |q|/2\ge 1/k$; we find a lower bound for $(q_n)$. Since $x>1$, we must have $\,x^{1/k}\le x^{|q_n|}$.

Suppose $q$ is positive, so eventually all the terms in $(q_n)$ would be positive rational number (otherwise $(q_n)$ cannot be a Cauchy sequence). Thus, $\,x^{1/k}\le x^{q_n}$ and then $x^{1/k}\le \text{lim}_{n \rightarrow \infty}\, x^{q_n}= x^q$. Since $x^{1/k}>0$, we're done. Similarly, when $q$ is negative.

Now, when $x<1$. We set $y= 1/x$ and use the same argument as above. Thus $y^q$ is a positive real number. We thus have $\text{lim}_{n\rightarrow \infty} (x^{q_n})= \text{lim}_{n\rightarrow \infty}\, 1/(1/(x^{q_n}))=1/ \text{lim}_{n\rightarrow \infty}\, y^{q_n} = 1/y^q$. Thus, $x^q$ must be positive.

(b) Since $x>y$ implies $x^{q_n}>y^{q_n}$, when we apply the limit we have $x^{q}\ge y^{q}$. We need to show that $x^{q}\not= y^{q}$.

For the sake of contradiction suppose $x^q=y^q$; the sequence $(x^{q_n}-y^{q_n})\rightarrow 0$, so it will suffice to show $(x/y)^{q_n} \rightarrow 1$ yields to a contradiction [since $x^{q_n}-y^{q_n}=x^{q_n}((x/y)^{q_n}-1)$]. By hypothesis we know $x>y$ so $(x/y)>1$. Since $q>0$, it is eventually bounded away from zero; then, for a sufficient large $N$, we have $q_n\ge c$ where $c\in \mathbb{Q}>0$. This would imply $(x/y)^{q_n}\ge (x/y)^c$ (since $x/y>1$ ), thus $(x/y)^{q}\ge (x/y)^c$. But $x^c>y^c$ because is a positive rational number. Hence $(x/y)^{q}>1$ which is a contradiction.

(c) Suppose $x>1$ and $q>r$, so the sequence $(q_n-r_n)$ is positively bounded away from zero, so, there is some $N$ so that $q_n-r_n\ge c>0$.Then $q_n>r_n$ for every $n\ge N$, and since $x>1$, we have $x^{q_n}> x^{q_n}$. Thus $x^q\ge x^r$ when we apply the limit in the inequality. Now we need to show that $x^q \not= x^r$. Suppose for contradiction that $x^q = x^r$, so $x^r (x^{q-r}-1)=0$, since $x^r>0$ by (a), we must have $x^{q-r}-1=0$ or $\,x^{q-r}=1$. We know that $q_n-r_n\ge c >0$ for each $n\ge N$. So, $q-r\ge c$ and thus $x^{q-r}\ge x^c>1$, a contradiction.

Suppose $x^q>x^r$. So, $x^q-x^r>0$ this means that the sequence $(x^{q_n}-y^{r_n})$ is eventually bounded away from zero, i.e., $x^{q_n}-x^{r_n}\ge c>0$. So, $x^{q_n}>x^{r_n}$ which implies $q_n\ge r_n$ and then $q\ge r$. Suppose for contradiction that $q=r$, so $(q_n-r_n)\rightarrow 0$. Now since $(r_n)$ is a convergent sequence it has some upper bounded. Let $M$ be some upper bound, then $r_n\le M$ and so $x^{r_n}\le x^M$ (since $x>1$). Then we have $x^{r_n} (x^{q_n-r_n}-1)\le x^{M} (x^{q_n-r_n}-1)$

Let $\epsilon= c/2$. Now since $x^{1/k}\rightarrow1$, so is eventually $\epsilon/ x^{M}$ -close to $1$. Thus, $d(x^{1/k},1)\le \epsilon /x^{M}$ for each $k\ge K$. Furthermore, $(q_n-r_n)\rightarrow 0$ and hence for a sufficient large $n$, it's possible $d(q_n,r_n)\le 1/K$. Thus, $x^{r_n} (x^{q_n-r_n}-1)\le x^{M} (x^{q_n-r_n}-1)\le x^{M} (x^{1/K}-1)\le \epsilon$. But, then $x^{q_n}-x^{r_n}\le c/2$, contradicting that is eventually bounded away from zero. Thus $q> r$

If $x<1$, then $y=1/x>1$. So, using the first part we have $y^r>y^q \iff r>q$. Then $1/x^r> 1/x^q$ and because both are positive [by the part (a)], then this implies $x^q>x^r$, as desired.

So, I have no idea of how to do (b). I know that $x^q\le y^q$ but I don't know how to show $x^q\not= y^q$. Any suggestion?

  • $\begingroup$ Guys, any suggestion, please? $\endgroup$ – Jose Antonio Nov 12 '13 at 3:20
  • $\begingroup$ It is a tricky estimate... one idea that I have is to look consider $x,y\in[a,b]$ then the rational power functions $x\mapsto x^n$ and $x\mapsto x^{1/n}$ are continuous on a compact set, and hence uniformly continuous. A uniform bound may help to give a contradicition. $\endgroup$ – Moss Nov 12 '13 at 3:29

If $x^{q_i}$ converge to the same thing as $y^{q_i}$ then $x^{q_i}-y^{q_i}=y^{q_i}((x/y)^{q_i}-1)$ converge to zero. But $y^{q_i}((x/y)^{q_i}-1)$ is easy to bound away from zero, as $q_i$ can be taken to be increasing.

  • $\begingroup$ I've done some changes in the original post. I hope this work. Thanks for you help :) $\endgroup$ – Jose Antonio Nov 12 '13 at 5:06

First, I am concerned about your definition. There are lots of sequences {$q_n$} $\rightarrow$ q. So just saying that $q_n \rightarrow q$ does not in itself imply that you can define $x^q =lim_{n \rightarrow \infty}x^{q_n} $ If you are going to use this kind of definition you have to show that if {$s_n$} is any other sequence that converges to q that $x^q =lim_{n \rightarrow \infty}x^{s_n} $ In other words, your $x^q$ must be invariant under any change in the choice of sequence going to q.

Assuming you fix that up and since you are happy with part a:

Part b. let 0 < x < y. Then y = x + b where b > 0. So $y^q = (x + b)^q = x^q +[\text{ positive stuff }] > x^q$. If you are uneasy about expanding $(x + b)^q$ in this way, you can expand $(x + b)^{q_n}$ and take the limit.

Part c.We have x > 1 and $x^q > x^r$. Suppose q < r. Then r + q + m where m > 0 and

$x^r = x^{(q+m)} = x^q \cdot x^m > x^q$ which is a contradiction to our hypothesis.

  • $\begingroup$ Yes, in other part of the book the author shows that this definition of exponentiation is well-define in the sense that the sequence always converges, if $(q_n)\rightarrow q$ and $q\in \mathbb{R}$, and which has the same limit for any other sequence which converges to $q$. Do you think the other parts are correct? (Thanks for you answer, I'll read carefully) $\endgroup$ – Jose Antonio Nov 12 '13 at 6:15
  • $\begingroup$ The binomial expansion is not define yet in the book, I think is two chapter later is when the author define a proof any factor about it. $\endgroup$ – Jose Antonio Nov 12 '13 at 6:32
  • $\begingroup$ @JoseAntonio I am glad your book gets around to correctly defining $x^q$. Re binomial theorem, instead do this: $(x+b)^q = x^q(1+b/x)^q \ge x^q,$ since x, b > 0 and 1 + x/b >1 Re your original proof part a seems correct; Patrick's approach of letting $q_n \rightarrow q$ where $q_n \le$ q is better. Note that depends on your later theorem that all sequences $q_n$ give you the same $x^q$. $\endgroup$ – Betty Mock Nov 13 '13 at 19:03
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    $\begingroup$ I admit I did not read b and c because they seem so cumbersome compared to using the known properties of the exponential functions. If you must submit this kind of primitive proof, let me know, and I will read through it. $\endgroup$ – Betty Mock Nov 13 '13 at 19:04
  • $\begingroup$ Thanks for all your entire help. Yes is one interesting aspect of the Tao's notes, much more of the proof are very "primitive". I really love this book, for the beautiful and careful explanation, and because is the only one (at least which I've seen) which develop everything of the very very low level. Thanks. $\endgroup$ – Jose Antonio Nov 14 '13 at 4:01

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