Real Exponentiation (help with an exercise)

Question

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 $x^q>x^r \iff q<r$.

Proof:

(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?

Answer 1

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.

Answer 2

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.