Fix $b>1$.

Problem 6(a): Let $m$, $n$, $p$, $q$ be integers such that $n>0$, $q>0$, and $r = m/n = p/q$. Then I've managed to prove that $$b^{m/n} = b^{p/q}.$$ So we can reasonably define $b^r$ as $$b^r \colon = \sqrt[n]{b^m}.$$

From this definition, we can
Problem 6(b) prove that $$ b^{r+s} = b^r \cdot b^s, $$ where $r$ and $s$ are any rational numbers.

Problem 6(c): Now for a rational number $r$, let the set $B(r)$ be defined as follows: $$B(r) \colon= \{ \, b^t \colon \, t\in \mathbb{Q}, \, t \leq r \, \}. $$ Then it is clear that $$b^r = \sup B(r).$$ So for every real $x$, we can define $b^x$ as follows: $$b^x \colon= \sup B(x) = \sup \{ \, b^t \, \colon \, t \in \mathbb{Q}, \, t \leq x \, \}. $$

Problem 6(d): Using this definition, we are required to prove that, for every pair of real $x$ and $y$, the equation $$b^{x+y} = b^x \cdot b^y$$ holds.

My work:

If $r$, $s \in \mathbb{Q}$ such that $r \leq x$ and $s \leq y$, then $r+s \in \mathbb{Q}$ also and $r+s \leq x+y$ so that we can write $$ B(x) \cdot B(y) = \{ \, b^r \cdot b^s \, \colon \, r \in \mathbb{Q}, s\in \mathbb{Q}, \, r \leq x, s\leq y \, \} \subseteq \{ \, b^t \, \colon \, t \in \mathbb{Q}, \, t \leq x+y \, \} = B(x+y),$$ and hence $$b^x \cdot b^y = \sup B(x) \cdot \sup B(y) = \sup \{ \, b^r \, \colon \, r \in \mathbb{Q}, \, r \leq x \, \} \cdot \sup \{ \, b^s \, \colon \, s \in \mathbb{Q}, \, s \leq y \, \} = sup \{ \, b^{r+s} \, \colon \, r \in \mathbb{Q}, \, s\in \mathbb{Q}, \, r \leq x, \, s \leq y \, \} \leq \sup \{ \, b^t \, \colon \, t \in \mathbb{Q}, \, t \leq x+y \, \} = \sup B(x+y) = b^{x+y}.$$ Here we have used the following definition: Given two non-empty sets $U$ and $V$, say of real numbers, we define the set $U \cdot V$ as follows: $$U \cdot V \colon= \{ \, u \cdot v \, \colon \, u \in U, \, v \in V \, \}.$$

And we have also used the fact that if $W$ and $Z$ are two non-empty bounded above subsets of the set of positive real numbers such that $W \subseteq Z$, then we must have $$ \sup W \leq \sup Z.$$

So far, we have shown that $$b^x \cdot b^y \leq b^{x+y}.$$ Now how to prove the reverse inequality using the machinery developed above? That is, how to prove that $$b^{x+y} \leq b^x \cdot b^y?$$

  • $\begingroup$ Use a sequence $(t_n) \to x+y$ where $t_n\in\mathbb Q$ to reduce it to the same identity with $x,y\in\mathbb Q$. $\endgroup$ – AlexR Nov 3 '14 at 18:05
  • $\begingroup$ AlexR, sequences and their convergence are to be discussed later on in Rudin's text; so we should avoid them for this problem in the interest of rigor, I suppose. $\endgroup$ – Saaqib Mahmood Nov 3 '14 at 18:21
  • $\begingroup$ You can find one solution here: minds.wisconsin.edu/handle/1793/67009 $\endgroup$ – user155861 Jan 10 '15 at 12:59
  • 2
    $\begingroup$ This solution is not satisfactory @wellynaught. Because, if you read it, that solution "cheats". It consider $B(x)=\{b^t|t<x;t\in \Bbb{Q}\}$. While the book (Rudin) defines $B(x)=\{b^t|t\leq x;t\in \Bbb{Q}\}$. Is exactly this equality that is causing trouble. If it were only "$<$", we would be done. $\endgroup$ – Ders Jan 10 '15 at 13:13
  • $\begingroup$ Under Rudin's framework, I would explore his section on using Dedekind cuts, which should "snap in" nicely with this definition of irrational exponentiation. $\endgroup$ – Emily Jan 16 '15 at 21:49

By definition $b^{x+y} = \sup B(x+y)$, where $B(x+y)$ is the set of all numbers $b^t$ with $t$ rational and $t<x+y$. Then any rational number $t < x+y$ can be written as $r+s$ where $r,s$ rational and $r<x,s<y$ (Think about why this is true). Hence we can write $B(x+y)$ as the set of all numbers $b^rb^s$ with $r<x,s<y$, this means that $B(x+y)$ is the set of all products $uv$ where $u\in B(x)$ and $v\in B(y)$.

Since any such product is less than $\sup B(x) \sup B(y)$, we can write $M= \sup B(x) \sup B(y)$ is an upper bound for $B(x+y)$. On the other bound, suppose $0<c<\sup B(x) \sup B(y)$, then $c/(\sup B(x)) < \sup B(y)$. Let $m=\frac{1}{2} \frac{c}{\sup B(x)}+\sup B(y)$. Then $c/(\sup B(x)) < m < \sup B(y)$, and $\exists u \in B(x), v \in B(y)$ s.t. $c/m < u, m<v$ Hence we have $c=(c/m) \times m < uv \in B(x+y)$, and so $c$ is not an upper bound of $B(x+y)$. It follows that $\sup B(x) \sup B(y)$ is the least upper bound of $B(x+y)$. Then we have this equality $b^{x+y}=b^xb^y$ as required.

  • $\begingroup$ Sorry, I don't understand. Since $m= \frac{1}{2} \frac{c}{\sup B(x)}+\sup B(y)$, clearly $m\gt \sup B(y)$. But you say $m\lt \sup B(y)$. $\endgroup$ – Hongyan May 6 '18 at 9:27

I am stuck in the same question. Maybe we can think together. :)

I've divided into two cases: $x+y\notin \Bbb{Q}$ and $x+y\in\Bbb{Q}$.

Case 1:$x+y\notin \Bbb{Q}$

Well, as you've done, we have that $B(x)\cdot B(y)\subset B(x+y)$. Ok.
So, let $b^t\in B(x+y)$. Then $t\in \Bbb{Q}$ and $t<x+y$, since $x+y\notin \Bbb{Q}$. So, we have $t-y<x$ and, therefore, $\exists r\in(t-y,x)\cap\Bbb{Q}$. Set $s=t-r\in\Bbb{Q}$. Thus, we have $$t=r+s<x+y, \text{ with }r<x \text{ and } s<y.$$ And so, $b^t=b^{r+s}=b^rb^s$, by item (b) of this exercise 6. But $b^r\in B(x)$, and $b^s\in B(y)$. So $b^t=b^rb^s\in B(x)\cdot B(y)$.

We've proved then that $B(x+y)\subset B(x)\cdot B(y)$ and, hence, $B(x+y)=B(x) \cdot B(y)$.
Then $$\sup B(x+y)=\sup B(x) \cdot B(y)$$ $$=\sup B(x) \cdot \sup B(y),$$ since $B(x),B(y) \subset [0, \infty )$.

But, what about the case $x+y\in \Bbb{Q}$? I haven't done great advances on it.. :( Can you guess anything?

  • $\begingroup$ @Saaqib Mahmuud $\endgroup$ – Ders Jan 10 '15 at 12:47
  • $\begingroup$ This does not really answer the question. If you have a different question, you can ask it by clicking Ask Question. You can also add a bounty to draw more attention to this question. $\endgroup$ – Paul Jan 10 '15 at 12:55
  • $\begingroup$ I've done it. I've added a bounty! ^_^ @Paul $\endgroup$ – Ders Jan 10 '15 at 12:57

Since $b^{x+y}=\sup\{b^t:t\in \mathbb Q,\ t\le x+y\}$, we know that $\forall \epsilon>0$, $\exists r\in\mathbb Q$, $r\le x+y$, such that $b^r>b^{x+y}-\epsilon$.

Claim: $\exists r'\in\mathbb Q$, $r'\lt x+y$, such that $b^{r'}>b^{x+y}-2\epsilon$.

If $x+y \notin \mathbb Q$, $r'=r$ satisfies the condition. Let's suppose $r=x+y\in\mathbb Q$, and consider $r'=r-\frac 1 n<r$. Since $b^r-b^{r'}=b^r(1-b^{-\frac 1 n})$, we only need to show that $\lim_{n\to\infty}b^\frac 1 n = 1$, and then choosing $n$ large enough, $r'$ is what we need to find. This is an easy exercise in calculus.

Now we can prove part(d) of the problem. Write $r'$ as $p+q$, where $p\le x$ and $q\le y$ are both rational. This is possible because $r'<x+y$ and rational numbers are dense. Hence, $b^x\cdot b^y\ge b^p\cdot b^q = b^{r'}>b^{x+y}-2\epsilon$. Since $\epsilon$ is arbitrary, $b^x\cdot b^y \ge b^{x+y}$.


If you know $$ b^{x} \cdot b^{y} \leq b^{x + y} \quad \text{for all real $x$, $y$,} \tag{1} $$ and you want to prove $$ b^{u + v} \leq b^{u} \cdot b^{v} \quad \text{for all real $u$, $v$,} \tag{2} $$ you can set $x = u + v$ and $y = -v$ (so that $u = x + y$), and read off $$ b^{u+v} \cdot b^{-v} \leq b^{u} \quad \text{for all real $u$, $v$.} $$ This reduces the desired inequality (2) to $$ b^{-v} \cdot b^{v} = 1 \quad \text{for all real $v$.} \tag{3} $$ At the very least, (3) looks simpler that (2), if it's not "obvious" (for some value of obvious). Particularly, $b^{-v} \cdot b^{v} = 1$ for all rational $v$, so if one knows exponentiation is continuous, the proof is complete.

  • $\begingroup$ If continuity is already known, nothing needs doing. $\endgroup$ – Eclipse Sun Jan 10 '15 at 14:56

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