# Proving Rudin 1.6d (Exponentiation rule for real numbers)

I have been working through some of the early problems in Baby Rudin to prepare for a class next year, but am stuck on part (d) of question 1.6.

Fix $b > 1$.

(a). If $m, n, p, q$ are integers, $n > 0, q > 0,$ and $r = \frac{m}{n} = \frac{p}{q}$, prove that: $$(b^m)^{1/n} = (b^p)^{1/q}.$$ Hence it makes sense to define $b^r = (b^m)^{1/n}$.

(b). Prove that $b^{r+s} = b^rb^s$ if $r$ and $s$ are rational.

(c). If $x$ is real, define $B(x)$ to be the set of numbers $b^t$, where $t$ is rational and $t\leq x$. Prove that: $$b^r = \sup B(r)$$ when $r$ is rational. Hence it makes sense to define: $$b^x = \sup B(x)$$ for every real $x$.

(d). Prove that $b^{x+y} = b^xb^y$ for all real $x$ and $y$.

I have thus far been able to show parts a,b,c with relatively easy concepts, but am struggling to find a solution for part d. Below is my work for parts a,b,c. Feel free to look them over for mistakes.

(a). Since $\frac{m}{n} = \frac{p}{q}$, we know: $mq = np = y$. Then, by $\textbf{Theorem 1.21 of the text}$, we know that $x^{nq} = b^y$ is unique. We shall demonstrate that $(b^m)^{1/n} = (b^p)^{1/q} = b^r$: $$((b^m)^{1/n})^{nq} = (b^m)^q = b^y$$ $$((b^p)^{1/q})^{nq} = (b^p)^n = b^y$$ Thus, $((b^m)^{1/n})^{nq} = b^y = ((b^p)^{1/q})^{nq}$, and so, $(b^m)^{1/n} = b^r = (b^p)^{1/q}$, as desired.

(b). First, let $r = \frac{m}{n}, s = \frac{p}{q}$ for $m,n,p,q \in \Bbb{Z}$. Then, $b^{r+s} = b^{m/n +p/q} = (b^{mq+np})^{1/nq}.$ Since $mq, np \in \Bbb{Z}$, we can say, $(b^{mq+np})^{1/nq} = (b^{mq}b^{np})^{1/nq}$. We get: $(b^{mq}b^{np})^{1/nq} = (b^{m/n}b^{p/q}) = b^rb^s$, as desired.

(c). We consider $B(r) = \{b^t \mid t \in \Bbb{Q} \ \& \ t \leq r\}$. For any $t$, $b^r = b^tb^{r-t} \geq b^t1^{r-t}$, since $b > 1$. Thus, $b^r$ is an upper bound of $B(r)$. Since $b^r \in B(r)$, we conclude that $b^r =\sup B(r)$, as desired.

(d). For this part, I have considered doing $b^x = \sup B(x)$, so $B(x) = \{b^t \mid t \leq x, t \in \Bbb{Q}\}$. Then, $b^xb^y = \sup B(x)\sup B(y) \geq b^rb^s = b^{r+s} = \sup B(r+s)$, for $r \leq x, \ s\leq y, \ r,s \in \Bbb{Q}$. Thus, $\sup B(x)\sup B(y) \geq\sup B(r+s)$, and since $r+s \leq x+y$, we have $\sup B(x+y) \leq \sup B(x)\sup B(y)$, which would set $b^xb^y$ as an upper bound for $b^{x+y}$.

Here I come across two issues, one in that I am not sure if this is in fact correct. Since we assumed $r,s$ were rational, I am not entirely sure if it is true that $\sup B(r+s) = \sup B(x+y)$ for $x,y \in \Bbb{R}$, as I think would have to be the case for me to then claim that since $r+s \leq x+y$, $b^xb^y$ is an upper bound for $b^{x+y}$. Is this true, how would one prove this?

My second issue is that, given the above is true, I can't figure out how one would proceed to demonstrate that $\sup B(x+y)$ is an upper bound for $b^xb^y$.

Any help would be greatly appreciated!

• If $s\in S$ is an upper bound on $A\subset S$, then $s\ge \sup S$, by definition.
– user418131
Commented Sep 17, 2018 at 13:26
• However, the problem is that if $x+y\in \mathbb{Q}$ and both $x$ and $y$ are not rational, then by the current definition for $b^x, x\in\mathbb{R}$, we cannot find $r\le x$ and $s\le y$ in $\mathbb{Q}$ such that $r+s=x+y$
– user418131
Commented Sep 17, 2018 at 13:28
• We can conclude at this stage, however, that $b^{x+y}\ge b^xb^y$ from your calculations, as opposed to the converse, which you proved
– user418131
Commented Sep 17, 2018 at 13:36
• @AnotherJohnDoe Can you explain how I could deduce that from what I have now? Commented Sep 17, 2018 at 13:38
• Ok, consider the following. We know that $b^x=\sup B(x)$ and $b^y=\sup B(y)$. So then the set of products $b^rb^s$ has supremum $b^xb^y$. If you refer to my first comment, we showed that $b^{x+y}$ greater than or equal to each of these products, and thus we have $b^{x+y}\ge b^xb^y$
– user418131
Commented Sep 17, 2018 at 13:55

I don't think the question has really been answered yet, and it's been bothering me for almost a week. I have a partial solution, which I present below (it's really too long for a comment.) I hope others can help fill in a critical gap. To prove $$b^xb^y=b^{x+y}$$, we prove the two inequalities $$b^xb^y\leq b^{x+y}$$ and $$b^{x+y}\leq b^xb^y$$. Theoretical Economist provided an outline for a proof of the first inequality above which I believe is correct. (I can provide all the details if anyone is interested.)

The difficulty is the second inequality. I have a solution for the case where $$x+y$$ is irrational which I present below. The case where $$x$$ and $$y$$ are irrational, but $$x+y$$ is rational is proving, at least for me, to be quite difficult and I don't yet have a solution. Note that this can occur, for instance, if, say $$x=\sqrt{2}$$ and $$y=2-\sqrt{2}$$. As far as I'm able to determine, none of the proof outlines presented above work in that case. (This was noted by a commenter above.) So here is my solution for the proof of $$b^{x+y}\leq b^xb^y$$ when $$x+y$$ is irrational:

I will prove it by assuming the contrary and deriving a contradiction. So assume that: $$b^xb^y This means that: $$\sup B(x)\cdot\sup B(y) < \sup B(x+y)$$ This means there must exist a $$b^t\in B(x+y)$$ which is strictly greater than $$b^xb^y$$, for otherwise $$b^xb^y$$ would be an upper bound for $$B(x+y)$$ which would contradict the fact that $$\sup B(x+y)$$ is a least upper bound. So we have, then, that $$b^xb^y Now, since $$x+y$$ is irrational, and $$t$$ is rational, we must have a strict inequality, $$t, so that: $$b^xb^y Since $$t, then $$t-x and since $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$, we can find a rational $$p$$ such that: $$t-x < p < y$$ Now, let $$q=t-p$$. Then it follows from $$t-x that $$q. So we have found rationals $$p$$ and $$q$$ such that $$t=p+q$$, $$p and $$q. Now, $$b^t=b^{p+q}=b^pb^q but we've already asserted that $$b^xb^y so we've arrived at at contradiction, proving that $$b^{x+y}\leq b^xb^y$$. Without the assumption that $$x+y$$ is irrational, this proof would not have worked.

I've also tried an approach similar to those suggested above, but they involve finding rationals $$r$$ and $$s$$ which satisfy inequalities like: $$b^{r+s}\leq b^t$$ for all rationals $$t\leq x+y$$, where $$r\leq x$$ and $$s\leq y$$. But such rationals $$r$$ and $$s$$ do not exist when $$t=x+y$$, which can occur if $$x+y$$ is rational, and $$x$$ and $$y$$ are irrational. So again, I don't think a complete solution to this problem has yet been offered. I would love to be corrected on this point, however.

I have been thinking about how to prove the inequality when $$x+y$$ is rational, but all my thoughts involve concepts beyond Chapter 1 in Rudin where this problem appears. One could invoke continuity or something like that, I suppose, having proven the identity for irrational numbers, but that isn't available in Chapter 1. A later problem in Chapter 1 introduces the existence of logarithms, which I suppose could be used to show that the set $$\{b^q | q\in\mathbb{Q}\}$$ is dense in $$\mathbb{R}$$ and thereby find a $$b^t$$ which satisfies the strict inequality in my proof, but again, we'd be using concepts introduced later than the problem. So can anyone offer a complete proof which uses only concepts found in Chapter 1 in Rudin?

I have been working through the first chapter of Baby Rudin and originally didn't think too much about Exercise 1.6 (d). I had written something like "if $$x + y \in \mathbf{Q}$$ then the result is clear", but on re-reading I realised this is only clear if $$x$$ is rational! Just working with the supremum definition, I think this is the hardest case actually, when $$x + y \in \mathbf{Q}$$ and $$x, y$$ are irrational, $$x = 1 - \sqrt{2}$$ and $$y = \sqrt{2}$$ for example. After banging my head against this for a while, I turned to Google and Math SE for some help, but I wasn't happy with any of the "solutions" I found (see here for example). I believe I have a full solution now, which I will post below.

Claim: $$b^{x + y} = b^x b^y$$ for all real $$x$$ and $$y$$.

Proof: We will show that both of the assumptions $$b^{x+y} < b^x b^y$$ and $$b^x b^y < b^{x+y}$$ lead to contradictions. First, suppose that $$b^{x+y} < b^x b^y$$, i.e. $$\sup B(x + y) < \sup B(x) \cdot \sup B(y)$$. This assumption is equivalent to $$\frac{\sup B(x + y)}{\sup B(y)} < \sup B(x)$$, so that $$\frac{\sup B(x + y)}{\sup B(y)}$$ is not an upper bound for $$B(x)$$. Then there must exist some rational $$r$$ such that $$r \leq x$$ and $$\frac{\sup B(x + y)}{\sup B(y)} < b^r \iff \frac{\sup B(x + y)}{b^r} < \sup B(y).$$ This demonstrates that $$\frac{\sup B(x + y)}{b^r}$$ is not an upper bound for $$B(y)$$, so there must exist a rational $$s$$ such that $$s \leq y$$ and $$\frac{\sup B(x + y)}{b^r} < b^s \iff \sup B(x + y) < b^r b^s = b^{r + s}.$$ This is a contradiction since $$r + s \leq x + y \implies b^{r+s} \in B(x + y) \implies b^{r+s} \leq \sup B(x + y).$$ Now suppose that $$b^x b^y < b^{x+y}$$. We shall make use of the following inequality: $$\forall n \in \mathbf{N} \quad\quad b^{1/n} \leq 1 + \frac{b-1}{n}.$$ This can be seen by taking $$a = b^{1/n}$$ in the inequality $$\textstyle{\forall a \geq 1 \quad\quad a^n - 1 = \left( \sum_{j=0}^{n-1} a^j \right) (a - 1) \geq n(a - 1)}.$$ (This is actually Exercise 7 (a)/(b) of Baby Rudin.) By assumption $$b^{x+y} - b^x b^y > 0$$, so by invoking the Archimedean property of $$\mathbf{R}$$ we may obtain a positive integer $$n$$ such that $$n(b^{x+y} - b^x b^y) > (b - 1) b^x b^y \implies \frac{b^{x+y}}{b^x b^y} > 1 + \frac{b-1}{n} \geq b^{1/n} \implies b^x b^y b^{1/n} < b^{x+y}.$$ The density of $$\mathbf{Q}$$ in $$\mathbf{R}$$ implies that there exist rational numbers $$r$$ and $$s$$ such that $$x - \frac{1}{2n} < r \leq x$$ and $$y - \frac{1}{2n} < s \leq y$$, which implies that $$x + y < r + s + \frac{1}{n}$$. It follows that $$b^{x+y} \leq b^{r+s+1/n} = b^r b^s b^{1/n} \leq b^x b^y b^{1/n} < b^{x+y},$$ i.e. $$b^{x+y} < b^{x+y}$$, a contradiction. $$\square$$

Your approach for part (d) actually looks mostly fine, modulo some minor details. Let's try to prove the reverse inequality.

Fix $x,y\in R$, and let $s,t \in Q$ such that $s \le x$ and $t \le y$. Clearly, we have that

$$b^s b^t = b^{s+t} \le b^{x+y}.$$

If we take the supremum over $s$ and then over $t$, we would find that

$$b^x b^y \le b^{x+y},$$

which is what we needed to show.

In case you're not sure if taking the supremum as above is valid, you may want to convince yourself of the following fact:

Let $A \subset R$, and $c > 0$, and define $cA = \{ cx : x \in A \}$. Then, $\sup cA = c\cdot\sup A$.

• I don't think proving $b^{x+y}\le b^xb^y$ is quite so straightforward
– user418131
Commented Sep 17, 2018 at 13:59
• @AnotherJohnDoe I think the OP's original attempt is reasonably close to what they need to do. I was hoping they could use my answer to figure out what needs to be done to fix their attempt. However, if you disagree, perhaps you should post an answer of your own. Commented Sep 17, 2018 at 14:04