Proving Rudin 1.6d (Exponentiation rule for real numbers)

Question

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!

Answer 1

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: $$b^xb^y<b^{x+y}.$$ 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 $$ b^xb^y<b^t\leq b^{x+y}.$$ Now, since $x+y$ is irrational, and $t$ is rational, we must have a strict inequality, $t<x+y$, so that: $$ b^xb^y<b^t<b^{x+y}.$$ Since $t<x+y$, then $t-x<y$ 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 $t-x<p$ that $q<x$. So we have found rationals $p$ and $q$ such that $t=p+q$, $p<y$ and $q<x$. Now, $$b^t=b^{p+q}=b^pb^q<b^xb^y,$$ but we've already asserted that $b^xb^y<b^t$ 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?

Answer 2

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 $

Answer 3

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, $\sup cA = c\cdot\sup A$.