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Here's problem $6$ in Chapter $1$ in the book Principles of Mathematical Analysis by Walter Rudin, $3$rd edition:

Fix a real number $b$, such that $b > 1$.

$(a)$ If $m, n, p, q$ are integers, $n > 0$, $q > 0$, and $r = m/n = p/q$, then (I've managed to show that)

$$\left( b^m \right)^{1/n} = \left( b^p \right)^{1/q}.$$

Hence it makes sense to define

$$b^r \colon= \left(b^m\right)^{1/n}.$$

$(b)$ (I've also managed to show that) for any rational numbers $r, s$, we have $$b^{r+s} = b^r b^s.$$

$(c)$ For any real number $x$, define the set $B(x)$ as follows:

$$B(x) \colon= \left\{ \ b^t \ \colon \ t \ \mbox{ is rational}, \ t \leq x \ \right\}.$$

We can then prove that

$$b^r = \sup B(r)$$

when $r$ is a rational number.

Hence it makes sense to define

$$b^x \colon= \sup B(x) = \sup \left\{ \ b^t \ \colon \ t \ \mbox{ is rational}, \ t \leq x \ \right\}$$ for every real number $x$.

$(d)$ How to prove (USING THE MACHINERY DEVELOPED HERE) that $$b^{x+y} = b^x b^y$$ for all real numbers $x$ and $y$?

I've already posted this question. Here's the link. However, none of the answers there seems to work for me.

My Work

Let $X, Y$ be any two non-empty subsets of $\mathbb{R}$. Then we have the following facts:

$(1)$ If $X \subset Y$ and if $Y$ is bounded above in $\mathbb{R}$, then so is $X$ and we have the inequality $$\sup X \leq \sup Y.$$

$(2)$ If $X, Y$ are non-empty subsets of the set of non-negative real numbers and if both $X$ and $Y$ are bounded above, then so is the set

$$\{ \ xy \ \colon \ x \in X, \ y \in Y \ \};$$ moreover, we have the identity

$$\sup \{ \ xy \ \colon \ x \in X, \ y \in Y \ \} = \sup X \cdot \sup Y.$$


$$\begin{align} b^x b^y &= \sup B(x) \sup B(y) \\ &= \sup \{ \ b^r \ \colon \ r \in \mathbb{Q}, r \leq x \ \} \sup \{ \ b^s \ \colon \ s \in \mathbb{Q}, s \leq y \ \} \\ &= \sup \{ \ b^{r+s} \ \colon \ r \in \mathbb{Q}, \ s \in \mathbb{Q}, \ r \leq x, \ s \leq y \ \}. \end{align}$$

But if $r \in \mathbb{Q}$, $s \in \mathbb{Q}$, $r \leq x$, and $s \leq y$, then the sum $r+s \in \mathbb{Q}$ and $r + s \leq x + y$ also. So

$$ \{ \ b^{r+s} \ \colon \ r \in \mathbb{Q}, \ s \in \mathbb{Q}, \ r \leq x, \ s \leq y \ \} \subseteq \{ \ b^t \ \colon \ t \in \mathbb{Q}, \ t \leq x+y \ \}.$$ Therefore,

$$\begin{align} b^x b^y &= \sup \{ \ b^{r+s} \ \colon \ r \in \mathbb{Q}, \ s \in \mathbb{Q}, \ r \leq x, \ s \leq y \ \} \\ &\leq \sup \{ \ b^t \ \colon \ t \in \mathbb{Q}, \ t \leq x+y \ \} \\ &= \sup B(x+y) = b^{x+y}. \end{align}$$

Now we have to show that

$$b^x b^y \geq b^{x+y}.$$

Case I. When $x+y$ is irrational:

If $t$ is a rational number such that $t \leq x+y$, then since $x+y$ is irrational, we can also conclude that $t < x+y$.

So we can find a rational number $r$ such that

$$ t-y < r < x.$$


$$t-r < y < x+y -r.$$

Let's take $s \colon = t-r$. Then $s \in \mathbb{Q}$. And $s < y$.

Thus, we have written $t$ as $t = r + s$, where $r, s \in \mathbb{Q}$, with $r < x$, $s < y$.

So, when $x + y \in \mathbb{R} - \mathbb{Q}$, then we have

$$\begin{align} b^{x+y} &= \sup B(x+y) = \sup \{ \ b^t \ \colon \ t \in \mathbb{Q}, \ t \leq x+y \ \} \\ &= \sup \{ \ b^t \ \colon \ t \in \mathbb{Q}, \ t < x+y \} \\ &= \sup \{ \ b^{r+s} \ \colon \ r \in \mathbb{Q}, \ s \in \mathbb{Q}, \ r< x, \ s < y \} \\ &= \sup \{ \ b^r b^s \ \colon \ r \in \mathbb{Q}, \ s \in \mathbb{Q}, \ r< x, \ s < y \} \\ &= \sup \{ \ b^r \ \colon \ r \in \mathbb{Q}, \ r < x \ \} \sup \{ \ b^s \ \colon \ s \in \mathbb{Q}, \ s < y \} \\ &\leq \sup \{ \ b^r \ \colon \ r \in \mathbb{Q}, \ r \leq x \ \} \sup \{ \ b^s \ \colon \ s \in \mathbb{Q}, \ s \leq y \} \\ &= \sup B(x) \sup B(y) \\ &= b^x b^y. \end{align}$$

Case II. What if $x + y \in \mathbb{Q}$?

In this case, we can conclude that both $x$ and $y$ are irrational.

How do we proceed with showing the remaining reverse inequality in this particular case?


marked as duplicate by Alex M., Claude Leibovici, N. F. Taussig, SchrodingersCat, Kamil Jarosz Feb 1 '16 at 15:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


This is another instance of proving stuff with one hand tied behind your back, limiting knowledge of the reals to what is in Rudin ch. 1. For this reason I write out a complete proof, even though most of it is given by the OP. We assume $b>1$ real.

Lemma 0. If $p>0$ is rational, $b^p>1$.

Lemma 1. If $s<x+y$, $s$ rational, then $s=p+q$ with $p$ and $q$ rational, $p<x$, $q<y$.

Proof. Pick $s'$ rational with $s<s'<x+y$. Pick $p$ rational with $x-(s'-s)<p<x$. Let $q=s-p$. We then have $q=s-p <s-x+(s'-s)<s-x+x+y-s=y$ as desired.

Lemma 2 (Rudin ch. 1 ex. 1.7(c)). If $t>1$ and $n>(b-1)/(t-1)$ then $b^{1/n}<t$.

Lemma 3. For any $x$ we have $b^x=\sup \{b^r:r<x, r\in\mathbb{Q}\}$.

Proof. If $x$ is irrational this follows from the definition and the fact that $r\ne x$ for rational $r$, so the set of rational $r\le x$ is really the set of rational $r<x$. If $x$ is rational, by Lemma 0 $b^x/b^r=b^{x-r}>1$ so $b^x$ is an upper bound of $\{b^r: r<x, r\in\mathbb{Q}\}$. Suppose there were a smaller upper bound $u$. Choose $n$ so that $b^{1/n}<b^x/u$, using Lemma 2. We then have $b^{x-1/n}>u$, with $x-1/n$ rational and $<x$, contradiction.

Lemma 4. If $X$ and $Y$ are sets of reals, $\sup X\cdot\sup Y = \sup \{xy:x\in X, y\in Y\}$. (Stated to be proven by the OP.)

Now we're ready for the main result, $b^{x+y}=b^x b^y$.

$$b^{x+y}=\sup\{b^r: r\in\mathbb{Q}, r<x+y\}=\sup\{b^{p+q}:p,q\in\mathbb{Q}, p<x, q<y\} =\\ \sup\{b^p:p\in\mathbb{Q}, p<x\}\cdot\sup\{b^q:q\in\mathbb{Q}, q<y\}=b^xb^y$$

The first and last equalities follow from Lemma 3. The second follows from Lemma 1. The third follows from Lemma 4. Fingers crossed I haven't made a mistake.

  • $\begingroup$ Lemma 4 is wrong. $X$ and $Y$ need to be sets of positive reals, with $\sup X \ge 1 $ and $sup Y \ge 1$ Luckily the sets we're dealing with satisfy those properties! $\endgroup$ – Quantaliinuxite Jul 4 '17 at 7:47
  • $\begingroup$ Oops. Very sorry about that. $\endgroup$ – ForgotALot Jul 4 '17 at 16:26

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