Defining the Real Exponential

What does $a^x$ mean? If $a = 0$ , then $0^x = 0$ no matter what $x$ we choose (except perhaps $x = 0$ , but that’s a debate for another day). Suppose $a > 0$ . We want the exponential function to satisfies these two properties.

Definition 1. Let $a > 0$ be a real number and $K$ be a set. Define the predicates $\phi,\psi$ on $K^2$ by

$\begin{aligned} \phi(m,n) &:= (a^{m+n} = a^ma^n), \\ \psi(m,n) &:= ((a^m)^n = a^{mn}).\end{aligned}$

With some effort, we can define $a^x$ reasonably for rational $x$ . We can even use these properties to prove most of the laws of exponents.

The real challenge begins now: what is $a^x$ for real $x$ ? For instance, what does ${10}^\pi$ mean? It is not at all obvious how we can formulate this notion rigorously.

We have defined $a^0 := 1$ , and for any $x > 0$ , the definition of $a^x$ yields the definition $a^{-x} := 1/{a^x} > 0$ .

Lemma 1. For rational numbers $r < s$ ,

$a^r < a^s$ if $a> 1$ and
$a^r > a^s$ if $0 < a < 1$ .

Proof. Suppose $a > 1$ . We first prove the special case $r = 0$ .

Thus, it suffices to prove the case $r = 0$ . Write $s = m/n$ for positive integers $m,n$ . Then $a^{m/n} = \sqrt[n]{a^m}$ , so that $a^m > 1^m = 1$ implies that

$a^s = a^{m/n} = \sqrt[n]{a^m} > \sqrt[n]{1} = 1$

by definition of the $n$ -th root. For the general case, if $r < s$ , then

$\displaystyle \frac{a^s}{a^r} = a^{s-r} > 1 \quad \Rightarrow \quad a^r < a^s.$

If $0 < a < 1$ , then $a^{-1} = 1/a > 1$ . By the first result,

$a^{-r} = (a^{-1})^r < (a^{-1})^s = a^{-s}.$

Taking reciprocals, $a^r = 1/{a^{-r}} > 1/{a^{-s}} = a^s$ .

Lemma 2. Fix $x > 0$ . Then there exists an increasing sequence $\{x_n\}$ of rational numbers such that $x_n \to x$ .

Proof. By the density theorem, find a rational number $x_1 \in (x/2,x)$ .

Inductively, find a rational number $x_{n+1} \in (\max\{x_n, x-1/n\}, x)$ . Then

$x_n \leq \max\{x_n , x-1/n\} < x_{n+1}< x$

and

$\displaystyle x-\frac 1n \leq \max\left\{x_n, x-\frac 1n\right\} < x_{n+1} < x \quad \Rightarrow \quad |x_{n+1} - x| < \frac 1n \to 0.$

By the squeeze theorem, $|x_{n+1} - x| \to 0 \iff x_{n+1} \to x$ .

Returning to a motivating question: what does ${10}^\pi$ mean? Intuitively, it is the limit of the sequence

$(10^3, 10^{3.1}, 10^{3.14},\dots) \approx (1000, 1259, 1380, \dots).$

Each term is well-defined, since it is of the form $a^x$ , where $x$ is rational. This example then suitably motivates our main discussion.

Theorem 1. Fix $x > 0$ . Then the limit

$\displaystyle a^x := \lim_{n \to \infty} a^{x_n}$

exists, where $\{x_n\}$ any increasing sequence of rational numbers such that $x_n \to x$ . In partiular, $10^{\pi} \approx 1385.456$ .

Proof. By Lemma 2, such a sequence $\{x_n\}$ exists.

Suppose $a > 1$ . By Lemma 1, $\{a^{x_n}\}$ is increasing and even bounded above, since we can find a rational number $x_0 \in (x,x+1)$ so that $a^{x_n} \leq a^{x_0}$ . By the monotone convergence theorem, $a^{x_n} \to L > 1$ exists.

Suppose $\{y_n\}$ is any increasing sequence with $y_n \to x$ , and $a^{y_n} \to M > 1$ . We need to prove that $L = M$ . Since

$y_n < x \quad \iff \quad x - y_n > 0,$

there exists $n_k \in \mathbb N^+$ such that

$0 < x-x_{n_k} < 1/n_k < x-y_n \quad \Rightarrow \quad y_n < x_{n_k}.$

Taking $n \to \infty$ yields $n_k \to \infty$ , so that $L \leq M$ . A symmetric argument yields $M \leq L$ , as required.

For $0 < a < 1$ , since $1/a > 0$ , the previous argument together with limit laws yields

$\displaystyle \frac 1{a^x} = \lim_{n \to \infty} \frac 1{a^{x_n}} \quad \Rightarrow \quad a^x = \lim_{n \to \infty} a^{x_n}.$

With any definition, we need to check that our basic rules apply.

Theorem 2. $\phi \wedge \psi$ on $\mathbb R^2$ .

Proof. It suffices to prove $\phi \wedge \psi$ on $\mathbb R_+^2$ , since the final result follows from defining $a^{-x} := 1/{a^x}$ .

For $\phi(x, y)$ , we consider increasing rational sequences $x_n \to x$ , $y_n \to y$ . Then $x_n + y_n \to x + y$ is increasing too, and

$\displaystyle a^{x+y} = \lim_{n \to \infty} a^{x_n+y_n} = \lim_{n \to \infty} (a^{x_n} a^{y_n}) = \lim_{n \to \infty} a^{x_n} \cdot \lim_{n \to \infty} a^{y_n} = a^x a^y.$

For $\psi(x, y)$ , we will first prove the result for real $x$ and rational $y$ . Consider any increasing rational sequence $x_n \to x$ . Then $x_n y \to xy$ is an increasing rational sequence, and by the (sequential) continuity of taking rational exponents (i.e. the map $x \mapsto x^y$ ),

$\displaystyle \begin{aligned}\displaystyle (a^x)^y &= \left(\lim_{n \to \infty} a^{x_n}\right)^y =\lim_{n \to \infty} (a^{x_n})^y = \lim_{n \to \infty} a^{x_n y} = a^{xy}. \end{aligned}$

For the general result, consider any increasing rational sequence $y_n \to y$ . Then

$\displaystyle \begin{aligned}\displaystyle (a^x)^y &= \lim_{n \to \infty} (a^x)^{y_n} = \lim_{n \to \infty} a^{xy_n}. \end{aligned}$

To complete the proof, find an increasing rational sequence $x_m \to x$ . In particular,

$a^{x_n y_n} \leq a^{xy_n} \leq a^{xy}.$

Since $x_ny_n \to xy$ , the left-hand side simplifies to $a^{x_ny_n} \to a^{xy}$ . By the squeeze theorem, $a^{xy_n} \to a^{xy}$ so that $(a^x)^y = a^{xy}$ , as required.

Once we have $\phi \wedge \psi$ on $\mathbb R^2$ , we obtain all the laws of exponents, and consequently, the laws of logarithms. However, the latter assumes that $a^x$ is bijective as a function of $x$ . We prove this final result here, with a technical lemma en route whose proof is isomorphic to that of Theorem 1.

Lemma 3. Fix $a > 0$ and $x > 0$ . Then the limit

$\displaystyle a^x := \lim_{n \to \infty} a^{x_n}$

exists, where $\{x_n\}$ any decreasing sequence of rational numbers such that $x_n \to x$ .

Proof. By hypothesis, $\{-x_n\}$ is an increasing sequence of rational numbers such that $-x_n \to -x$ . Using the definition $a^{-x} = 1/a^x$ , we have

$\displaystyle \frac 1{a^x} = a^{-x} = \lim_{n \to \infty} a^{-x_n} = \lim_{n \to \infty} \frac 1{a^{x_n}}.$

Using limit laws, $a^{x_n} \to a^x$ , as required.

Theorem 3. The function $f : \mathbb R \to \mathbb R^+$ defined by $f(x) = a^x$ is bijective.

Proof. We can show that $f$ is injective by showing that $r<s$ implies $f(r)<f(s)$ :

$\displaystyle \frac{ f(s) }{ f(r) } = \frac{ a^s }{ a^r } = a^{s-r}.$

Since $s > r$ implies $a^{s-r} > 1$ by Lemma 1, we have $f(r) < f(s)$ , as required.

To show that $f$ is surjective, we will first need the continuity of $f$ . Let $x_n \to x$ be any real-valued sequence. Find a rational sequence $r_n$ such that $|x_n - r_n| < 1/n$ . This will imply that $r_n \to x$ as well.

Define $r_n^{\pm}$ such that $r_n^- \leq x_n \leq r_n^+$ , where $\{r_n^-\}$ is increasing and $\{r_n^+\}$ is decreasing. By definition, $a^{r_n^-} \to a^x$ . By Lemma 3, $a^{r_n^+} \to a^x$ . The squeeze theorem yields

$f(x_n) = a^{x_n} \to a^x = f(x),$

establishing continuity.

Now we aim to show that $f$ is surjective, that is: for any $y > 0$ , there exists $x \in \mathbb R$ such that $f(x) = y$ .

For $y = 1$ , choose $x = 0$ to obtain $f(0) = a^0 = 1$ , as required. Suppose $y \neq 1$ .

Fix $y > 1$ . Find $n \in \mathbb N$ such that $a^n > y$ . Then

$a^{-n} = 1/{a^n} < 1/y < y.$

Since $f$ can be shown to be continuous on $[-n,n]$ , by the intermediate value theorem, there exists $x \in (-n, n)$ such that $a^x = y$ . For $y < 1$ , apply the previous result to $1/y > 1$ , then use the laws of exponents.