how to define $a^x$ for $a>0$ for all $x \in \mathbb{R}$ and how is the continuity of $a^x$ is proved without that definition

Question

I am trying to understand how to define the exponential function $a^x$ for any $a>0$ and any $x\in\mathbb{R}$ I know the basic definition of the exponential function for natural numbers, which is $a^n=a⋅a⋅…⋅a $(n times). I also know how to extend this definition to rational numbers by using the fact that $a^{\frac{n}{m}}= \sqrt[m]{a^n }$​, where $m$⋅​ denotes the $m$-th root also I know the definition of the negative power of rational numbers . But I am curious how to define the function for all real numbers, not just rational ones. obviously we will define $f(x+y) =f(x) f(y)$ for all real numbers $x,y$ but here I have no Idea how to extended to all Real numbers. I have heard that there are some ways to do this using limits, continuity, but I needed this so I can prove that $a^x$ is continuous function so I am confused of how to use calculus to make the definition of $a^x$ by using $e^x$ as continuous function and that would make me ask why is $e^x$ continuous in the first place one can show that $e^x$ is continuous at $ \mathbb{R}$ if it is continuous at zero but how we will establish that if we didn't define it at irrationals?

. Could someone please explain to me the general definition of an exponential function and why it works?

Answer 1

As has been pointed out in the comments, one definition of exponentiation comes through defining $a^r:=\exp(r\ln(a))$, where the exponential and natural log are already defined and shown to be continuous through other means.

However, perhaps a more intuitive approach comes from showing the usual definition for rational exponents has a unique continuous extension to the reals.

To see this, let's suppose $a>1$, as we can deal with $a<1$ in a similar manner (or we can define $a^r=(\frac{1}{a})^{-r}$ when $a\in (0,1)$.)

Note that $f\colon \mathbb Q \to \mathbb R$ given by $f(q)=a^q$ is certainly increasing in $q$, since if $q_1<q_2$, then for some $n\in\mathbb N$, $nq_1<nq_2$ are integers, and then we have $a^{nq_1}< a^{nq_2}$, and taking $n$-th roots gives us $a^{q_1}<a^{q_2}$.

Moreover, we can see $f$ is continuous at $0$. Since it is increasing, it is enough to show $\sup_{q<0} a^q=\inf_{q>0}a^q = 1$. This holds, since if the second equation fails, then we have some $\eta>1$ with $a^{\frac{1}{n}}\geq\eta$ for each $n\in\mathbb N$, whereby $a\geq\eta^n$ for all $n$, contradicting $\lim \eta^n=\infty$ for $\eta>1$, and you get a similar contradiction if the supremum fails to equal $1$.

Next, we argue that $f$ is uniformly continuous on bounded sets. To see this, note that by continuity at $0$, for each $\epsilon>0$, we have a $\delta>0$ for which $$|q|<\delta\implies |a^q-1|<\epsilon\text{.}$$

Then if $q_1<q_2$ are in $[-M,M]$, and $q_2-q_1<\delta$, we have $$|a^{q_2}-a^{q_1}|= a^{q_1}|a^{q_2-q_i}-1|\leq a^{q_1}\epsilon\leq a^M\epsilon.$$ Thus given $\epsilon_0>0$, we may let $\epsilon=\frac{\epsilon_0}{a^M}$, and choose the $\delta$ corresponding to $\epsilon$.

Since $f$ is uniformly continuous on bounded sets, and $\mathbb Q$ is dense, there is a unique continuous extension $\bar{f}$ to $\mathbb R$, and so we may define $a^r=\bar{f}(r)$.

Remark

From monotonicity, continuity and density of $\mathbb Q$ we have $a^r=\sup_{q<r} a^q =\inf_{q>r} a^q$, where the supremum and infimum are taken over rationals. We could also have started from one of these as a definition, and then established continuity through monotonicity, and showing the supremum and infimum coincide.

Answer 2

If we begin by constructing the real numbers using [Dedekind cuts][1] then each real number is a set of rational numbers and we're interested in the supremum of that set. So for example $\pi = \{q \in \mathbb{Q} | q < \pi \}$ is the cut defining $\pi$ and $\sqrt{2} = \{q \in \mathbb{Q} | q < \sqrt{2}\}$ is the cut defining $\sqrt{2}$. Note that every set is just rational numbers. Now we define $x^y$ by $\{q^r | q\in x, r \in y\}$. Note that this is just rational powers of rational numbers and we now need to verify that $x^y$ is a cut. It's not hard to see that $x^y$ is non-empty, downward closed, not all of $\mathbb{Q}$, and does not have a greatest element using the fact that $x$ and $y$ are cuts. Basically all the complications disappear since you're always considering sets of rational numbers. I'll leave it to you to fill in the details.