I simply want a proof that $e^x$ is continuous.

I have never really been able to find something satisfying these points:

  1. $e$ is defined to be the limit $\lim_{n\to\infty}\left(1+{1\over n}\right)^n$. This is not essential, and another reasonable definition may be used if that significantly simplifies things. EDIT: In fact, it does not really matter what the base is taken to be. Any number, or a general $a^x$, would be fine.

  2. For positive integers, we define $a^n:=a\times a\times\cdots\times a$ where there are $n$ multiplications. We then let $a^0=1$ and $a^{-n}=1/a^n$. For roots, $a^{1/n}$ is the (principal) number $x$ such that $x^n=a$. For general rationals, $a^{b/c}=(a^{1/c})^b$ is the $c$'th root brought to the $b$'th power. And for irrational numbers $x$, $a^x=\lim{a^n}$ where $n$ is a sequence of rational numbers which converge to $x$.

In other words, I want a proof of the exponential's continuity derived from the "arithmetic" definition itself. The reason I ask is because as a student I never had an intuition for why a number brought to two close exponents would have to yield similar answers. For example, why would $$2^{1/2}\approx 2^{47/99}$$ I thought? After all, one has a $99$'th root to a $47$'th power, and the other is just a square root.

I now know that I was looking for a proof of continuity... and I'm still looking. Please do not use anything advanced - the kid in me still groans when I see people 'define' $e^x$ as a power series, because I know students everywhere (like I was) are getting confused.

Also, I want the proof to be convincing, as in I can sleep well at night, but it need not be headache-inducingly thorough. For example, I didn't derive some of the laws of exponents. I didn't explain what I mean by an irrational number, or prove that the limit must exist and be unique, etc. These you should take as granted - what I want is a straightforward proof of continuity based on the "actual" (!) definition that could be explained to an advanced calculus student frustrated about this question and ready to push themselves to learn - in other words, to my past self.

  • $\begingroup$ It looks to me like $e$ isn't important here; a proof that didn't mention $e$ at all might actually be better. $\endgroup$ Apr 21, 2014 at 5:34
  • $\begingroup$ @user2357112 Correct. I considered doing $2^x$ instead, but I guess I assumed $e^x$ would be simplest because it is most easily inverted by $\ln x$ and I didn't think it would make a difference what the base was otherwise. I will add a note though. $\endgroup$
    – user142299
    Apr 21, 2014 at 5:40
  • $\begingroup$ I suggest you change the title of the question to reflect what you are really asking about: why is for fixed base $a>0$ the function $\frac bc\mapsto a^{b/c}=(\sqrt[c]a)^b$ a continuous function of the rational number $\frac bc$. (If it is, then by completeness of the real it can be interpolated to a unique continuous function of a real argument.) This will avoid getting answers along completely different lines (people often don't take care to read the details of a long question; I've been guilty of this often too). $\endgroup$ Apr 21, 2014 at 6:01
  • $\begingroup$ @MarcvanLeeuwen But I thought nobody can resist reading my blissfully floral English! :) I really want a proof of continuity for the whole function (irrationals too) though, and I'm getting lots of interesting answers so I think I'm going to leave it as is. Thanks for suggestion and your post, cheers $\endgroup$
    – user142299
    Apr 21, 2014 at 16:46

8 Answers 8


Fix $a>1$ (the case $a=1$ is trivial, and $0<a<1$ can be handled by switching to $a^{-1}>1$). I'll assume standard facts about $n\mapsto a^n$ for $n\in\Bbb Z$ are known (it is increasing and satisfies the law of exponents). Now to argue the continuity of $\frac bc\mapsto a^{b/c}=(\sqrt[c]a)^b$ on the set of rational numbers $\frac bc$, which is dense in$~\Bbb R$, one can proceed by the following steps.

  1. $a^{b/c}$ is well defined, namely independent of the representation of the rational number $\frac bc$. This means that whenever $\frac bc=\frac pq$, which by definition means $bq=cp\neq0$, one should have $(\sqrt[c]a)^b=(\sqrt[q]a)^p$. Taking positive integer powers is an injective operation ($x^n=y^n$ implies $x=y$), so we raise both sides to the power $cq$ (assumed positive), so that we must prove $(\sqrt[c]a)^{bcq}=(\sqrt[q]a)^{pcq}$; now we can cancel each root against part of the exponent, and there remains to prove $a^{bq}=a^{cp}$, which is clear.

  2. The function $\frac bc\mapsto a^{b/c}$ is strictly increasing: if $\frac bc<\frac pq$ then $a^{b/c}<a^{p/q}$. Given the previous point, one may assume the fractions have a common denominator $c=q$. Raising both sides to this power (which is also a strictly increasing operation) means it suffices to show $a^b<a^p$; again this is clear.

  3. This increasing function has no jump discontinuities*: for any nonempty open interval$~I$ of positive real numbers, there is some $\frac bc$ with $a^{b/c}\in I$ (a jump discontinuity would give such interval without images). If $I=(x,y)$ put $\gamma=y/x\in\Bbb R_{>1}$, and let $n\in\Bbb N$ be sufficiently large that $\gamma^n>a$. Then $y^n>ax^n$ so there is some integer power $a^m$ of $a$ with $x^n<a^m<y^n$. Taking $n$-th roots (again a strictly increasing operation) it follows that one has $x<a^{m/n}<y$, in other words $a^{m/n}\in I$.

  4. Any increasing function$~f$ defined on a dense set and without jump discontinuities is continuous. Given $x$ in the domain of $f$, and $\epsilon>0$, find values $x_<,x_>$ with $f(x_<)\in(f(x)-\epsilon,f(x))$ and $f(x_>)\in(f(x),f(x)+\epsilon)$ by point$~$3, and take $\delta=\min(x-x_<,x_>-x)$. Then for all $x'$ with $|x-x'|<\delta$ one has $x'\in(x_<,x_>)$ so $f(x')\in(f(x_<),f(x_>))$ by the increasing property of$~f$, and $(f(x_<),f(x_>))\subset(f(x)-\epsilon,f(x)+\epsilon)$, so that $|f(x)-f(x')|<\epsilon$.

This does not yet entirely answer your question, for which it is necessary to show that any function like in 4. can be uniquely interpolated to a continuous increasing function defined on $\Bbb R$. For $x_0$ not in the domain $D$ of $f$ (here an irrational number) one wants to have $f(x_0)=\lim_{x\to x_0, x\in D}f(x)$. To show that the limit exists, one uses that $\sup\{\,f(x)\mid x\in D_{<x_0}\,\}=\inf\{\,f(x)\mid x\in D_{>x_0}\,\}$ (by the absence of jump discontinuities) which is going to be the value of the limit; now the proof that the limit converges can be done by finding values in $D$ sufficiently close to the limit value and some easy but boring estimations, which I will not do here (as per the headache-avoiding request). Also the proof that the extended function is still is continuous and increasing is straightforward.

*This terminology is is not quite proper, as the function on $\Bbb Q \to\Bbb R$ sending $x\mapsto x+[x^3>2]$ (the brackets meaning $1$ if the condition holds, $0$ otherwise) is continuous on its domain, so it is somewhat unfair to accuse it of having a jump discontinuity, as the above implies. The proper term for what I call "function without jump discontinuities" would be something like "function the closure of whose image is connected" (assuming an increasing function defined on a dense set in both cases), not very attractive.

  • $\begingroup$ For (2), I don't like the assumption that both have the same denominator. You should just write $$a^{b/c}<a^{p/q}\iff a^{bq}<a^{pc}$$ and $$\frac{b}{c}<\frac{p}{q}\implies bq<pc$$ As I wrote in the question, I was specifically tripped up by very 'different' fractions that were close leading to very close exponentials. Other than that, excellent answer. +1 upvote cheers! $\endgroup$
    – user142299
    Apr 21, 2014 at 16:43
  • $\begingroup$ @NotNotLogical: Yes, one can do that. But since I just proved that one can change the representation of a rational exponent without changing the power, one might as well use that fact to make the denominators equal (to the least common multiple of the original ones); this is what I meant by one may assume. But again, one does not have to. $\endgroup$ Apr 21, 2014 at 16:58
  • $\begingroup$ Right, but then the relevant fact is that $bq<pc$ (or whatever the new numerators are), not that $b<p$ which is not necessarily true. No big deal though. And I remember that I had another question: how did you get from $y^n>ax^n$ to $\exists m\in\mathbb{Z}: x^n<a^m<y^n$? I didn't understand that. $\endgroup$
    – user142299
    Apr 21, 2014 at 17:06
  • $\begingroup$ @NotNotLogical: Well the point of saying "we can assume $c=q$" rather than "we can make $c=q$", is that the assumption then implies that $b,p$ also have been modified, so $b<p$ refers to the new numerators, not the old ones. This is a rather standard (and quite practical) way of setting up an argument, but it takes some getting used to. As for the added question, since $a^k$ grows without bound, there is some $k\in\Bbb Z$ with $a^k>x^n$; there is also $k\in\Bbb Z$ with $a^k<x^n$ (for similar reasons). Then $m=\min\{\,k\in\Bbb Z\mid a^k>x^n\,\}$ is well defined, and satisfies the condition. $\endgroup$ Apr 21, 2014 at 19:18

$\mathbf{Result \; 1}$: If $a > 0 $, then $\lim_{n \to \infty } a^{\frac{1}{n}} = 1 $.

If you want a solution to this problem, let me know.

$\mathbf{Result \; 2} $: If $a> 0$, then $f(x) = a^x$ is continuous in $\mathbb{R}$

$\mathbf{Solution}:$ Let $a>0$, and let $\epsilon >0$ .We will employ the sequential characterization of continuity. This means that for any given sequence $(x_n) \subseteq \mathbb{R}$ with $x_n \to x$, then $f(x_n) \to f(x)$. To show this, let $(x_n) \subseteq \mathbb{R}$ be an arbitrary sequence such that $x_n \to x$. Notice by arithmetic,

$$ a^{x_n} - a^x = a^x(a^{x_n-x} - 1) $$

By previous result, we can obtain an $N \in \mathbb{N}$ such that for all $n > N$,

$$ |a^{\frac{1}{n}} - 1 | < \frac{\epsilon}{a^x} $$

Similarly, since $x_n \to x$, we can select $K \in \mathbb{N}$ such that for all $n > K$,

$$ |x_n - x| < \frac{1}{N} $$

Therefore, if we put $M = \max\{N,K\} $, we have

$$ |f(x_n) - f(x) | = |a^{x_n} - a^x| = a^x|a^{x_n-x} - 1| < a^x \frac{ \epsilon}{a^x} = \epsilon$$

Hence, by definition, $f(x_n) \to f(x) $. In particular, $f(x) = a^x$ is continuous as desired.

$\mathbf{Proposition}:$ $\exp(x) $ is continuous on $\mathbb{R}$.

$\mathbf{Solution}:$ Apply $\mathbf{Result \; 2}$ to $a = e$, and we are done.

  • $\begingroup$ Result 1 only shows that $$\lim_{n\to 0^+}a^n=1$$ correct? $\endgroup$
    – user142299
    Apr 21, 2014 at 5:51
  • 3
    $\begingroup$ I don't see how you show that $|x_n-x|<\frac{1}{N}$ implies anything about $a^{x_n-x}$. $\endgroup$ Apr 21, 2014 at 5:53

Consider the function $f:(0,\infty)\to\mathbb R$ such that $f(x)=\int_1^x\frac{\mathrm dt}{t}$. This is continuous, differentiable and, in fact $f'(t)=1/t$ for all $t\in(0,\infty)$. In particular, it is clear that $f'(t)>0$ for all such $t$ so that $f$ is strictly increasing in its domain. Some work will show that $f$ is not bounded neither above not below —indeed, and for example, the first is equivalent to the statement that the integral $\int_1^\infty\frac{\mathrm dt}{t}$ diverges.

Now we have

If $h:(a,b)\to(c,d)$ is a continuous, strictly increasing function which is surjective, then there is an inverse function $h^{-1}:(c,d)\to(a,b)$ and it is itself also continuous and strictly increasing.

This applies to our $f$, and shows that it has an inverse function $f^{-1}:\mathbb R\to(0,\infty)$ which is continuous. Since this is of course the exponential function, we have shown what we wanted.

Later. Notice that one can define the exponential functio in this way —and it is done so classically, too. In any case, It is very easy to check that it is the exponential. Let $h=f^{-1}$. A change of variables shows that $f(xy)=f(x)+f(y)$, so applying inverse functions we see that $h(x+y)=h(x)h(y)$. It follows that $(h(x+y)-h(x))/y=h(x)(h(y)-1)/y$ and, since we know that $h$ is differentiable, it follows that $h'(x)=h'(0)h(x)$ for all $x$. The inverse function theorem allows us to compute $h'(0)=1$, so $h'(x)=h(x)$ and $h(0)=1$. One can now compute the derivative of $h(x)e^{-x}$, see that it is identically zero, adn evaluate his function at $0$ to conclude that $h(x)=e^{-x}$.

  • 2
    $\begingroup$ "Since this is of course the exponential function" - this part needs expansion. It's not clear that the inverse of this integral is in fact the same function as the one defined in terms of powers and roots. $\endgroup$ Apr 21, 2014 at 6:23
  • $\begingroup$ @user2357112, the OP observed that any reasonable definition would please him, and defining the exponential as the inverse of the primitive of $1/x$ is essentally the best way to do it. $\endgroup$ Apr 21, 2014 at 16:29
  • $\begingroup$ @MarianoSuárez-Alvarez I meant any definition of the number $e$. I said that I wanted the function itself to be defined in terms of the "arithmetic" definition. Read the point (2) in the post. $\endgroup$
    – user142299
    Apr 21, 2014 at 16:48
  • $\begingroup$ I had misread that, in fact. I posit that that arithmetical definition is the worst possible alternative. $\endgroup$ Apr 21, 2014 at 16:51
  • $\begingroup$ @MarianoSuárez-Alvarez Haha, well I guess we'll just have to agree to disagree about that :) $\endgroup$
    – user142299
    Apr 21, 2014 at 17:10

Okay, I'll be a little stubborn here and I'll try to convince you that defining $e^x$ as a power series is actually better. (Respecting your second wish, I'll not enter in a lot of technicalities)

First, let me say something: People usually think that mathematicians do things in order to complicate simple things. The fact is that most of times, the mathematician is actually simplifying things. As a rule of thumb, in order to simplify something, you need to make a theory, and the more you want to simplify, the more abstract or hard to grasp the theory will be.

Now, okay, let's make a function called $\xi (x)$, defined by:

$\displaystyle \xi (x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ (Here, we adopt $0^0=1$ to avoid notational complications)

Now, don't be afraid of this definition (I didn't even say that this is the exponential! This is a auxiliary function, for now)

The important things are:

  1. $\xi$ is continuous. (This follows from it being a power series)
  2. $\xi (x+y) = \xi(x)\xi(y)$. (This follows from the Cauchy Product Formula)
  3. $\xi$ is defined for ALL reals. (This is a simple observation, but will be extremely useful)
  4. $\xi(0)=1$, $\xi(1)=e$

And now, let's take a look at the function you call "exponential". Let's call it $\exp$. We have:

  1. $\exp(x+y)=\exp(x)\exp(y)$ for all $n \in \mathbb{N}$ (I hope this is clear)

  2. $\exp(0)=1$, $\exp(1)=e$

Now, for $n \in \mathbb{N}$, we have that $\exp(n)=\exp(1)...\exp(1)$ $n$ times=$\exp(1)^n$

For $-n$ negative integer, we have that $1=\exp(0)=\exp(n-n)=\exp(n)\exp(-n)$, hence $\displaystyle \exp(-n)=\frac{1}{\exp(n)}=\frac{1}{\exp(1)^n}=\exp(1)^{-n}$

For $a=\frac{p}{q} \in \mathbb{Q}$, $\exp(p)=\exp(q.\frac{p}{q})=\exp(\frac{p}{q}+\frac{p}{q}+...+\frac{p}{q})$ $q$ times $=\exp(p/q)^q$. Hence, $\exp(\frac{p}{q})=\exp(p)^{1/q}=(\exp(1)^p)^{1/q}$

Now, note the following:

  • I computed the values of $\exp(x)$ for all $x$ rational using only properties that $\xi$ AND $\exp$ have in common. Hence, $\xi$ coincides with $\exp$ in the rationals.
  • By the previous observation, we have immediately that $\exp$ is continuous in the rationals, as a restriction of a continuous function is continuous.

Okay, now, since we have that $\xi$ is defined for ALL reals, we could very well say:

Well, make $\xi$ be what we will call $\exp$ for the reals, then.

But we can do more. Since $\exp$ coincides with $\xi$ in a dense set ($\mathbb{Q}$), if we want it to extend continuously, the ONLY way is for the extension to be $\xi$ (maybe this is not clear for you, but it is a theorem, that is stated as follows, and holds in a way more general situation:)

Two continuous functions that are equal in a dense subset of $\mathbb{R}$ are equal at the whole domain.

Now, notice the irony. You began asking why $\exp$ is continuous, and we derived quite "easily" that $\exp$ is continuous in the rationals, and the motivation came as to whether we could have a continuous "$\exp$" in the reals.

Just to close up, you may feel like I'm cheating in requiring a continuous extension. Well, I'm not. Following your intuition, I'd say that the only things that you consider the exponential surely satisfies is $\exp(x+y)=\exp(x).\exp(y)$, $\exp(0)=1$ and $\exp(1)=e$. Well, there are a LOT of non-continuous functions satisfying this in $\mathbb{R}$, coinciding with the values of $\exp$ in the rationals. And, by the way, your way of defining $\exp$ for reals as limits of sequences is blalantly, explictly requiring it to be continuous in $\mathbb{R}$.

EDIT: Appendix: Why considering THIS series:

Whatever $\exp$ will be, we want that $\exp(x+y)=\exp(x).\exp(y)$

Now, let's say we should guess a power series for $\exp(x)$


Well, $\displaystyle \exp(x+y)=\sum_{n=0}^{\infty}a_n(x+y)^n=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_n\frac{n! x^ky^{n-k}}{k!(n-k)!}=\sum_{n=0}^{\infty}a_nn!\sum_{k=0}^{n}\frac{ x^ky^{n-k}}{k!(n-k)!}$ , by the binomial theorem.

But, hey, look! The right hand side is begging for the cauchy product formula. If that $a_n.n!$ were $1$... then, the right hand side would be $\displaystyle \left(\sum_{k=0}^{\infty}\frac{x^k}{k!}\right)\left(\sum_{m=0}^{\infty}\frac{y^m}{m!}\right)$. So, why not? Let $a_n.n!$ be $1$, then! So, we have $\displaystyle a_n=\frac{1}{n!}$. Now, look back at the series for $\exp$, and perceive our luck.

  • $\begingroup$ I posted this based on OP's response "Now you have the not-so-easy task of explaining to your frustrated high schooler why that definition is the same as what they thought was the definition of exponentiation... :) and yes, that's exactly what I didn't want, but no worries." $\endgroup$
    – Aloizio Macedo
    Apr 21, 2014 at 15:28
  • $\begingroup$ This is a fairly good response. My two difficulties are: 1) I really wanted a "pre-analysis" proof that a strong calculus student could understand. Proving 1,2,3 of your "important things" requires building up a substantial amount of theory about infinite series. I wouldn't have known what you were talking about when I was at that stage in my education. 2) You have nicely proved that the series is equivalent to $\exp x$. However, where did you get it? It looks like it got pulled out of a hat. I think the best way is: prove continuity, prove derivative, and derive Taylor series from that. $\endgroup$
    – user142299
    Apr 21, 2014 at 17:02
  • $\begingroup$ Answering to 1): I agree with you, these claims require a substantial amount of theory, but to be formally and correctly understood. A "calculus" way of proving things and being comfortable with them often rely on intuitive or heuristic arguments (at least in the educational system of my country). That said, a power series intuitively continuous, and the Cauchy Product Formula is also intuitive (just look at multiplication of polynomials). That it is defined for all reals is just a problem of convergence, which can be "justified" by the factorial growing hugely. $\endgroup$
    – Aloizio Macedo
    Apr 21, 2014 at 17:28
  • $\begingroup$ Answering to 2): From my perspective, I don't need to know where something came from, just why it works if it is exposed (and in a lot of times, where things come from are actually miraculous and a surge of "inspiration"). This is HUGELY personal, so I'll depict some reasons why to consider that series: 1 - As you said, if you naively start a theory of exponential function, you can arrive at its derivative and compute the taylor series. I don't like this explanation, though. 2 - $e=\sum_{n=1}^{\infty}\frac{1}{n!}$ This is, for me, a good enough reason to consider the series. ... $\endgroup$
    – Aloizio Macedo
    Apr 21, 2014 at 17:32
  • $\begingroup$ (continuing...) 3 - You want the $additive-to-multiplication$ property to hold. Looking at the Cauchy product formula, it seems quite reasonable that something with factorials should appear, because we will want to use the Binomial Theorem to expand things. $\endgroup$
    – Aloizio Macedo
    Apr 21, 2014 at 17:35

Your idea of defining exponential function $a^{x}$ is intuitive but has many complications. The definition assumes that $a^{p}$ is defined for all $a > 0$ and $p \in \mathbb{Q}$. If $x \in \mathbb{R} - \mathbb{Q}$ and if $x_{n}$ is a sequence of rationals with $\lim_{n \to \infty}x_{n} = x$ then you define $a^{x} = \lim_{n \to \infty}a^{x_{n}}$. This requires us to first justify that the definition is unambiguous. This further implies that we must establish

1) if $x_{n}$ is a sequence of rationals with $\lim_{n \to \infty}x_{n} = x$ then the limit $\lim_{n\to\infty}a^{x_{n}}$ exists.

2) if $x_{n}, y_{n}$ are sequences of rationals with $\lim_{n \to \infty}x_{n} = \lim_{n \to \infty}y_{n} = x$ then $\lim_{n \to \infty}a^{x_{n}} = \lim_{n \to \infty}a^{y_{n}}$

Both the above results although seem trivial but are not so easy to prove. Assuming that you have overcome the above stumbling blocks in your definition it is easy to show that $a^{x}$ is continuous for all $x$. This requires us to prove the exponential rule $a^{x + y} = a^{x}a^{y}$ which is easy according to the definition adopted here. And then it needs to be established that $\lim_{x \to 0}a^{x} = 1$.

The limit can be established provided we establish the weaker inequality $a^{x} \leq a^{y}$ for $x < y, a > 1$. For $a < 1$ the inequality sign is reversed. This is also easily done by the definition used here. (Hint: If $a > 1$ and $x < y$ then put $h = y - x > 0$ and show that $a^{h} \geq 1$ and then multiply the inequality by $a^{x}$)

Next we need to first show that $a^{1/n} \to 1$ as $n \to \infty$. Let $a > 1$ and we put $a^{1/n} = 1 + b$ so that $a =(1 + b)^{n} > 1 + nb$ so that $$0 < b < \frac{a - 1}{n}$$ Letting $n \to \infty$ we see that $b \to 0$ and hence $a^{1/n} = 1 + b \to 1$.

We can now show that $a^{x} \to 1$ as $x \to 0$. We will consider only the case for $x \to 0^{+}$. Let $\epsilon > 0$ be given. Then there is a positive integer $m $ such that $1 - \epsilon < a^{1/n} < 1 + \epsilon$ for $n \geq m$. Let $\delta = 1/m > 0$. If $0 < x < \delta$ then $0 < x < 1/m $ and then $1 \leq a^{x} \leq a^{1/m}$ so that $1 - \epsilon < a^{x} < 1 + \epsilon$ for $0 < x < \delta$. It follows that $a^{x} \to 1$ as $x \to 0^{+}$.

Now we can see that $$\begin{aligned}\lim_{x \to b}a^{x} &= \lim_{h \to 0}a^{b + h}\\ &= \lim_{h \to 0}a^{b}a^{h} = a^{b}\cdot 1 = a^{b}\end{aligned}$$ so that $a^{x}$ is continuous for all $x$. The proofs above assume that $a > 1$. The case $a < 1$ can be handled by noting that $1/a > 1$ and applying the results on $1/a$.

Update: If you are only concerned why $2^{1/2}\approx 2^{47/99}$ it is easy to provide a simple proof and avoid all limit machinery mentioned above. You ideally want $a^{x}$and $a^{y}$ to be nearly equal if $x, y$ are nearly equal and let's keep $x, y$ rational to avoid all limit arguments. By the way this is in reality the proof for stumbling block 2) which I mentioned in start of my answer.

We have $a^{x} - a^{y} = a^{y}(a^{x - y} - 1)$ so the difference $a^{x} - a^{y}$ will be small if $a^{x - y} - 1$ is small. Let $x - y = h$ and keep $x > y$ otherwise we interchange roles of $x, y$. Then we see that $h$ is small (as $x$ is near to $y$) and we want to make $a^{h} - 1$ small. We have the inequality $$\frac{a^{h} - 1}{h} < a - 1$$ for rational $a, h$ with $a > 1, 0 < h < 1$. See its proof here. Hence we have $0 < a^{h} - 1 < h(a - 1)$. Now we can see that $h$ is small so $h(a - 1)$ is small and therefore $a^{h} - 1$ is small and therefore $a^{x} - a^{y}$ is small.

  • $\begingroup$ When you say that $0<x<1/m$ implies $1\le a^x\le a^{1/m}$ 1) aren't you assuming that the function is strictly increasing? and 2) why the loose inequality instead of strict, since the first inequality is strict? I feel like assuming monotone increasing is a big assumption to make - it feels almost tantamount to assuming continuity. $\endgroup$
    – user142299
    Apr 21, 2014 at 16:55
  • $\begingroup$ Also, when you say $(1+b)^n>1+nb$ you appear to be using the generalized binomial theorem. This relies on the proof of the derivative of $x^n$ for general $n\in \mathbb{R}$, and this in turn relies on the derivative of $e^x$ which in turn relies on the continuity of $e^x$ being known. Is there a non-circular way to prove that part? $\endgroup$
    – user142299
    Apr 21, 2014 at 16:56
  • $\begingroup$ @NotNotLogical: monotone nature of $a^{x}$ is easy to prove, especially the non-strict version. There is strict inequality if exponents are rational and that can be easily proved. Taking limits we get weak inequality for irrational exponents. $\endgroup$
    – Paramanand Singh
    Apr 22, 2014 at 3:06
  • $\begingroup$ @NotNotLogical: the $n$ represents positive integer. $\endgroup$
    – Paramanand Singh
    Apr 22, 2014 at 3:08

(Imho) the easiest (but maybe to theory burdened) way, I guess, is to define $e^x =\lim_{n\rightarrow\infty}\sum^n_{k=0}\frac{x^k}{k!}$, to show that the series converges locally uniformly and to use the fact that (locally) uniform convergence of continuous functions produces a continuous function.

  • 2
    $\begingroup$ A power series definition of $e^x$ is the exact kind of definition the OP wants to avoid. $\endgroup$ Apr 21, 2014 at 5:48
  • $\begingroup$ Now you have the not-so-easy task of explaining to your frustrated high schooler why that definition is the same as what they thought was the definition of exponentiation... :) and yes, that's exactly what I didn't want, but no worries. $\endgroup$
    – user142299
    Apr 21, 2014 at 5:50
  • $\begingroup$ @NotNotLogical You see, while it might look daunting in the beginning, if you choose a different approach defining $e^x$ you will face several other questions which are not so easy to answer. To begin with, what exactly is $e^x$ for nonrational $x$? Then for complex $x$? Why should $e^{x+y}=e^x e^y$ hold in general? You get all this rather easily with the power series definition. The task of explaining why this is the same as what you thought is rather easy, too -- it is not too difficult to see, that the functional equation together with the value for $x=1$, say, has only one solution. $\endgroup$
    – Thomas
    Apr 21, 2014 at 8:03
  • $\begingroup$ @Thomas That's not originally a definition, it was a theorem. It was proved from the actual definition of what exponentiation means. I agree that the series 'definition' makes proofs easy, but it also leaves the task of explaining why this is equivalent to using the original definition from which the series was derived. $\endgroup$
    – user142299
    Apr 21, 2014 at 16:07

Let $a \gt 1$.

Aspiring math students will be able to accept the following assertions.

(1) $f: \Bbb Q \to \Bbb R$ defined by $f(x) = a^x$ is a strictly increasing function
(2) $f(u+v) = f(u)f(v)$

These two facts are used to develop the following:

Proposition 1: The decreasing sequence $a^{+{\frac{1}{2^n}}}$ converges to $1$ as $n\to\infty$.
Proof: Exercise, but see just keep hitting that square root key.

Proposition 2: The increasing sequence $a^{-{\frac{1}{2^m}}}$ converges to $1$ as $m\to\infty$.
Proof: Exercise.

Proposition 3: The function $f$ is continuous at 0.
Proof: Exercise.

Proposition 4: The function $f$ is continuous at all points in $\Bbb Q$.
Proof: Exercise.

Proposition 5: For any $q \in \Bbb Q$, $f(q) = sup(f(p))$ with all $p \lt q$.
Proof: Exercise.

Now extend the function $f$ to any real number $x$ by taking the supremum $sup(f(p))$ with $p \in \Bbb Q$ and $p \lt x$. We use the same notation, $f$, for the new function.

Proposition 6: $f: \Bbb R \to (0, +\infty)$ is strictly increasing and surjective.
Proof: Exercise.

Proposition 7: $f(x) = a^x$ is continuous.
Proof: See Prop 6 implies Prop 7.

Proposition 8: The extension $f$ also satisfies (2).
Proof: Exercise (use properties of limits).

The OP mentioned that they had no intuition for why
$2^{1/2}\approx 2^{47/99}$
But, we can rewrite it with a common denominator,
$2^{99/198}\approx 2^{94/198}$
$2^{94/198}2^{5/198}\approx 2^{94/198}$
Now the term on the left is larger, so divide both sides by it,
$1 \approx \frac{2^{94/198}}{2^{94/198}2^{5/198}} = 2^{-5/198} \approx .983 $

So in general, if $0 \lt x \lt y$ and $x$ and $y$ are close you can get the relative (percent) difference between $a^x$ and $a^y$ with

$\frac{a^x}{a^y} = a^{x-y}$


$f$ is said to be continuous at $a$ if $\lim_{h\to 0}f(a+h)=f(a)$.

In our case:

$\lim_{h\to 0}f(a+h) =\lim_{h\to 0}e^{a+h} = \lim_{h\to 0}e^ae^h = e^a$

This follows because $e^0=1$.

so $e^x$ is continuous at $a$ for all $a \in \mathbb R$

Doesn't get any simpler than that.

  • 1
    $\begingroup$ "This follows because $e^0 = 1$." No it doesn't, it only follows if $e^x$ is continuous at $0$. $\endgroup$ Apr 21, 2014 at 6:02
  • $\begingroup$ Please read the whole question before answering $\endgroup$ Apr 21, 2014 at 6:03

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