In this book I'm using the author seems to feel a need to prove

$e^{u+v} = e^ue^v$


$\ln(e^{u+v}) = u + v = \ln(e^u) + \ln(e^v) = \ln(e^u e^v)$

Hence $e^{u+v} = e^u e^v$

But we know from basic algebra that $x^{a+b} = x^ax^b$.

Earlier in the chapter the author says that you should not assume $e^x$ "is an ordinary power of a base e with exponent x."

This is both a math and pedagogy question then, why does he do that?

So 2 questions really

  1. Do we need to prove this for such a basic property?
  2. If we don't need to, then why does he do it? Fun? To make it memorable? Establish more neural connections? A case of wildly uncontrolled OCD?

Also I've always taken for granted the property that $x^{a+b} = x^a x^b$. I take it as an axiom, but I actually don't know where that axiom is listed.

  • 9
    $\begingroup$ How is the function $e^x$ defined in the text? $\endgroup$
    – Alf
    Jan 16, 2011 at 16:00
  • 9
    $\begingroup$ @bobobobo: How is $\text{ln}(x)$ defined? $\endgroup$
    – Isaac
    Jan 16, 2011 at 16:12
  • 16
    $\begingroup$ @bobobobo: in other words, I think your confusion here is linguistic. You know there is an operation called exponentiation which satisfies x^{a+b} = x^a x^b, and the author is defining an operation which he calls exponentiation, so you assume it also satisfies x^{a+b} = x^a x^b. But you can't assume this. If I call something red, and you call something red, we aren't necessarily talking about the same color. We have to prove it, e.g. by agreeing on a unit of length and measuring the wavelengths of the appropriate type of light. Only then can we agree on the meaning of a statement like... $\endgroup$ Jan 16, 2011 at 17:19
  • 18
    $\begingroup$ -1 for asking a question of the form "Why did the author of my book do this?" without identifying the book in question. This makes things unnecessarily difficult and can waste a lot of people's time as they are forced to guess what the text might say. $\endgroup$ Jan 17, 2011 at 3:43
  • 12
    $\begingroup$ @bobobobo: there's absolutely nothing wrong with being a "math noob", and of course everyone who knows some advanced mathematics was at one point a beginner. I removed my downvote and added an upvote since you identified the book. @Jonas: "There's nothing wrong with inexpensive math books." No kidding. I'll complement your observation with: "There's something wrong with expensive math books." (Namely, they're expensive!) $\endgroup$ Jan 19, 2011 at 0:24

6 Answers 6


Mathematicians have a habit of using the same notation to denote many different concepts. To justify their overloading, they tend to point to similarities between properties of those concepts. You may think of the exponential as one concept, but it is actually a large family of related concepts:

  • The exponentials $x^n$ where $x$ is an element of an arbitrary monoid and $n$ a non-negative integer,
  • The exponentials $x^n$ where $x$ is an element of an arbitrary group and $n$ an integer,
  • The exponentials $x^n$ where $x$ is an element of an arbitrary group and $n$ a rational (which are not guaranteed to exist in general, and are also not guaranteed to be unique in general),
  • The exponential $x^n$ where $x$ is an element of a topological group where $x^n$ for $n$ rational is unique, and $n$ a real number (defined by limits as in Rudin),
  • The exponential $e^x = 1 + x + \frac{x^2}{2} + ... $ where $x$ is an element of a topological ring containing $\mathbb{Q}$ and the series converges,
  • etc.

It is easy to be fooled into thinking that these are all the same concept because they define essentially the same operation on $\mathbb{R}$, but 1) this is not obvious and requires proof, and 2) they generalize in different directions, so should be regarded as different in full generality. Note that the condition for $x^n$ to be defined where $x$ is an element of a topological group and $n$ a real number is quite strong and rarely satisfied.

Note also that your proof using logarithms runs into unnecessary subtleties when $u, v$ are taken to be complex numbers, since in this setting the logarithm is not uniquely defined. Nevertheless, the exponential law still holds in this setting; it is one of the basic properties that mathematicians point to to justify calling something an exponential in the first place.

In higher mathematics, the exponential further generalizes to:

So it is good to keep in mind that "exponential" does not just refer to one thing.

  • 7
    $\begingroup$ You've probably lost the reader when talking about group theory and topology - you certainly lost me. Best to stick to the level of the question, in my view. $\endgroup$
    – Noldorin
    Jan 16, 2011 at 16:17
  • 26
    $\begingroup$ @Noldorin: answers are not just for the questioner. The reason questions stick around after the OP accepts an answer is that questions and answers are supposed to be searchable and preserved for future users, who may have a very different background from the OP but may be wondering the same thing. This is an enormous difference between the internet and your grandmother's living room. Like I said, since I figured several other people would be giving answers at the appropriate level, I thought I would try something different. And if your grandmother asked about the properties of silicon... $\endgroup$ Jan 16, 2011 at 16:38
  • 8
    $\begingroup$ Saying "A,B,C are several concepts in higher mathematics that differ from your usual view of D, so you shouldn't take the properties of D for granted" is not itself an argument in 'higher mathematics' even though it refers to it. If Qiaochu used technical details of the latest set-theoretic constructions of group exponential functions by ultrafilters, I might agree with you; but since he just refers to the idea as an example I'd say it's at the appropriate level. $\endgroup$ Jan 16, 2011 at 16:48
  • 19
    $\begingroup$ @Noldorin: I can't help but feel like you're missing the point. The existence of people who would appreciate low-level answers does not say anything about the existence of people who would appreciate high-level answers. I guessed that several other people would provide simple answers, and they did - three of them. This question didn't need a fourth one, at least not until the OP clarified what his definitions were. Meanwhile, the comments in this answer apply regardless of what definition of exponentiation the OP is using. I don't know why you think I wrote this answer because I have a $\endgroup$ Jan 16, 2011 at 17:12
  • 8
    $\begingroup$ This discussion reminds me of Mordell's review of Lang's book on Diophantine geometry (projecteuclid.org/DPubS/Repository/1.0/…) and Lang's later review of Mordell's book on Diophantine equations (projecteuclid.org/DPubS/Repository/1.0/…). $\endgroup$
    – KCd
    Jan 17, 2011 at 4:49

In fact, such a proof is often necessary, which is why many authors write the function $e^x$ as $\exp(x)$ until they establish that it's just a "normal" exponent. For instance, if the original definition is given as

$$\exp(x) = \lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n,$$

then proving that $\exp(x + y) = \exp(x) \exp(y)$ is non-obvious, and certainly necessary.

  • $\begingroup$ This is very true, though defining the exponential function in this way and proceeding from there is slightly masochistic! $\endgroup$
    – Noldorin
    Jan 18, 2011 at 16:29
  • 7
    $\begingroup$ @Noldorin: why? It is an excellent intuitive definition from the point of view of differential equations: it's exactly what you get if you use Euler's method to approximate the solution to y' = y with step size 1/n and take n to infinity. Certainly it's better motivated from first principles than the power series definition, even if the latter is easier to prove things about. $\endgroup$ Jan 18, 2011 at 16:39
  • $\begingroup$ @Quiaochu: Oh, it's a lovely definition; only, it makes the following proof less trivial than other definitions. :) $\endgroup$
    – Noldorin
    Jan 18, 2011 at 16:39
  • $\begingroup$ I remember in Analysis I my instructor defined $e^x$ by its Taylor series. It is easy to check with this definition that this function is increasing, $f(1)=e$ and $f(x+y)=f(x)f(y)$. It follows imediatelly from these three simple properties that this is what we would call $e^x$... I liked that approach. $\endgroup$
    – N. S.
    Oct 12, 2011 at 23:09

This may seem like a pretty pointless proof, at least on the surface, but I suspect there's some subtlety in the way the author's defined things here. (It may even appear circular at first, considering that the logarithm is often introduced as the inverse of the exponential function. Saying that, it can be derived the other way round, and this can sometimes be enlightening.)

A rigorous proof for integer exponents is very straightforward indeed, and follows simply from the definition of the exponential function (of arbitrary base). For arbitrary exponents, things get slightly more complicated. I present a more complete proof below.

So, let us suppose that the author began by defining the (natural) logarithm function,

$$\ln a = \int_1^a \frac{dx}{x} .$$

We can then prove the addition property of logarithms, $\ln (ab) = \ln a + \ln b$, by considering

$$ \ln (ab) = \int_1^{ab} \frac{1}{x} \; dx = \int_1^a \frac{1}{x} \; dx \; + \int_a^{ab} \frac{1}{x} \; dx =\int_1^{a} \frac{1}{x} \; dx \; + \int_1^{b} \frac{1}{t} \; dt = \ln (a) + \ln (b) $$

(See this Wikipedia page for reference.)

The exponential function can of course be defined as the inverse of the logarithm, i.e.

$$exp(ln(a)) = a$$

Now, to prove the property of exponentials, $e^{u+v} = e^u e^v$, we start as follows.

$$\text{Let}\ u = \ln a, v = \ln b .$$

Then, using this property of logarithms and the definition of the inverse, consider

$$e^{u+v} = e^{\ln a + \ln b} = e^{\ln (ab)} = ab = e^u e^v .\ \square$$

That should hopefuly be straightforward enough to follow. There are of course other equivalent definitions of $exp$ and $ln$. (You can for example define the Taylor series of $exp$, use the Cauchy product, and then simplify, but that's slightly trickier.)

  • 1
    $\begingroup$ It's also quite common to define $ln(x)$ as the integral $\int_1^a \frac{1}{x} \, dx$, which might be the case here. $\endgroup$
    – Alf
    Jan 16, 2011 at 16:01
  • 1
    $\begingroup$ @Jon Yes, he did define $ln(x)$ like that previously $\endgroup$
    – bobobobo
    Jan 16, 2011 at 16:07
  • $\begingroup$ @Jon: Didn't notice that comment, but I was just writing that in my update as you made it. ;) $\endgroup$
    – Noldorin
    Jan 16, 2011 at 16:12
  • 2
    $\begingroup$ @bobobobo: Then this is your answer. It would have been great if you had given that definition of $\ln(x)$ in your question after Isaac in his comment asked for it. $\endgroup$ Jan 18, 2011 at 16:16

The necessity of a proof depends very much on how each of those things is defined and the domain over which the variables may vary. For example, if $e^x$ is defined by the power series $\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{n!}$ for all $x \in \mathbb{C}$, then there is a very real need to prove that $e^{x + y} = e^x e^y$, and the proof is non-trivial. On the other hand, if $e$ is some constant defined elsewhere and $x$ and $y$ are positive integers, then there's essentially nothing to prove.

I would also like to add that the law doesn't hold everywhere. If $A$ and $B$ are square matrices (or endomorphisms of a vector space) that do not commute, then in general $\exp(A + B) \ne \exp(A) \exp(B)$. This fact allows us to study certain non-commutative groups using the tools of analysis and differential geometry — this is the field of study known as Lie theory.

  • $\begingroup$ @ Zhen:"On the other hand, if e is some constant defined elsewhere and x and y are positive integers, then there's essentially nothing to prove." Why is it there is nothing to prove? How do we know that $x^a x^b = x^{a+b}$ in general $\forall x,a,b$? My line of thought is if I can prove that $e^{a_1 + b_1} = e^{a_1}e^{b_1}$ and $(e^{c})^d = e^{cd}$ then I can choose $a_1 = a \log_e(x)$ and $b_1 = b \log_e(x)$ and prove that $x^ax^b = x^{a+b}$. Or you could prove it for natural numbers, extend it to integers, to rationals to reals and to complex. In either case,there is some proof needed right? $\endgroup$
    – user17762
    Jan 17, 2011 at 16:31
  • 1
    $\begingroup$ @Sivaram: If $x^a x^b = x^{a+b}$ for any number $x$ and any positive integers $a$ and $b$ essentially by definition of the notation. The only assumption is that multiplication is associative, so in fact, as Qiaochu alludes to, this is true for $x$ in any arbitrary monoid. $\endgroup$
    – Zhen Lin
    Jan 17, 2011 at 17:22

One advanced reason for giving a proof is that in some related contexts such a formula breaks down. On the 2-adic numbers, the power series $A(x) = \exp(2x^2 - 2x)$ when expanded out and written in standard form turns out to converge on ${\mathbf Z}_2$, so in particular $A(1)$ is defined. One can prove $A(1)^2 = 1$, but although naively one may expect $A(1) = 1$ by just plugging 1 directly into the original definition I gave for $A(x)$, in fact $A(1) = -1$. (The point here is that the naive calculation $A(1) = \exp(2 - 2) = \exp(2)\exp(-2) = 1$ is wrong since $\exp(2)$ doesn't make sense 2-adically.)


One definition of $e^x$ is $\displaystyle \lim_{n \rightarrow \infty}(1 + \frac{x}{n})^n$. From this definition, it doesn't automatically follow that $e^x e^y = e^{x+y}$.

In fact, it doesn't even follow immediately that $e^x = \displaystyle \lim_{n \rightarrow \infty}(1 + \frac{x}{n})^n = (\displaystyle \lim_{n \rightarrow \infty}(1 + \frac{1}{n})^n)^x = (e)^x$. What this means is $e^x$ is just a short hand notation for the limit which after some analysis we realize it as $(e)^x$.

By limit arguments, we can now show that $\displaystyle \lim_{n \rightarrow \infty}(1 + \frac{x}{n})^n = 1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}$, $\forall x \in \mathbb{R}$.

Now $e^x \times e^y = (1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}) \times (1 + \sum_{k=1}^{\infty} \frac{y^k}{k!})$.

Now we need to realize that we can rearrange the terms in the series and multiply terms of the two series since both of them converge absolutely.

Hence $e^x \times e^y = (1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}) \times (1 + \sum_{k=1}^{\infty} \frac{y^k}{k!}) = 1 + (x+y) + (\frac{x^2 + 2xy + y^2}{2!}) + (\frac{x^3 + 3x^2y + 3xy^2 + y^3}{3!}) + \cdots$

Now make use of the binomial theorem to get

$$e^{x} e^{y} = e^{x+y}$$

PS: Though I have taken care to make sure the line of thought is right, you need to be careful when writing down the argument as to when you can interchange terms in an infinite series, multiply out two infinite series etc etc.

  • 4
    $\begingroup$ One definition of e^x is what you have written. $\endgroup$ Jan 17, 2011 at 17:57
  • $\begingroup$ @Qiaochu: I prefer to define $e^x$ as $\displaystyle \lim_{n \rightarrow \infty}(1 + \frac{x}{n})^n$ (at least for complex numbers) and then derive all the other formulas like the infinite series etc from this definition. $\endgroup$
    – user17762
    Jan 17, 2011 at 18:53
  • 1
    $\begingroup$ yes, but your use of the word "the" is misleading. The crux of this question is that it is not trivial to show that multiple definitions are equivalent. Your second argument doesn't work, either; like I mentioned in the comments, you still have to show that this function exists. $\endgroup$ Jan 17, 2011 at 19:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .