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Let me explain it better: after this question, I've been looking for a way to put famous constants in the real line in a geometrical way -- just for fun. Putting $\sqrt2$ is really easy: constructing a $45º - 90º - 45º$ triangle with unitary sides will make me have an idea of what $\sqrt2$ is. Extending this to $\sqrt5$, $\sqrt{13}$, and other algebraic numbers is easy using Trigonometry; however, it turned difficult working with some transcendental constants. Constructing $\pi$ is easy using circumferences; but I couldn't figure out how I should work with $e$. Looking at enter image description here

made me realize that $e$ is the point $\omega$ such that $\displaystyle\int_1^{\omega}\frac{1}{x}dx = 1$. However, I don't have any other ideas. And I keep asking myself:

Is there any way to "see" $e$ geometrically? And more: is it true that one can build any real number geometrically? Any help will be appreciated. Thanks.

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If by "geometrical" you mean compass and straightedge, then no it is not possible since $e$ is a transcendental number. – tomcuchta Jun 18 '12 at 0:07
Yes, kind of it. But $\pi$ is a trancendental number, and it can be "seen" in a circumference of radius $\frac{1}{2}$. – Ian Mateus Jun 18 '12 at 0:09
Then I would ask you to define the constraints of when something is "seen" :). I could say that from your circle you can "see" $e$ because a circle is parametrized in the complex numbers as $e^{i \theta}$. – tomcuchta Jun 18 '12 at 0:11
$e$ is the value of $w$ such that the region under the reciprocal curve, $y=1/x$, from $x=1$ to $x=w$, has area $1$. (That is, $\int_{1}^{w} \frac{1}{u} du = 1$.) Unlike your $\int \log$ example, this approach doesn't implicitly use $e$ to define the region's upper boundary. – Blue Jun 18 '12 at 0:16
@tomcuchta In my question, "seen" is when a constant appears in a geomatrical construction, using compass and straightedge -- as you said! I'm asking it because of the fact that there is a transcendental number $\left(\pi\right)$ which appears in this construction. I'm thinking in words to put it in a clear way, but I couldn't find them. It is kind of "intuition". – Ian Mateus Jun 18 '12 at 0:20
up vote 39 down vote accepted

For a certain definition of "geometrically," the answer is that this is an open problem. You can construct $\pi$ geometrically in terms of the circumference of the unit circle. This is a certain integral of a "nice" function over a "nice" domain; formalizing this idea leads to the notion of a period in algebraic geometry. $\pi$, as well as any algebraic number, is a period.

It is an open problem whether $e$ is a period. According to Wikipedia, the answer is expected to be no.

In general, for a reasonable definition of "geometrically" you should only be able to construct computable numbers, of which there are countably many. Since the reals are uncountable, most real numbers cannot be constructed "geometrically."

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I read your last paragraph as implying that $e$ is not, or may not be computable. I am sure you did not mean to imply this. – MJD Jun 18 '12 at 1:23
@Mark: I do not see where I make that implication. Claiming that all geometrically constructible numbers are computable is not claiming that all non-geometrically constructible numbers are not computable. – Qiaochu Yuan Jun 18 '12 at 1:36
It's because you didn't indicate that you were no longer talking about $e$. – MJD Jun 18 '12 at 1:37
@Mark: I think it was quite clear from the words "in general" that Qiaochu was answering the last question of the OP and not talking about $e$ in particular. – Hans Lundmark Jun 18 '12 at 6:19

The area beneath the reciprocal function $$\frac{1}{x}$$ from $x=1$ to $x=e$ is $1$. Though this isn't really geometric like you want, it is still a clear way to see $e$ physically.


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I would call this a way of recognizing $e$, not constructing $e$. – Greg Martin Jun 18 '12 at 4:22
I'd say it gives us a way to construct an approximation to $e$ much like inscribing/circumscribing polygons in a circle allows us to construct an approximation to $\pi$. – hardmath Jun 18 '12 at 11:12

Debeaune asked Descartes this problem in a letter in 1638:

Consider a curve $y=f(x)$. Lets consider the tangent line $t(x)$ through the point $(x_{0},y_{0})$ which would look like $t(x) = y_{0} + f'(x_{0})\cdot(x-x_{0})$. What curve has the property that, every such tangent line intersects the $x$ axis at $x_{0}-1$, i.e., $$ t(x_{0}-1)=0 $$ What curve can do this? Only $y=C\exp(x)$...where $C$ is some nonzero constant.

For a thorough derivation, see

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I would call this a way of recognizing $e$, not constructing $e$. – Greg Martin Jun 18 '12 at 4:21
What's the distinction between the two? Using calculus to solve this problem explicitly constructs the curve $y=\exp(x)$, and this construction is unique when we specify $(0,1)$ lies on the curve. – Alex Nelson Jun 18 '12 at 4:28
How would you use this (nice) mathematical fact to construct $e$ (say a line segment of length $e$) from scratch? You can use any computer you want, as long as it doesn't already know what $e$ is. Does your method improve on constructing infinitely many line segments of lengths $1/n!$ and laying them end to end? – Greg Martin Jun 18 '12 at 19:36
But isn't that going backwards, starting with one definition of $e$ then trying to concoct geometric problem? Debeaune's problem naturally gives us the limit definition for $\exp(x)$, from a given geometric problem, upon using basic differential calculus...I still fail to see how this is "recognizing" $e$ instead of "constructing" $e$... – Alex Nelson Jun 19 '12 at 2:04
@GregMartin: Such a construction of $e$ is easy. Have your computer graph any exponential function, say $y=f(x)=2^x$. Now pick some $x_0$, draw the tangent through $(x_0,f(x_0))$ on the graph, intersect with $y=0$ to find $x_1$; now $f(x_0)/f(x_1)=e$, where $f(x_0)$ and $f(x_1)$ are lengths of segments easily found in the figure. – Marc van Leeuwen Oct 11 '12 at 7:37

You can't build any real number geometrically. They aren't even all computable. If you don't want to consider functions, you could (this is kind of cheating) look at $\lim_{n\rightarrow\infty} (1+\frac 1 n)^n$ as the volume of a suitably sized hypercube as the dimension increases.

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Sorry, but I would ask you why one can't build every real number geometrically. – Ian Mateus Jun 18 '12 at 0:14
It can get somewhat hairy, but basically any geometric construction is a finite algorithm (series of steps) of some kind. And by most real numbers are not computable, meaning that they cannot be constructed with any such algorithm, geometric or not. – Robert Mastragostino Jun 18 '12 at 0:17
Going to take look at it. Thanks. – Ian Mateus Jun 18 '12 at 0:22
It's a tricky argument. Paying careful attention to how one means the words, I believe it's actually consistent for there to be a construction of every real number, despite there not being an (internal) bijection between (internal) constructions and real numbers within a set theoretic universe. Look at Skolem's paradox to see how such arguments can go horribly awry. – Hurkyl Jun 18 '12 at 0:44
So, you can't build $\pi$ geometrically? Or $1$? Or $\sqrt{2}$? Your first sentense is exactly the same as 'you can build no real number geometrically'. Please, be more careful with your wording – Yuriy S Mar 2 at 17:20

I'll take my comment at the answer of Alex Nelson and make it into a separate answer, although it is really just the same answer reformulated. But I'll start pre-empting the predictable comment "I would call this a way of recognizing $e$, not constructing $e$" by countering that, even if one proclaims that constructing a point at a given distance along a curved line is a valid operation, one still cannot construct $\pi$ with ruler and compass either: one can only recognise it as the distance around a circle of diameter $1$ needed to get to the point diametrically opposite to the starting point.

Obviously we need some non-ruler-and-compass ingredient to construct $e$. I'll take this to be the graph of some exponential function, together with its unique asymptote: say in some coordinate system in which that asymptote is the $x$-axis we are given the set of points $(x,a^x)$ for some $a>1$ (neither the unit length of the coordinate system nor the value of $a$ need to be known; the $x$-axis is of course determined by the graph, but I'd have difficulty giving a construction of it).

Here is the construction: pick a point $P$ on the graph, find the point of intersection $Q_0$ of the tangent line to the graph at $P$ with the $x$-axis. Then taking perpendiculars to the $x$-axis through $P$ and $Q_0$ which intersect the $x$-axis in $P_0$ respectively the graph in $Q$, one has $\frac{PP_0}{QQ_0}=e$.

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I captured most of what you said but I don't know why the last procedure you have described in the last paragraph results in (e)? Thanks. – NoChance Oct 11 '12 at 10:51
@EmmadKareem: There are two way to see this. The lazy way is to argue that up to horizontal scaling and translation, all graphs of exponential function are the same, so it suffices to check for the graph of $x\mapsto e^x$. The assiduous way is to verify that the tangent at $(x_0,a^{x_0})$ has slope $a^{x_0}\ln a$, therefore passes through $(x_1,0)$ where $x_1=x_0-\frac1{\ln a}$, so that $a^{x_1}=a^{x_0}a^{-1/\ln a}=a^{x_0}e^{-1}$ – Marc van Leeuwen Oct 11 '12 at 13:26
+1, thank you for your explanation. Very interesting. – NoChance Oct 11 '12 at 13:52

Another approach might be finding a polar curve such that it's tangent line forms a constant angle with the segment from $(0,0)$ to $(\theta,\rho(\theta))$. The solution is the logarithmic spiral, defined by

$$\rho =c_0 e^{a\theta}$$

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I would call this a way of recognizing $e$, not constructing $e$. – Greg Martin Jun 18 '12 at 4:22

Plot the curves $y = a^x$ for $a > 0$. As $a$ gets larger, you find the slope of the tangent line is larger. If $a < 1$, this slope is negative. There is exactly one value for which the slope is 1, and that is $e$.

You can use this to define $e$ and derive the fact that $e=\sum_{n\ge 0}1/n!$.

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I would call this a way of recognizing $e$, not constructing $e$. – Greg Martin Jun 18 '12 at 4:22
Do you mean the slope at $x=0$? – Henry Oct 11 '12 at 6:59
Yes, Henry, at zero. – ncmathsadist Oct 11 '12 at 19:35

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