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 $45^\circ$-$90^\circ$-$45^\circ$ 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.

  • 14
    $\begingroup$ If by "geometrical" you mean compass and straightedge, then no it is not possible since $e$ is a transcendental number. $\endgroup$
    – tomcuchta
    Jun 18, 2012 at 0:07
  • 13
    $\begingroup$ Yes, kind of it. But $\pi$ is a trancendental number, and it can be "seen" in a circumference of radius $\frac{1}{2}$. $\endgroup$
    – Ian Mateus
    Jun 18, 2012 at 0:09
  • 9
    $\begingroup$ 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 $e^{i \theta}$. $\endgroup$
    – tomcuchta
    Jun 18, 2012 at 0:11
  • 2
    $\begingroup$ $e$ 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. $\endgroup$
    – Blue
    Jun 18, 2012 at 0:16
  • 2
    $\begingroup$ @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". $\endgroup$
    – Ian Mateus
    Jun 18, 2012 at 0:20

14 Answers 14


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."

  • 7
    $\begingroup$ 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. $\endgroup$
    – MJD
    Jun 18, 2012 at 1:23
  • 6
    $\begingroup$ @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. $\endgroup$ Jun 18, 2012 at 1:36
  • 4
    $\begingroup$ It's because you didn't indicate that you were no longer talking about $e$. $\endgroup$
    – MJD
    Jun 18, 2012 at 1:37
  • 9
    $\begingroup$ @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. $\endgroup$ Jun 18, 2012 at 6:19

The area beneath the reciprocal function $$x\mapsto\frac{1}{x}$$ 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.


  • 8
    $\begingroup$ I would call this a way of recognizing $e$, not constructing $e$. $\endgroup$ Jun 18, 2012 at 4:22
  • 6
    $\begingroup$ 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 $\pi$. $\endgroup$
    – hardmath
    Jun 18, 2012 at 11:12
  • 1
    $\begingroup$ This is a far cry from sqrt and $pi$ $\endgroup$
    – sixtytrees
    Jul 8, 2016 at 23:58
  • $\begingroup$ @Argon A simpler equivalent alternative would be to draw a line segment and mark the length as $e$... $\endgroup$ Sep 6, 2021 at 22:27

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 http://pqnelson.wordpress.com/2012/06/03/exponential-function/

  • $\begingroup$ I would call this a way of recognizing $e$, not constructing $e$. $\endgroup$ Jun 18, 2012 at 4:21
  • $\begingroup$ 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. $\endgroup$ Jun 18, 2012 at 4:28
  • 1
    $\begingroup$ 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? $\endgroup$ Jun 18, 2012 at 19:36
  • 1
    $\begingroup$ @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. $\endgroup$ Oct 11, 2012 at 7:37
  • 1
    $\begingroup$ @Marc: this is a great idea ... but how does one construct the curve $y=2^x$, without using the exponential function? Again, I don't consider a limiting process a construction (else we can just construct $e = \sum 1/n!$ directly). $\endgroup$ Oct 18, 2012 at 7:46

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.

  • $\begingroup$ Sorry, but I would ask you why one can't build every real number geometrically. $\endgroup$
    – Ian Mateus
    Jun 18, 2012 at 0:14
  • 5
    $\begingroup$ It can get somewhat hairy, but basically any geometric construction is a finite algorithm (series of steps) of some kind. And by en.wikipedia.org/wiki/Computable_number most real numbers are not computable, meaning that they cannot be constructed with any such algorithm, geometric or not. $\endgroup$ Jun 18, 2012 at 0:17
  • $\begingroup$ Going to take look at it. Thanks. $\endgroup$
    – Ian Mateus
    Jun 18, 2012 at 0:22
  • $\begingroup$ 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. $\endgroup$
    – user14972
    Jun 18, 2012 at 0:44
  • $\begingroup$ 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 $\endgroup$
    – Yuriy S
    Mar 2, 2016 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 $\frac{PP_0}{QQ_0}=e$.

  • $\begingroup$ 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. $\endgroup$
    – NoChance
    Oct 11, 2012 at 10:51
  • $\begingroup$ @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 $a^{x_1}=a^{x_0}a^{-1/\ln a}=a^{x_0}e^{-1}$ $\endgroup$ Oct 11, 2012 at 13:26

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

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

  • $\begingroup$ I would call this a way of recognizing $e$, not constructing $e$. $\endgroup$ Jun 18, 2012 at 4:22
  • 1
    $\begingroup$ @GregMartin This is definitely a construction and not just recognizing $e$. If you really want the value $e = 2.71828\ldots$, then just take $c_0 = 1$ and $a\theta=1$. Ensuring a constant angle for the tangent is a concrete procedure, just like involutes, evolutes, roulette, etc. Compass-and-straightedge is an idealization (of a physical process) that can never be achieved and only approximated; so are these. Confining "geometric construct" to merely compass-and-straightedge is an unnecessarily narrow view. $\endgroup$ Dec 2, 2017 at 22:15

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 $e=\sum_{n\ge 0}1/n!$.

  • $\begingroup$ I would call this a way of recognizing $e$, not constructing $e$. $\endgroup$ Jun 18, 2012 at 4:22
  • $\begingroup$ Do you mean the slope at $x=0$? $\endgroup$
    – Henry
    Oct 11, 2012 at 6:59
  • $\begingroup$ Yes, Henry, at zero. $\endgroup$ Oct 11, 2012 at 19:35

The answer is "yes, you can," if you think beyond Euclidean geometry and consider geometric quantities appearing in hyperbolic geometry. In the Euclidean geometry, the number $\pi$, of course, you "see" by looking at circumferences and areas of circles. In hyperbolic geometry, the circumference of the circle of radius $r$ is $$ L_h(r)= 2\pi \sinh(r), $$ see for instance here.

Hence, for the unit circle, we get $$ L_h(1)= \pi\left(e - \frac{1}{e}\right).$$ This gives you a mixture of $e$ and $\pi$. In order to "see" $e$ by itself, we can look at the ratio of hyperbolic lengths: $$ \frac{L_h(r+1)}{L_h(r)}= \frac{\sinh(r+1)}{\sinh(r)}. $$ We got rid of $\pi$ but did not get $e$ (yet). To get $e$, you look at the asymptotics of this ratio as $r\to \infty$. In the Euclidean case, we would get $$ \lim_{r\to\infty} \frac{2\pi (r+1)}{2\pi r}= 1 $$ (another universal constant), while in the hyperbolic case, the limit is $$ \lim_{r\to\infty} \frac{\sinh(r+1)}{\sinh(r)}= \lim_{r\to\infty} \frac{e^{r+1} - e^{-(r+1)}}{e^r - e^{-r}}= e. $$ If you do not regard limits as geometric operations (I do think of limits geometrically), then instead think of this as the supremum $$ \sup_{r\ge 1} \frac{L_h(r+1)}{L_h(r)}= e. $$

This is how to "see" $e$ geometrically.

Incidentally, there are many ways to "see" $\pi$ in hyperbolic geometry. My favorite is that $\pi$ is the supremum of areas of hyperbolic triangles. The bottom line is that you can "see" more in hyperbolic geometry than in Euclidean geometry.


From a geometrical viewpoint, Euler's constant $e$ is equal to the sum of the volume of all standard $d$-simplexes, i.e. over all positive integers $d$.

Alternatively, one can states that $e$ is equal to the sum of the first hyperoctant hypervolume of all $d$-dimensional hypertetrahedrons with standard unit vectors as vertices, i.e. over all non-negative integers $d$.


There is another fomulation of Euler’s Number that is geometric in nature (using hypercubes):

Here is Euler's Number expressed a series of hypercubes each divided by n!, where n is the dimension of the hypercube.

$$e^{x} = \sum_{n=0}^{\infty} \frac{x^{k}}{k!} = \frac{x^{0}}{0!} + \frac{x^{1}}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + \cdots $$

You'll notice a similarity to elementary problems in combinatorics: from an arbitrary set {a,b,c,d,e} where each element is in [0,x], how many ways can you form coordinates in $\mathbb{R}$3? It's just a combination with repetition problem. There are

$$\binom{n + 1 - r}{r}$$

such ways. Similarly, each term can be thought of the set of combinations of coordinates with repetition in each hypercube space.


Here is another way to "see" (not construct) $e$ geometrically.

In a semicircle, draw zigzag line segments that, together with the base, form $n$ equal-width isosceles triangles whose top vertices lie on the semicircle. Here is an example with $n=6$.

enter image description here

$e$ is the diameter of the semicircle such that the product of the lengths of the line segments converges to a positive number as $n\to\infty$.

The limit can be expressed as $\lim\limits_{n\to\infty}\prod\limits_{k=1}^n e^2\left(\left(\frac{1}{2n}\right)^2+\frac14-\left(\frac{k}{n}-\frac{1}{2n}-\frac12\right)^2\right)$, which turns out to equal $2$.


Don't knowing if this answer could met your expectatives, but I have made a similar question here, and so far I have found this:

Through the properties $\lim_{x \to 0} (1+x)^{\frac{1}{x}} = e$ and $\lim_{x \to 1} (1+x)^{\frac{1}{x}}= 2$, and using that $\log(1+x)=\int_1^{1+x} \frac{du}{u}$, you could use:

$$\int_0^1 \frac{\partial}{\partial x}\left[2x-(1+x)^{\frac{1}{x}} \right]\,dx = e$$ $$ \iff \int\limits_0^1 \left(2-(1+x)^{\frac{1}{x}}\left(\frac{1}{x(1+x)}-\frac{\log(1+x)}{x^2}\right)\right) \,dx = e$$

wolfram alpha

where the constant $e$ is not explicetly involved, but its hidden that it still requires to find the limit $\lim_{x \to 0} (1+x)^{\frac{1}{x}} = e$ so I don't know if its valid (under the integral is just a point of zero meassure, so maybe the problem is avoided, but I don't really know), but I found it interesting enough to be shared, maybe someone gets an idea to improve it.

It is possible to use also terms of the form $q(x)=\left(1+\tan\left(\frac{\pi x}{2}\right)\right)^{\cot\left(\frac{\pi x}{2}\right)}$ to find that $$\int_0^1 1-q'(x)\ dx=e$$ but is almost the same function but even more complicated, and also has the same limitations due the logarithm appearing on the derivative.

Another related ones: $$\int_0^1 \frac{\partial}{\partial x}\left[2x+(2-x)^{\frac{1}{1-x}} \right]\,dx = e$$

$$\int_0^1 \frac{\partial}{\partial x}\left[2x+\frac{1}{2}\left((2-x)^{\frac{1}{1-x}}-(1+x)^{\frac{1}{x}}\right) \right]\,dx = e$$

So at least the area below these functions adds up to $e$.


EDIT: To my answer (under the line) I am gonna add pictures I have draw- I construct an aprroximation of $e$ into 5th iteration on the Number Line, where:

$$L_0=a_0=1 $$ $$L_{n+1}=L_n+a_{n+1} ; a_{n+1}=\frac{a_n}{n+1} $$

enter image description here Altenative we can write:

$$L_2=2.5 ;a_2=0.5 $$ $$L_{n+1}=L_n+a_{n+1} ; a_{n+1}=\frac{a_n}{n+1} $$

Here I tried to approximate between $[2;3]$ (into 6th step): enter image description here enter image description here

Old Answer:


A) We gonna use the Taylor series for $e$ and multiply by $A$:



In special case: $A=1$ (leght unit)

We can lead $Ae$ to this form:

$Ae=A(1+\frac{1}{1}(1+\frac{1}{2}(1+\frac{1}{3}(1+\frac{1}{4}(1+...)...)...)$ (Formula 1)

B) Optional: How a line segment can be devide via ruler and compass: Intercept theorem

C) (Formula 1) will be explained as an Algorithm plus picture .

---The Algorithm---

Short :


  • Have a line segment $a_0$ with a length of $A$ .


I) Duplicate $a_{n-1}$ and divide the new segment into $n$ equal parts. Pick one element as $a_n$.

II) Add $a_{n-1}$ to the older line segments: $L_{n-1}=a_{n-1}+\displaystyle\sum_{i=0}^{n-2}a_i$

Long :

  1. Have a line segment $a_0$ with a length of $A$ .

1.1. Duplicate $a_0$ and divide the new segment into $1$ equal parts .

1.2. Pick one element as $a_1$.

1.3. Add $a_0$ to the older line segments: $L_0=a_0$

2.1. Duplicate $a_1$ and divide the new segment into $2$ equal parts .

2.2. Pick one element as $a_2$.

2.3. Add $a_0$ to the older line segments: $L_1=a_1+a_0$

3.1. Duplicate $a_2$ and divide the new segment into $3$ equal parts .

3.2. Pick one element as $a_3$.

3.3. Add $a_2$ to the older line segments: $L_2=a_2+a_1+a_0$




n.1. Duplicate $a_{n-1}$ and divide the new segment into $n$ equal parts .

n.2. Pick one element as $a_n$.

n.3. Add $a_{n-1}$ to the older line segments: $L_{n-1}=a_{n-1}+(a_{n-2}+...+a_2+a_1+a_0)$

---Picture--- enter image description here

  • $\begingroup$ Wish the downvoter had left a comment why. $\endgroup$ Sep 9, 2018 at 15:13
  • 2
    $\begingroup$ I didn't downvote but perhaps it was because of the use of many colors, specially the use of light green in your handmade calculations. You should check this https://math.meta.stackexchange.com/questions/4195/on-the-use-of-color-in-equations it basically mention the complications for colorblind people, when using colors. $\endgroup$
    – user486983
    Oct 1, 2018 at 22:35

From the hyperbolic identity:

          cosh(x) + sinh(x) = e^x  ...in general,

we have

          cosh(2) + sinh(2) = e^2  ...in particular.

If we take individually, the square of the square root of each of the first two terms of the second equation, we have a Pythagorean identity, where the sides are:

F = √(cosh(2)), G = √(sinh(2)), and the hypotenuse of the right-angled triangle, H = e.

I hear you say, not as good as π, the ratio of the circumference of a circle to its diameter, since one can measure the circumference and the diameter, by some means. One needs the formula or Taylor series for cosh(2) and sinh(2).

(My original edit here was pointless. It involved the additional identity:

         1  +  sinh^2(x)  =  cosh^2(x)

which is also a Pythagorean identity, but since it can be derived from the previous case, and vice versa, it was pointless in calculating a value for 'e').

N.B. 'e', the base of natural logarithms, was created as a valid idea in calculus, to make differentiation, and therefore integration, easier. It is not a geometrical constant, but a mathematical constant, and is the limit of:

                      (1 + t)^(1/t), as t tends to 0

which occurs in the differentiation of log(x).

  • $\begingroup$ How geometrically it is possible to construct $\cosh(2)$ and $\sinh(2)$ exactly? (Edit: I just realised I replied to years-old answer!) $\endgroup$ Aug 3, 2021 at 0:42

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