# Proofs without words

Can you give examples of proofs without words? In particular, can you give examples of proofs without words for non-trivial results?

(One could ask if this is of interest to mathematicians, and I would say yes, in so far as the kind of little gems that usually fall under the title of 'proofs without words' is quite capable of providing the aesthetic rush we all so professionally appreciate. That is why we will sometimes stubbornly stare at one of these mathematical autostereograms with determination until we joyously see it.)

(I'll provide an answer as an example of what I have in mind in a second)

• I hope I am not alone in being (usually) unable to appreciate "proof by picture"... – Suvrit Jul 8 '11 at 21:14
• @Suvrit: I hope I am not alone in being most often unable to appreciate "proof by word" until I've read it at least twenty times and wrestled with it for many days per page! – Selene Routley Jul 9 '11 at 12:11
• My opinion is that almost every proof-without-words is improved by a few well-chosen words. – Joel David Hamkins Feb 12 '12 at 0:47
• @goblin, I am afraid that you have completely misunderstood the concept. The idea is pictures which have the rather amazing capability of immediately suggesting on the mind of the viewer the idea of a proof. How on earth you managed to get from the rather well-known idea involved in this question to «proofs without logic» is a mystery to me. – Mariano Suárez-Álvarez Jan 23 '15 at 3:55
• If you cannot tell the difference between a proof-tree and a proof without words in the tradition of, say, the AMM Monthly, then that is clearly a limitation of yours. I would rather you start a meta thread, or a blog, instead of further polluting this thread with what is clearly rather orthogonal chatter. – Mariano Suárez-Álvarez Jan 23 '15 at 22:52

The cover of Peter Winkler's first book is a great proof without words of a statement which I'll leave you to guess, regarding the combinatorics of tiling a hexagon with rhombi.

EDIT: I think the guessing game isn't helpful. The statement is that when tiling a perfect hexagon with the appropriate kind of rhombi of various orientations, the number of tiles in each orientation is the same. The image is slightly misleading in its use of color; there ought to be just three colors, corresponding to the three orientations.

• I'd be more impressed by this if I knew what statement was supposedly being proven by this illustration. That rhombus tilings are in 1-1 correspondence with 3d orthogonal surfaces (Thurston 1990, dx.doi.org/10.2307/2324578)? – David Eppstein Dec 14 '09 at 23:06
• That rhombus tilings are equinumerous to plane partitions which fit in a box. – Mariano Suárez-Álvarez Dec 15 '09 at 0:11
• Also, there are equal numbers of rhombi of each orientation in any tiling, and in fact, any tiling can be obtained from any other one by rotating "unit" hexagons formed by three rhombi. – Darsh Ranjan Dec 15 '09 at 2:35
• What do the colors represent? In particular, there are two colors for "upward-facing" rhombi (red and light gray) and two colors for "right-facing" rhombi (brown and dark gray), and I don't see why. – Michael Lugo Dec 15 '09 at 3:03
• @Xodarap sorry, doesn’t ring a bell. – Alon Amit Nov 2 '20 at 18:36

In an attempt to push the bar towards the non-trivial, I'll mention the proof that the boundary complex of every polytope is shellable. The proof is virtually word-free but requires an actual movie rather than a still image: imagine yourself in a spaceship, taking off in a straight line from one of the facets, away from the polytope. Every once in a while a new facet is visible to you; under assumptions of general position, this provides a shelling of the complex (obviously, you need to fly off to projective infinity and come back on the other side).

This was assumed by Euler but first proved only in 1970 by Brugesser and Mani, who said that the idea came to him in a dream. More details here (search for "shellability") or here.

• Why are there so many words and so few pictures in this answer? – David Eppstein Dec 14 '09 at 23:07
• Because I couldn't a way to draw this, let alone animate, in a reasonable time. I trust that the description is helpful in imagining what the actual wordless proof is. – Alon Amit Dec 15 '09 at 5:17
• I want a video! – Emil Jan 16 '10 at 22:45
• @AlonAmit The difference between what we want and what you gave us is explained in this video. – Simply Beautiful Art Nov 15 '16 at 23:07

Conway and Soifer tried to set a record for least number of words in a mathematical paper. I've reproduced it here in its entirety.

Can n2 + 1 unit equilateral triangles cover an equilateral triangle of side > n, say n + ε?
John H. Conway & Alexander Soifer
Princeton University, Mathematics
Fine Hall, Princeton, NJ 08544, USA
conway@math.princeton.edu asoifer@princeton.edu

n2 + 2 can:

• "This paper is worth publication, provided the authors add some words of explanation on their construction." – Pietro Majer Apr 10 '13 at 10:29

(I'd post this as a comment to Mariano Suárez-Alvarez, but I've not enough rep). From a ME thread.

$$\sum_{k=1}^n (-1)^{n-k} k^2 = {n+1 \choose 2} = \sum_{k=1}^n \; k = \frac{(n+1) \; n}{2}$$

I wonder why Apostol's proof of the irrationality of $$\sqrt{2}$$ - which is as visual as a proof can be (in my opinion) - has not been mentioned: One can literally see at a glance that it proves what it's supposed to prove: the impossibility of a isosceles right triangle with integer side length (by infinite descent):

Note that it's not a proof completely without words. It helps a lot to read the comments of the author:

Each line segment in the diagram has integer length, and the three segments with double tick marks have equal lengths. (Two of them are tangents to the circle from the same point.) Therefore the smaller isosceles right triangle with hypotenuse on the horizontal base also has integer sides.

But through own thinking one could come up with this by oneself (having in mind what's to be proved).

• I think you mean to refer to the impossibility of an integer isosceles right triangle, since clearly one can have isosceles triangles with integer sides, such as an equilateral unit triangle. – Joel David Hamkins Jan 22 '20 at 16:19
• @JoelDavidHamkins: Happy to hear from you again after all those years;-) I made the correction. Thanks for the hint. – Hans-Peter Stricker Jan 22 '20 at 16:21
• Oh yes, I've enjoyed many of your questions on MO over the years. – Joel David Hamkins Jan 22 '20 at 16:29
• So happy to hear that! But your enjoyment could not have been greater than mine was about your answers. – Hans-Peter Stricker Jan 22 '20 at 16:32
• It might be worth noting that one doesn't need to introduce the circle here; drawing the line between the top vertex of the triangle and the interior point on the bottom edge shows that the kite-shaped quadrilateral is bilaterally symmetric (the two triangles it's split into are congruent by side-angle-side equivalence) – Steven Stadnicki Nov 2 '20 at 18:03

A line that bisects the right angle in a right triangle also bisects a square erected on the hypotenuse:

I just saw this proof, which is of course not mine.

Proof with words: The [area of a] circle

Similar in concept to the above video:

This should really be a comment on Marco Radeschi's answer from Feb 22 involving the area formula for spherical triangles, but since I'm new here I don't have the reputation to leave comments yet.

In reply to Igor's comment (on Marco's answer) wondering about an analogous proof for the area formula of hyperbolic triangles: there is one along similar lines, and you're rescued from non-compactness by the fact that asymptotic triangles have finite area. In particular, the proof in the spherical case relies on the fact that the area of a double wedge with angle $$\alpha$$ is proportional to $$\alpha$$; in the hyperbolic case, you need to replace the double wedge with a doubly asymptotic triangle (one vertex in the hyperbolic plane and two vertices on the ideal boundary) and show that if the angle at the finite vertex is $$\alpha$$, then the area is proportional to $$\pi - \alpha$$. That follows from similar arguments to those in the spherical case (show that the area function depends affinely on $$\alpha$$ and use what you know about the cases $$\alpha=0,\pi$$).

Once you have that, then everything follows from the picture below, since you know the area of the triply asymptotic triangle and of the three (yellow, red, blue) doubly asymptotic triangles.

(That picture is slightly modified from p. 221 of this book, which has the whole proof in more detail.)

Q: Can you tile with ?

 

• I don't think this is clear enough to be self-contained, although I have something in mind to fix it. Do you mind if I try? – Jason Dyer Dec 19 '09 at 18:30
• I have edited and put in my modification of the image. – Jason Dyer Dec 19 '09 at 22:16
• can someone explain this in words? – 1.. Feb 15 '12 at 6:51
• It's easier if you do use words. If you take away opposite squares, you have more of one color than another... – Todd Trimble Aug 27 '12 at 3:42
• @Turbo, When you tile with dominoes, each domino covers two adjacent squares. Because every two adjacent squares contain 1 black and 1 white, the original question is equivalent to the question: Can you tile with black-and-white dominoes where each domino's colors match the colors of the two squares it covers. (So we have replaced the original question with a more constrained question, but know that the answers must be identical). Yet when tiling with black-and-white dominos that match what they cover, it's clear you can only cover boards with an equal number of black and white squares. – Jonah Sep 24 '17 at 6:54

The area under a cycloid is three times the area of the generating circle.

• How do you know each of the red bars of the football shape has exactly the width as the corresponding chord on the circle at the same height? – Jonah Sep 24 '17 at 7:02
• @Jonah: Consider a circle moving rightwards, starting at the left corner of the football. If we put a dot on the bottom of this circle moving clockwise at the same speed as the circle, we can think of the dot as painted on the circle, and the circle as rolling on the bottom line. But if the dot instead moves counterclockwise, we can imagine the dot on a wheel rolling on the top line. This produces the cycloids in the diagram. But for a fixed position of the circle, the dots will be at the same altitude, and so their distance will be the width of the circle at that altitude. – RavenclawPrefect Nov 9 '20 at 8:23

In the movie category, I'm surprised that no-one has yet posted a link to Moebius Transformations Revealed.

• But what does that movie prove? – Mariano Suárez-Álvarez Nov 8 '10 at 3:50
• @Mariano: it doesn't prove anything, but then again neither do any proofs without words. They merely give us insight into the proof, and in that respect, any movie has even more potential than a simple image. I think we will soon see very innovative approaches in movie-proofs. – Thierry Zell Nov 8 '10 at 3:59
• Very beautiful. I suppose it proves the usefulness of abstraction in obtaining unity – William Feb 12 '12 at 0:08

From the book "Proofs without words", there are ton of others too but this one I had trouble proving in UG, so like it most.

• I like using $\int_{-1}^01+r+r^2+\dots+r^{n-1}dr=\int_{-1}^0\frac{1-r^N}{1-r}dr$, but this way is very much visual. – Simply Beautiful Art Nov 15 '16 at 23:24

Given three mutually tangent circles in the plane, there exist exactly two circles tangent to all three.

• The first homotopy group of SO_3 has an order 2 element (that's a classic).

• The surface area of a quarter of the unit sphere is Pi via Gauss-Bonnet (My source is Ariel Shaqed - it should have been a classic, but no one I asked seems to knew it). The sphere is what you reach with a straight hand while standing still. Hold a Pencil in your hand, that's your tangent vector. Now parallel transport the pencil on a quarter sphere: it points in the opposite direction. QED

• For SO(3) has order 2 element: gregegan.customer.netspace.net.au/APPLETS/21/21.html – Dan Piponi Dec 14 '09 at 15:29
• Place a glass on the open palm of your hand. You can, with a bit of practice, rotate the glass twice (but not once) around the vertical axis without spilling any liquid from it, and return to your original position. Each part of your body goes through a loop in SO_3. Moving from the shoulder via the arm to the glass, you get a homotopy essentially proving the theorem. I have seen dancers from somewhere in south-east Asia incorporating this move into their dance. – Harald Hanche-Olsen Dec 14 '09 at 21:21
• Why are there so many words and so few pictures in this answer? – David Eppstein Dec 14 '09 at 23:07
• @David: well, you can think if this answer (or of Harald's comment, which gets my emphatic upvote) as a script for the choreography which, when acted out, is a proof without words :P – Mariano Suárez-Álvarez Dec 15 '09 at 0:00
• It doesn't feature Feynman, but here's a video of a human doing the plate trick (just after 1 minute in): youtube.com/watch?v=CYBqIRM8GiY – Harrison Brown Dec 15 '09 at 2:42

Have a look at this document from an MIT-instructor: http://mit.edu/18.098/book/extract2009-01-21.pdf

The pathspace of any topological space is contractible.

Pf (as given in my homotopy theory class): slurp spaghetti.

Proof of the associativity law $f * (g * h) = (f * g) * h$ in the fundamental groupoid of a topological space:

You can find more of these diagrams in J. P. May's A Concise course in algebraic topology.

I like the tiling proof of the Pythagorean Theorem. The left image is credited to Al-Nayrizi and Thābit ibn Qurra (9th century) and the right by Henry Perigal (19th century).

• I'm having trouble seeing a triangle (of the appropriate dimensions) in the Perigal tiling. – Gerry Myerson Aug 20 '12 at 22:46
• Gerry, slide the red square to the left by half the side length of a white square. The segment connecting the two lower red corners is the hypotnuse, and the legs have lengths which are the widths of the two tiling squares. Yes, I know that sounds confusing. – Marc Chamberland Aug 21 '12 at 0:58
• Thanks, Marc, not confusing at all. But I think if you have to add that to see that there's a triangle there, it's not really a proof without words. Well, at least, for me it's not a proof without words. – Gerry Myerson Aug 21 '12 at 5:33

I was somehow challenged by the idea: do more abstract topics allow proofs without words? I came out with this example. Of course, it is disputable if it is really "without words", since some words of explanation should be given (here they are: vertical segments represent Banach spaces and subspaces; connecting segments between two of them represent a linear operator. Italic letters $a,b,c,d,e$ are the dimensions of the corresponding linear subspaces).

This proves the Minkowski version of the Pythagorean theorem:

$c^2 = a^2 - b^2$

Also elementary, but here is a proof that

$$C_n = \binom{2n}{n} - \binom{2n}{n+1} = \frac{\binom{2n}{n}}{n+1},$$

where $$C_n$$ is the $$n$$th Catalan number.

Sorry for the link; new users may not use image tags.

Here's the image:

• Do you have an explanation for the picture? I looked at it, and looked at it, and don't get it. – Willie Wong Mar 11 '10 at 16:38
• Sorry for not noticing your question (much) earlier. The differences between adjacent terms in Pascal's triangle form another triangle which obeys the same generation rules. In my picture of that triangle, the yellow squares count some of the downward paths on a square grid which has been rotated $45^\circ$, namely those that never fall to the left of the top square. One definition of $C_n$ is that it is the number of such paths which terminate at the bottom corner of an $n \times n$ grid. – Tobias Hagge Oct 26 '10 at 5:46

Can you tile an 8x8 chessboard with one corner cut off with dominoes of dimension 3x1?

This is a simple way to show that choosing a useful coloring can make a proof trivial.

This proof was also a result of the Conjecture and Proof class in the Budapest Semesters in Mathematics. It was one of the first problems encountered there, hence not that hard :)

$S^2 \vee S^1 \vee S^1$ is homotopy equivalent to the Klein bottle with self-intersection.

• What a masterpiece :)! – justadzr Dec 5 '19 at 11:28

Posted this quite a while ago on math.SE as an answer to https://math.stackexchange.com/q/733754/214353; I think it has a place here too.

$\hskip7em\phantom{\ }$

To be rigorous, it maybe only proves that $f(z):=\displaystyle\lim_{n\to\infty}\left(1+\frac zn\right)^n$ is $1$ at integer multiples of $2\pi i$ (perhaps also that $f$ is $2\pi i$-periodic), but does not give any insight into why $f(z)=\text{something}^z$; still...

I've seen it much better rendered once somewhere on the web but could not find it again, so tried to reconstruct it myself

There is a beautiful proof of the fact that a checkerboard with sides $2^{n}$, and one square removed can be tiled with $L$-shaped pieces formed by three squares. Given that a checkerboard of sides $2^{n-1}$ can be so tiled, then a square checkerboard of sides $2^{n}$ can be tiled by filling in the quarter in which the removed piece lies, and then placing an extra $L$-shaped tile with one square in each of the remaining three quarters.

• I first learned this in Dan Velleman's book "How to prove it." I'm not sure if he originated it or not. – Jim Conant Feb 7 '11 at 2:36

.
This is an example I did when I was in high school.
Let it be a unit disc, consider the length of horizontal line, we know Yellow=$2\cos \frac{3}{7}\pi$, Yellow+Green=$-2\cos \frac{5}{7}\pi$, Red+Green=$2\cos \frac{1}{7}\pi$.
Then 1=Red=Red+Green-(Green+Yellow)+Yellow=$2(\cos \frac{3}{7}\pi+\cos \frac{5}{7}\pi+\cos \frac{1}{7}\pi).$

• I don't understand it. Seems like it does require some words... – Johannes Hahn Nov 11 '15 at 18:25
• Sorry about that. Actually, I don't really know how to explain it well. – Guo Qi Nov 12 '15 at 6:08

I've tried to read carefully through all the posts here, and I think no one has mentioned the excellent book by R.B. Nelsen which contains many more such examples. I give here an explicit citation:

Nelsen, Roger B. Proofs without words: Exercises in visual thinking. No. 1. MAA, 1993.‏

Here is an example which I have recently used:

• Nelson's book was already mentioned eight years ago (but without any illustrations): mathoverflow.net/a/8849 (and was the second answer to question). – jeq Aug 3 '17 at 14:16

Here you can find Grace Lin's proof without Words that The Product of the Perimeter of a Triangle and Its Inradius Is Twice the Area of the Triangle (see the figure below)

The proof originally appeared in the 1999 October issue of Mathematics Magazine.

• Beautiful animation! Is this rendered or hand-drawn? The colors/shader are fantastic... – Steven Stadnicki Nov 2 '20 at 17:24
• A computer rendering by Jos leys and his YouTube channel – C.F.G Nov 2 '20 at 17:30

There is a proof of Erdös-Mordell Inequality 'without words' which an impressive one. Please follow the link http://forumgeom.fau.edu/FG2007volume7/FG200711.pdf

• How is that "without words"? :-/ – Andrea Ferretti Mar 5 '10 at 17:23
• If you observe carefully on the graphs, you don't even need to write a word. – Sunni Mar 5 '10 at 19:15