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)

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    I hope I am not alone in being (usually) unable to appreciate "proof by picture"... – Suvrit Jul 8 '11 at 21:14
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    @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! – WetSavannaAnimal aka Rod Vance Jul 9 '11 at 12:11
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    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
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    There is no such thing as a "proof without logic," and since words are usually the best tool for conveying logical relations, I'm going to have to reject the idea of "proof without words." Sorry, -1. – goblin Jan 23 '15 at 3:14
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    @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

78 Answers 78

A proof of the identity $$1+2+\cdots + (n-1) = \binom{n}{2}$$

proof without words

(Adapted from an entry I saw at Wolfram Demonstrations, see also the original faster animation)

This proof was discovered by Loren Larson, professor emeritus at St. Olaf College. He included it along with a number of other, more standard, proofs, in "A Discrete Look at 1+2+...+n," published in 1985 in The College Mathematics Journal (vol. 16, no. 5, pp. 369-382, DOI: 10.1080/07468342.1985.11972910, JSTOR).

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    Wow ! – Dinakar Muthiah Dec 19 '09 at 22:56
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    @Johann, people who thing that mathematics is about deducing theorems from axioms have such a mistaken idea of what the mathematical activity is thar their judgment is more or less irrelevant :D – Mariano Suárez-Álvarez Jun 29 '10 at 13:05
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    Am I the only one who doesn't understand this "proof" at all? – mathreader Oct 17 '10 at 17:07
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    @mathreader - the yellow dots are the sum of the first n numbers. Choosing two of the n+1 blue dots uniquely specifies a yellow dot in a bijective fashion. – Steven Gubkin Nov 11 '10 at 13:40
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    This beautiful proof warrants proper attribution. It was discovered by Loren Larson, professor emeritus at St. Olaf College. He included it along with a number of other, more standard, proofs, in "A Discrete Look at 1+2+...+n," published in 1985 in The College Mathematics Journal (vol. 16, no. 5, pp. 369-382). – Barry Cipra Oct 15 '11 at 2:17

Because I think proof by picture is potentially dangerous, I'll present a link to the standard proof that 32.5 = 31.5:

"proof" that 32.5 = 31.5

An animation of the above is:

enter image description here

(This work has been released into the public domain by its author, Trekky0623 at English Wikipedia. This applies worldwide.)

There does not seem to be any necessity for the particular 'path in the relevant configuration space' that was used by the author of the above animated gif. This may be seen as an argument against including an animation.

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    I think it is just as easy to introduce some kind of logical gap in a written proof as in a graphical one. – Steven Gubkin Mar 7 '10 at 23:41
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    @Steven: I think there is some truth to your claim, but I don't agree fully. First, we may notice that most proofs rely much more on writing than on pictures, and so mathematicians have developed a better radar for "written gaps". Second, there is a very strong sense in which written proofs may be formalized and checked by computer. Picture proofs, unless they share quite a bit of the "discrete" character of written proofs, usually are not amenable to such treatment. (And the notions of discreteness I can think of pretty much ensure that the picture proof could be turned into words.) – Pietro KC May 15 '10 at 20:22
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    +1 for "the standard proof that $32.5 = 31.5$." Made me laugh. :) – Quadrescence Oct 15 '10 at 20:10
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    @Pietro: “there is a very strong sense in which written proofs may be formalised”? Formalisation is a highly non-trivial task, and typically depends on quite a lot of mathematical background. What affects the difficulty is not whether the proof is written or graphical, but whether it’s detailed or highly abstracted. Formalising a good proof-by-picture is no harder than formalising a high-level written proof. Insofar as there’s a difference, I’d say it’s just that written proofs can be made detailed enough that formalising them is straightforward, whereas picture proofs perhaps can’t. – Peter LeFanu Lumsdaine Nov 29 '10 at 1:04
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    +1 , Here is the wiki page for this. – PKumar Oct 25 '14 at 7:17

The cardinality of the real number line is the same as a finite open interval of the real number line.

Proof without words

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    I suppose this picture can also be adapted to obtain the stereographic projection proof that a sphere is a manifold? – Kevin H. Lin Dec 14 '09 at 23:47
  • I could swear I've seen the exact picture you speak of somewhere. – Jason Dyer Dec 15 '09 at 1:37
  • Also, I drew the above image myself. Feel free to use in whatever you like. – Jason Dyer Dec 15 '09 at 1:38
  • @Jason Dyer: What software did you use? Inkscape? – davidk01 Dec 15 '09 at 1:48
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    I usually use Inkscape for my vector-based needs, but this was just done with my Smartboard presentation software. – Jason Dyer Dec 15 '09 at 14:21

There are a couple of Fibonacci identities, I think. For example

$F_0^2+F_1^2+\cdots+F_n^2=F_{n}F_{n+1}$, with $F_0=1$.

By puting together squares of side $F_n$, one at a time, you get a rectangle of dimension $F_nF_{n+1}$: The two squares of side 1, then the square of side 2, then the square of side 3 and so on.

Here is an image I found online


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    fantastic ! – Martin Brandenburg Apr 17 '10 at 23:30
  • Really exceptional! – Koundinya Vajjha Jul 24 '10 at 6:49
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    I think that there is a nice pictorial proof for this fact, but I don't think this is it. It's a proof for a specific $n$. To make it a general proof, the inductive step needs to be illustrated. – Max Mar 16 '11 at 14:08
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    @Max: The inductive step is easy to figure out, since the rectangle above contains the rectangles from previous steps. – Daniel Litt Mar 16 '11 at 20:01

It is known (see this other answer) that an 8x8 board in which squares at opposite corners have been removed cannot be tiled with dominoes, as the removed squares are of the same "colour". But what if two squares of different colours are removed? Ralph E. Gomory showed that it is always possible, no matter where the two removed squares are, and this is his proof:

Image of board

(Imagine A and B are the squares removed.) The image is from Mathematical Gems I by Ross Honsberger.

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    What I like about this example is that there seems to be no straightforward proof without the picture; the crux of the proof's idea is specifically this picture. – shreevatsa May 3 '16 at 21:21
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    Very nice. I guess the crux of the proof is that, when $mn$ is even, $P_m\times P_n$ is a Hamiltonian bipartite graph? – bof May 4 '16 at 4:15
  • Well, I'd call that a generalization, not the crux of the proof. :-) Staying concrete, for the question about the specific case of $m=n=8$, the crux of this proof (that this graph is Hamiltonian) is this picture. Similar pictures can be draw whenever $mn$ is even, sure. – shreevatsa Oct 17 '16 at 16:37

This is elementary as well, but one of my favorite ones :)

$1^2 + 2^2 + \dots + n^2 = \frac13n(n+1)(n+\frac12)$

(Author: Man-Keung Siu)

alt text

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    There's an analogous proof that the integral of n^2 from 0 to x is x^3/3. It can be obtained from this proof by smoothing out the stepped pyramids into actual pyramids. – Michael Lugo Dec 14 '09 at 16:47
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    I think very few people have enough spatial imagination to figure out what happens exactly in the area where the three pieces come together, or could easily depict the structure seen from the opposite end. For me the picture is not convincing at all (I'd rather say the formula convinces me the picture is correct than the other way round). However maybe playing with an actual model would be quite convincing. – Marc van Leeuwen Dec 12 '11 at 13:31
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    @Mark - I think if you just think about the width of each step at each level, you will be able to see that they do all fit together. Just counting back along a given row or column shows you that it all fits. – Steven Gubkin Feb 15 '12 at 15:10
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    A variant of Mike's construction for $\sum_{k=1}^n k^2$, easier to visualize (I'm going to try a proof-without-words, without pictures). Take $6$ copies of each parallelepiped of size $k \times k \times 1$. Glue them together so as to make the four lateral walls of a parallelepiped of (external) size $k \times (k+1) \times (2k+1)$. Do this for k from 1 to n, forming a collection of bracelets. Insert each one in the next, like matrioskas, getting a whole parallelepiped of size $n\times(n+1)\times(2n+1)$. – Pietro Majer Apr 10 '13 at 10:48
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    @Michael Lugo: The continuous version of this proof is "elementary geometry": the volume of a pyramid is one third of its height times the area of its ground surface! – nsrt Mar 18 '14 at 13:09

It's a long list of wonderful answers already, but I can't resist...

Question: Is it possible to find six points on a square lattice that form the vertices of a regular hexagon?

Proof without words:

alt text

Hint: A square lattice is invariant under rotation by π/2 around any lattice point. Use reductio ad absurdum.

Credit: I learned that proof from György Elekes during the Conjecture and Proof course in the Budapest Semesters in Mathematics, after constructing a proof of my own that used entirely too many words and made very laboured use of the fact that $\sqrt{3}$ is irrational. The picture here is my own creation (using Asymptote).

Follow-up: Can you find four points on a hexagonal lattice that form the vertices of a square? The proof is similar but not immediate.

There's a picture proof in the Princeton Companion, or alternatively on p. 340 of Hatcher, of the fact that the higher homotopy groups are abelian. Actually, here's a screenshot of the one in Hatcher (hopefully fair-use!):

Hatcher p. 340

Here $f$ and $g$ are mappings (with basepoint) of $S^n$ into some space for $n > 1$; the picture shows a homotopy between $f + g$ and $g + f$.

The above diagrams show an application of the interchange law, a more general expression of the Eckmann-Hilton argument, for double categories or groupoids. Here is a more general picture


which shows that the interchange law for a double groupoid implies the second rule $v^{-1}uv= u^{\delta v} $, where in the picture $a=\delta v$, for the crossed module associated to a double groupoid, taken from the book advertised here. There are many $2$-dimensional rewriting arguments which are essential to the results of this book.

This might be trivial but integration by parts has a nice proof without words:

enter image description here

(Got from: Roger B. Nelsen, Proof without Words: Integration by Parts, Mathematics Magazine, Vol. 64, No. 2 (Apr., 1991), p. 130; the original link is

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    @Daniel, I've turned the PDF into a PNG, and inserted the relevant part. I did keep the URL to the PDF for reference. Thanks, by the way! – Mariano Suárez-Álvarez Feb 7 '11 at 2:55
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    The same picture also gives an interesting formula for the integral of an inverse function! – Matt Noonan Jun 29 '11 at 0:57
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    I guess this proof works only when $f$ and $g$ are both increasing? – Greg Martin Nov 19 '15 at 19:16

I'm partial to the proof using Dandelin spheres that (certain) cross sections of cones are ellipses, where an ellipse is defined as the locus of points whose total distance to two foci is constant. It's particularly nice because it explains the foci geometrically, as well as the focus-directrix property with some more work.

Dandelin spheres touch the light blue plane that intersects the cone.

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    Yes, this one is beautiful. – Andrés E. Caicedo May 15 '10 at 18:47
  • Indeed. And there are also similar visual proofs for the hyperbola and the parabola – godelian Feb 18 '11 at 22:17
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    How does the picture explain the invariance of the total distance to two foci? I don't see it ; I haven't done geometry in a while though, I'm guessing it's some triviality... refresh my memory please? :) – Patrick Da Silva Jul 28 '13 at 18:35
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    @PatrickDaSilva: $PF1 = PP1$ because tangents to a circle/sphere have equal length. The total distance is thus equal to $PF1 + PF2 = PP1 + PP2 = P1P2$, which is constant. – aorq Jul 29 '13 at 8:53
  • @aorq : Riiiight. Thanks for the clarification! It is indeed very visual. – Patrick Da Silva Jul 29 '13 at 13:14

                          visual proof

This visual proof of $$\sum\limits_{n=1}^\infty \left (\frac{1}{2}\right)^{\,2n}=\frac{1}{3}$$ is from (Visible Structures in Number Theory, by Peter Borwein and Loki Jorgenson, The American Mathematical Monthly, vol. 108, no. 5, 2002, pp. 897-910).

Means inequalities:

alt text

The image was sent to me by James M. Lawrence, grazie! See also page 53 of "Proofs without words: exercises in visual thinking, Volume 2" for a very different layout of the same 4 inequalities.

Another one exists involving the sum $$1^3 + 2^3 + \cdots + n^3:$$

alt text

The second image is due to Brian Sears

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    I used the second proof (involving sum of cubes) in my class today after proving it by induction. A few were quite inspired by it! – Somnath Basu Feb 24 '12 at 18:42
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    2nd proof: It would be nicer if the small strips were above and to the left of the big square. – Günter Rote Feb 25 '13 at 22:56
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    The first proof could use some words. How is HM constructed? What is the small circle for? How does one prove that those segments have the claimed lengths? – Federico Poloni Apr 19 '14 at 16:04
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    For completeness, since a link is still missing: Mariano Suárez-Álvarez has given a beautiful improved version of the second image here: – Peter Heinig Feb 24 at 13:20

$$2 \pi > 6$$

hexagon inscribed in a circle

$ $ $ $ $ $ $ $

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    And similarly one proves that $\pi < 4$ by inscribing a circle in a square. – Michael Hardy Nov 16 '10 at 21:46
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    At first I was thrown off by this, because I was looking at area and not circumference. The area of an inscribed regular 12-sided polygon in the unit circle is also 3. – Todd Trimble Mar 12 '11 at 22:07
  • Helpful to remember that pi is ratio of circumference of the circle to diameter. – Talespin_Kit Dec 21 '16 at 15:54
  • Another case for $\tau$. – Daniel R. Collins Jul 25 at 16:53

Wikipedia has a few nice proofs of the pythagorean theorem. Elementary, but elegant.

Pythagorean Theorem, picture proof

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    Pythagoras' theorem is trivial? I had no idea … Seriously, I don't necessarily think that the existence of a very simple proof implies triviality. Such proofs are, after all, not so easily discovered. Anyway, this is my favourite proof of the theorem. – Harald Hanche-Olsen Dec 14 '09 at 20:58
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    The 20th President of the US, James Garfield, independently discovered the proof obtained by halving the right-hand diagram along a diagonal of the square of side length c. It requires you to write down an equation, though. That's my favorite proof, but mostly because of the corollary that B. Obama isn't the first geeky POTUS. – Harrison Brown Dec 15 '09 at 3:23
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    @HB: Um, Thomas Jefferson? – Pete L. Clark Mar 6 '10 at 3:23
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    A typical fake proof --- a simple statement as Pythagorean theorem is proved using much more advanced theorem on existence of area... – Anton Petrunin Nov 30 '10 at 20:26
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    A typical fake refutation. You don't need to define Lebesgue measure to do manipulations in geometry. All operations can be defined geometrically if I associate a number X with the segment of length X, and define $X \mapsto X^2$ as a function, mapping a segment to a square with such side. In fact, even many of infinite summations can be done geometrically, using the obvious topology and metric on shapes. Thanks to this formalistic tradition it took 100 years of pain to get from non-trivial Lebesgue construction to much more natural motivic integration. – Anton Fetisov Nov 13 '11 at 10:38

Duality between $\ell^1$ and $\ell^\infty$ norms.

p-norm 1 -> infinity

and the reverse animation

p-norm infinity -> 1

  • Mariano, thanks for fixing up my post! – Igor Khavkine Dec 16 '09 at 13:56
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    I... don't quite get it. I think I need a few more words: What's the dot representing in each picture? – Harrison Brown Dec 16 '09 at 15:01
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    The red line in xy-space satisfies the given equation. The dot gives the (a,b) coordinates of the same line in ab-space. The xy- and ab-spaces are linearly dual to each other. The resulting black and red shapes represent the unit balls in respective norms. – Igor Khavkine Dec 16 '09 at 15:34

If we have 3 circles on the plane with tangent lines, we can notice they have colinear intersection!

Made in Inkscape

To prove it, we can visualize the same configuration in 3D, the balls lay on a surface and rather than tangent lines we take cones: The colinearity comes from the fact that if we lay a plane ontop of this configuration it will intersect the table in a line!

This is from 'curious and interesting geometry' and the proof is attributed to John Edson Sweet. I really like this proof because it gives a vivid example of the general idea that sometimes, to solve a problem in the most simple way you need to view it as a part of some bigger whole.

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    You need to draw a 3D picture of this to get rid of the words! – Ian Agol Jun 28 '11 at 16:39
  • Don't we need the cones to all have the same slope? – benblumsmith Jan 23 '12 at 13:39
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    In this pretty solution there is another pretty geometric problem: Given three spheres there is a plane which is tangent to all three. – Rogelio Fernández-Alonso Jan 29 '12 at 16:51
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    Where is the picture? – Patrick Da Silva Jul 28 '13 at 18:29
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    This was one of my favorite proofs in this list... it's a shame that imageshack took this picture off to promote their site. – KalEl Nov 7 '14 at 22:36

Another proof of the sum of the first $n$ squares, relying on the knowledge of the formula for the sum of the first $n$ numbers:

$$1^2 + 2^2 + \dots + n^2 = n(n+1)(2n+1)/6$$

This one has a similar flavor to the fabled proof by Gauss of the sum of the first $n$ numbers. It's a good follow up for students after Gauss's proof.

  • There's probably a nice three-dimensional rendition of this that doesn't require writing down all those numbers. – Michael Lugo Aug 16 '12 at 23:03
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    +1 Superb. Is this original? If not, to whom is it attributed? – I. J. Kennedy Nov 22 '13 at 2:05
  • @MichaelLugo the 3D rendition is elsewhere on this page (you commented on it), but I do prefer this version. – adl Jan 12 '16 at 16:21

Here's a proof of the inequality of the arithmetic and geometric means in the form $$\frac{x_1^n}{n} + \cdots + \frac{x_n^n}{n} \geq x_1\cdots x_n.$$

Proof for $n=3$:

(there should be a figure here...)

The "figure" for general $n$ is similar, with $n$ right pyramids, one with an $(n-1)$-cube of side length $x_k$ as its base and height $x_k$ for each $k=1,\ldots,n$.

(I made this in Inkscape, a wonderful free-software vector drawing application. For the inequality and associated labels, I used the textext extension.)

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    And what exactly is a proof about this? – darij grinberg Nov 10 '10 at 23:40
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    The box has volume xyz and is contained in the union of the three square pyramids, which respectively have volumes x^3/3, y^3/3, and z^3/3. Thus xyz <= x^3/3 + y^3/3 + z^3/3. – Darsh Ranjan Nov 11 '10 at 3:41

I'm quite surprised no-one pointed out this one yet:

Theorem. The trefoil knot is knotted.


3-colored trefoil know $\square$

Some comments: a 3-colouring of a knot diagram D is a choice of one of three colours for each arc D, such that at each crossing one sees either all three colours or one single colour. Every diagram admits at least three colourings, i.e. the constant ones. We'll call nontrivial every 3-colouring in which at least two colours (and therefore all three) actually show up. It's easy to see (one theorem, more pictures!) that Reidemeister moves preserve the property of having a nontrivial 3-colouring, and that the unknot doesn't have any nontrivial colouring.

The picture shows a (nontrivial) 3-colouring of the trefoil.

EDIT: I've made explicit what "nontrivial" meant ― see comments below. Since I'm here, let me also point out that the number of 3-colourings is independent of the diagram, and is itself a knot invariant. It also happens to be a power of 3, and is related to the fundamental group of the knot complement (see Justin Robert's Knot knotes if you're interested).

  • This is wonderful. – Kevin H. Lin Jun 16 '11 at 19:28
  • What does "nontrivial" mean? – Tom Goodwillie Jun 28 '11 at 15:11
  • @Tom Goodwillie: I've edited, and added some remarks. Thank you. – Marco Golla Jun 28 '11 at 20:39

from Steven Strogatz's column:


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    Nice, but that reminds me of the "proof" of $2=\pi$ by approximating a straight line of length 2 by starting with a circle with this line as diameter, then two circles with one half of the line as diameter each, then for circles with on quarter of the line as diameter, ... One still has to find an argument that a geometric process converges at all and converges to the desired result. Both cannot be deduced purely from looking at a picture. – Johannes Hahn Nov 8 '10 at 11:27
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    Hmm, not sure, the point behind a proof by picture is that you do "get it," i.e., you see how the argument works in its full rigor. Now, either you do or you don't, but in this case I think it's all there. With circular arcs approximating a straight line you might notice upon observation that the arc length is independent of the iterations, which immediately discounts convergence... – AndrewLMarshall Nov 10 '10 at 6:21
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    By contrast, here you might observe that the difference between, say, how 2 circular wedges differ from their triangular counterparts in ratio, and how a wedge of twice the size differs from its triangular counterpart in ratio, does give on the order of geometric convergence. You can more or less just see that. – AndrewLMarshall Nov 10 '10 at 6:21
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    Wikipedia attributes this proof to Leonardo da Vinci. You can make establish rigorous convergence by using triangles that inscribe and circumscribe the wedges. – S. Carnahan Nov 11 '10 at 3:04
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    Hah, this is actually the proof appear in my primary school textbook. (I went to primary school in China, it was like 6th or 5th year) I'm amazed by this proof, but I'm not sure many kids can remember this though. – temp May 30 '12 at 2:40

Here is the very first piece of original mathematics I ever did, in high school:

The derivative of sine is cosine.

alt text

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    It looks like your image is no longer available... – I. J. Kennedy Nov 16 '10 at 19:09
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    Leibniz actually did this drawing. It's very nice because you can teach it to undergrads. You can do the same with any of the trig functions and their inverses. For tangent, you can extend the hypotenuse of the above triangle until it intersects the line tangent at the point $1$ (assuming this is the unit circle in the complex plane). Then you get a triangle with base $1$, height tangent, and hypotenuse secant. For cosecant and cotangent, you draw a tangent line from the point i. Then through similar triangles you can differentiate all these functions and their inverses. – Phil Isett Jul 8 '11 at 21:00
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    Yup, I have drawn pictures for all of those as well, but this always seemed like the simplest one. Never understood why this isn't in calculus books. – Steven Gubkin Jul 10 '11 at 15:41
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    @StevenGubkin If you still have it, could you put the picture back in? No one can see it. – Todd Trimble Oct 20 '15 at 16:05
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    This post is very much related to this. – Simply Beautiful Art Nov 17 '16 at 0:46

The sequence of pictures

intersection of 3 diangles intersection of 2 diangles intersection of 2 diangles intersection of 2 diangles

proves the area formula for spherical triangles $A=\hat{ABC}+\hat{BCA}+\hat{CAB}-\pi$.

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    Thomas Harriot first proved this formula in 1603, apparently by a similar argument, though I have not seen his picture(s). – John Stillwell Feb 22 '10 at 22:31
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    Haha, I'm happy to see these illustrations useful to someone! I created them some years ago, mainly to crystalize what I saw in my minds eye after finding some simple proofs of this identity online. The words accompanying these images can be found at Also, original MetaPost source can be obtained from this unfortunately obscure link: – Igor Khavkine Apr 26 '10 at 20:55
  • Incidentally, I tried to find a similar proof for the area formula for hyperbolic triangles. Unfortunately, that did not work due to non-compactness of hyperbolic space. If anyone knows whether such a proof exists, I'd be happy to see it. – Igor Khavkine Apr 26 '10 at 20:57
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    There is an analogous proof using the fact that although the hyperbolic plane has infinite area, a triply asymptotic triangle has finite area, so once you pick one of the two triply asymptotic triangles containing your triangle, you're in business. The relevant picture's in my answer posted separately (I posted it before I had the reputation to leave comments):… – Vaughn Climenhaga May 18 '10 at 19:04
  • This same proof also appears at the very opening of this paper: – Yaakov Baruch Jun 8 '16 at 14:44

Algebraic manipulations in monoidal categories can also be performed in a graphical calculus. And the best part is that this is completely rigorous: a statement holds in the graphical language if and only if it holds (in the algebraic formulation). See for example Peter Selinger's "A survey of graphical languages for monoidal categories". There are many instances, for example in knot theory studied via braided categories. The following specific example comes from Joachim Kock's book "Frobenius Algebras and 2D Topological Quantum Field Theories", and proves that the comultiplication of a Frobenius algebra is cocommutative if and only if the multiplication is commutative.

enter image description here

  • The link to Selinger's paper wasn't working – Yemon Choi Feb 15 '12 at 4:47
  • There are many proofs of similar flavor about 4-manifolds using the Kirby calculus. – Matt Brin Aug 15 '12 at 14:04
  • Picture is dead – BlueRaja Jun 28 '13 at 6:37
  • I'm voting this up because I like string diagrams, even though you don't mention them specifically. – Ryan Reich Oct 11 '13 at 14:48
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    this proof makes me wonder what is a 'picture' and what is – Jeremy Apr 19 '14 at 14:43

Sphere eversion

And here's a two-dimensional rendering of the sphere eversion: enter image description here

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    As pretty as it is, that is nowhere understandable as a proof. More as an illustration. – Willie Wong Mar 11 '10 at 16:44
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    @Willie: Suppose someone wrote down the equations/formulas for the sphere eversion in that video. It seems to me that checking that the formulas indeed give a sphere eversion would be a rather difficult and tedious task, whereas a video animation is, although not a rigorous proof, much more immediately convincing. – Kevin H. Lin Apr 6 '10 at 16:30
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    I just watched the video, which was excellent, but it had a lot of words in it. – Patricia Hersh Aug 19 '12 at 0:23

A classic one, from the late 19th century, that surprized Peano's contemporaries.

Question : "A curve that fills a plane ? You must be kidding"

Answer :

enter image description here

Well, of course a formal proof was necessary, but it is still one of my favorites.

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    How can you be sure that you're eventually covering all the points with irrational or transcendental coordinates? And giving a sequence of curves which fill more and more of the plane isn't the same as giving a single curve that does it all at once - it's not clear that such a limiting curve exists just looking at the pictures. – Michael Burge Sep 14 '10 at 8:47
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    Existence of the limits object is something that is very often forgotten. For example most Introductions to fractals give geometric descriptions of Koch's snowflake etc. via such an iteration but don't prove that there exists a limit of this iteration. – Johannes Hahn Sep 14 '10 at 9:22
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    Project: Fill the square one pixel at a time by following (an approximation to) this curve; then find some suitable baroque music accompaniment; then upload it to youtube. – Michael Hardy Nov 16 '10 at 21:51
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    If you look at the picture in detail you can see that you are defining a sequence of continuous functions that converge uniformly. It's also clear from the picture that the image is dense. Therefore the limiting function exists and its image (being dense and compact) is the whole square. Of course, this proof isn't 100% visual but the non-visual part -- the basic facts about uniform convergence and compactness -- can be regarded as background knowledge. So I think it's a nice example. – gowers Apr 10 '11 at 20:18
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    Remarkably, no picture nor mention to it was made in Peano's article, the construction being completely based on ternary expansions. The picture of a sequence converging to a square-filling curve appeared one year later in the paper by Hilbert. – Pietro Majer Nov 17 '11 at 14:14

As you probably already know — there are loads of these in Proofs without Words (and II) by Roger Nelson.

  • Yup. I was wondering if there are examples proving non elementary results (my example is not exactly non-elementary, I know...) – Mariano Suárez-Álvarez Dec 14 '09 at 6:14

$$\arctan \frac{1}{3} + \arctan \frac{1}{2} = \arctan 1$$

enter image description here

It's easy to generalize this to

$$ \arctan \frac{1}{n} + \arctan \frac{n-1}{n+1} = \arctan 1, \text{ for } n \in \mathbb{N}$$

which can further be generalized to

$$ \arctan \frac{a}{b} + \arctan \frac{b-a}{b+a} = \arctan 1, \text{ for } a,b \in \mathbb{N}, a \leq b $$

Edit: A similar result relating Fibonacci numbers to arctangents can be found here and here.

  • 3
    It needed quite a long time for me to understand this. But, well, then it is amazing! – Gottfried Helms Oct 28 '15 at 10:16
  • 1
    Very nice. It might be a bit more clear if the right triangle for arctan(1/2) and arctan(1/3) were colored in the first picture, and the triangle for arctan(1) was colored in the second picture. – Steven Gubkin Nov 17 '16 at 2:57

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.

  • 27
    Why are there so many words and so few pictures in this answer? – David Eppstein Dec 14 '09 at 23:07
  • 17
    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
  • 12
    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

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 heaxagon 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.

  • 3
    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, – 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
  • 2
    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
  • 1
    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
  • Sorry - I didn't want to spoil the fun right away. The statement being proven is the one indicated by Darsh Ranjan: however you tile a hexagon with rhombi, there's an equal number of tiles in each of the three orientations. The picture-proof asks you to believe that all such tilings can be regarded as the facets of a cubical arrangement, and the orientations correspond to the viewing angle. As far as I know, the colors are random and are just a distraction. – Alon Amit Dec 15 '09 at 5:15

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

n2 + 2 can:

figure 1

figure 2

  • 4
    "This paper is worth publication, provided the authors add some words of explanation on their construction." – Pietro Majer Apr 10 '13 at 10:29

protected by Scott Morrison Oct 11 '13 at 0:51

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