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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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where possible could people also either note the image source or explain/provide a link to a "how to" for constructing the associated diagram? I think that such would also be helpful for folks` – Carter Tazio Schonwald Dec 14 '09 at 23:57
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! – WetSavannaAnimal aka Rod Vance 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
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

68 Answers 68

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

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

alt text

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


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I am surprised that no one had cited the "proof" that the rationals are countable yet. See, for example, this picture

enter image description here

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Maybe it doesn't fit into the "non-trivial" category? – Campello Mar 19 '14 at 12:12
I think that the fact that the rationals are countable qualifies as non-trivial, when put in historical perspective – Geoff Robinson Mar 19 '14 at 19:02

Q: Can you tile chessboard less two opposite corner tiles with two tiles?

Tiling proof

$ $ $ $

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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
go ahead....... – Gil Kalai Dec 19 '09 at 18:59
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? – Turbo 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

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

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But what does that movie prove? – Mariano Suárez-Alvarez 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

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

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.

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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
  • 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

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For SO(3) has order 2 element: – 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-Alvarez 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): – Harrison Brown Dec 15 '09 at 2:42

The area under a cycloid is three times the area of the generating circle. enter image description here

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

enter image description here

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

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

link text

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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).


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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

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:

alt text

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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?

alt text

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 :)

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See also: MESE 220. – Benjamin Dickman Nov 12 '15 at 8:14

The pathspace of any topological space is contractible.

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

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There a proof of Erd˝os-Mordell Inequality 'without words' is an impressive one. Please follow the link

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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

The idea is to prove things in ways that are obvious to different parts of your brain, right? Anyone found any "auditory proofs"? Some candidates -

  1. Nyquist sampling theorem?

  2. sin[a] + sin[b] = 2sin[(a+b)/2]cos[(a-b)/2]. If you use at and bt instead of a and b, you can translate that to show how the addition of two sine tones close in frequency can also be perceived as a modulation or "vibrato" around the centre frequency. The factor of 2 might be hard, though you can add a gain instead of 2 and show that the difference is silence when the gain is 2 :)

  3. Sampling in frequency domain (comb filter) is periodicity in time domain?

Here are some "audio illusions" though, for your amusement -

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Are there more details on 1? 2 and 3 don't seem like proofs so much as examples, though maybe you are just putting them forth as challenges. – j.c. Mar 7 '10 at 11:54

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.

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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

Trig proof

A nice proof for trigonometric equation:


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This is an exhibit of the fact, but it isn't really a proof - it doesn't explain why those two functions sum to 1, just shows (arguably, just claims) that they do. You could replace the curve with any function $f$ with $f(\pi/2)=1$. – Steven Stadnicki May 16 '15 at 0:45
@StevenStadnicki In fact, I'm pretty sure that the function in the picture is not $\sin x$, the inflection point has a sharper third derivative than it should (although I'm sure this is just a limitation of the means by which the picture was drawn). There is certainly nothing geometrical constraining the shape of the diagram. – Mario Carneiro Jun 25 '15 at 18:53
However, this could be a nice proof of $\int_{0}^{\pi/2}\sin^2 x\,\mathrm{d}x = \int_{0}^{\pi/2}\cos^2 x\,\mathrm{d}x = \frac{1}{2}\left(\frac{\pi}{2}\cdot 1\right)$ – Machinato Jul 18 at 13:29

Rich Schwartz had on his site a great paper consisting of only a picture which proved that every right triangle admits a periodic billiard path. Unfortunately, he's since deleted it, so I can't post it here. (It shouldn't take too long for anyone interested to re-construct the proof, though.)

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I am guessing he did it by assembling four of the said right-triangles into a parallelogram. There is a path that bounces directly between the two longer sides. Mod out by the symmetry and you get a periodic path in the triangle. – Willie Wong Mar 11 '10 at 17:04

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

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Let $0\leq x,y,z,t\leq1$ Prove that $x(1-y)+t(1-x)+z(1-t)+y(1-z)\leq 2$.

Draw a 1x1 square and mark in consecutive sides disjoint segments starting at the vertexes of lengths $x,y,z,t$. Joining the consecutive end points of the intervals that are not vertexes of the square form four triangles, the area of the triangles is the left hand side divided by 2, the area of the square is the right hand side divided by 2.

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This proof without words has an awful lot of them! – I. J. Kennedy Apr 30 '11 at 17:39
Indeed, there is a proof with only eleven words: rearrangement and arithmetic-geometric inequalities. (Details are left to the reader.) – dvitek May 4 '11 at 0:13
I don't think drvitek's proof makes sense. – darij grinberg Jun 28 '11 at 14:55
+1 It is quite a nice idea, if you actually do the drawing. – rem Jun 17 '14 at 10:23

This proves the Minkowski version of the Pythagorean theorem:


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

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I suppose I should link to the physics stackexchange question that this came from:… – Ron Maimon Aug 23 '11 at 21:48

From Wikipedia: here is a "proof without words" of the Yoneda Lemma.

alt text

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This answer has already been proposed, and after some discussion it was more or less agreed that this is not a proof-without-words in standard sense of the term. – Mariano Suárez-Alvarez Oct 1 '11 at 23:45

Proof of the lantern relation (taken from the book: A Primer on Mapping Class Groups by Farb, B. and Margalit, D.)

Proof of the lantern relation

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enter image description here

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.

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enter image description here.
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=$2(\cos \frac{3}{7}\pi+\cos \frac{5}{7}\pi+\cos \frac{1}{7}\pi).$

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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

This is apparently not was intended, but I think it qualifies. From Principia Mathematica: the proof of 1+1=2 (I can't include the image bc I'm a new user, but perhaps an experienced user can edit this answer for me.)

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alt text

The composition of two continuous mappings is continuous.

Bloody thing won't let me embed the image...

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Here are some dynamic versions: (two of the summation formulas mentioned above)

Several belt, plate, and tangle trick animations:

A visual derivation of complex multiplication:

Pythagoras in the Isosceles case, based on the Yale tablet:

and the general case:

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There is an animation of the Dandelin Spheres construction depicted above in 3d-XplorMath in anaglyph 3d. Not really a proof, but visual descriptions of the relationship of cosine and sine curves and uniform circular motion: and epicycloids: – Bob Palais Feb 7 '11 at 4:04
All the links in this post appear to be broken. – I. J. Kennedy Apr 30 '11 at 17:21

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

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