# 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)

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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
I am actually quite fond of this question, David! I tend to make comments on answers that are not relevant, and they have a tendency to get deleted after that. –  Mariano Suárez-Alvarez Sep 16 '11 at 17:34
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

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

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I originally posted for http://math.stackexchange.com/a/614082/53259:

Here, $f: A \rightarrow B$ is in green and $\{S_i\} = S_i$ for $S = A, B$ and $i = 1, 2.$
Ignore $\{b_2\} = B_2$ in this picture. This picture proves that $f(A_1) - f(A_2) \neq f(A_1 - A_2)$.
Incidentally, the same picture works for http://math.stackexchange.com/questions/225333/is-this-proof-correct-for-does-fa-cap-fb-subseteq-fa-cap-b-for-all-fu

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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!):

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.

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Page 340 of Hatcher's book: math.cornell.edu/~hatcher/AT/AT.pdf –  Dan Piponi Dec 14 '09 at 18:27
This is sometimes called the Eckmann-Hilton argument: en.wikipedia.org/wiki/Eckmann%E2%80%93Hilton_argument –  Kevin H. Lin Dec 14 '09 at 20:46
I've heard that term, but I've never quite understood how the diagram is supposed to prove the more general abstract nonsense theorem. But if you can explain it, that's what community wiki's for! :D –  Harrison Brown Dec 14 '09 at 20:53
There are lots of places on the web where this is explained nicely: youtube.com/watch?v=Rjdo-RWQVIY , math.ucr.edu/home/baez/week258.html , ncatlab.org/nlab/show/Eckmann-Hilton+argument , etc.... –  Kevin H. Lin Dec 14 '09 at 23:50
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This other answer shows that an 8x8 board with opposite squares removed cannot be tiled with dominoes, as they are of the same "colour". But what if two squares of opposite 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.

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

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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.

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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.

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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
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

The sequence of pictures

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
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 planetmath.org/encyclopedia/AreaOfASphericalTriangle.html Also, original MetaPost source can be obtained from this unfortunately obscure link: images.planetmath.org:8080/cache/objects/5841/src/sph-tri.mp –  Igor Khavkine Apr 26 '10 at 20:55
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): mathoverflow.net/questions/8846/proofs-without-words/… –  Vaughn Climenhaga May 18 '10 at 19:04
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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.

http://utdallas.edu/~hagge/images/Catalan.pdf

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

Here's the image:

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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

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

Theorem. The trefoil knot is knotted.

Proof.

$\square$

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).

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The image isn't loading... –  Nate Eldredge Aug 16 '12 at 22:32

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.)

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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:

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.

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Why would you resist? –  Mariano Suárez-Alvarez May 20 '10 at 17:41
+1 for the "Conjecture & Proof" shout-out. Best, course, ever! –  Kevin O'Bryant Nov 10 '10 at 23:18
Igen, nagyon jó. –  Douglas Zare Feb 7 '11 at 5:08
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This is a "proof without words" by an equation, not a picture.

Three complex numbers $a,b,c$ in the complex plane form the vertices of an equilateral triangle if and only if $~a^2 + b^2 + c^2 = ab + bc + ca$:



$$\hspace{-3in} 2 |a^2 + b^2 + c^2 - ab - bc - ca|^2$$ $$= ( |a-b|^2 - |b-c|^2)^2 + ( |b-c|^2 - |c-a|^2)^2 + ( |c-a|^2 - |a-b|^2)^2 .$$

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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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The pathspace of any topological space is contractible.

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

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

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Means inequalities:

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 $$1^3 + 2^3 + \cdots + n^3:$$

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
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

For $0 \lt k \lt n$,

$$\binom{n}{k} = \frac{n}{n-k}\binom{n-1}{k}$$

How k-subsets of [n], marked dark green in the rows, come from k-subsets of [n-1] after n-fold duplication and rearrangement:

Exactly $n-k$ times:

By induction, a base case, and taking $k=n$ and $k=0$ for granted: $$\binom{n}{k} = \frac{n}{(n-k)} \frac{(n-1)!}{(n-1-k)!\ k!} = \frac{n!}{(n-k)!\ k!}$$

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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.

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

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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

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

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This is not entirely without words, but Byrne's edition of Euclid's elements has cut down the number of words to a bare minimum.

http://www.math.ubc.ca/~cass/Euclid/byrne.html

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This is not quite in the spirit of the question... –  Mariano Suárez-Alvarez Sep 16 '11 at 17:27
+1: Thanks for this wonderful and beautiful link (be it in the spirit of the question or not). –  Hans Stricker Sep 16 '11 at 18:03

Of cause, this is not intuitive and it isn't elementary at all, but when I was asked to give the shortest proof of $\pi_1(S^1)=\mathbb{Z}$ I could imagine I answered $S^1\cong\mathbb{R}/\mathbb{Z}$.

I have to apologize if this becomes an example of misunderstanding the question, but I wanted to state that in my opinion "Proof without words" doesn't need to mean "Proof with picture".

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This proves the Minkowski version of the Pythagorean theorem:

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

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Here's a proof of the area of a circle (or sector) which is different from the one posted previously.

EDIT: I was unable to embed the file, which is in pdf form. Here is a link:

http://wildpositron.files.wordpress.com/2011/04/sectorarea2.pdf

I discussed what goes into making the proof complete to show that the map preserves area on my blog here (it requires just another picture or two, but it's essentially still only a geometric argument):

http://wildpositron.wordpress.com/2011/04/05/calculating-the-area-of-a-sector/

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

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oops! didn't see the word "non-trivial" in there... –  Steve Flammia Dec 14 '09 at 15:40
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
@HB: Um, Thomas Jefferson? –  Pete L. Clark Mar 6 '10 at 3:23
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
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

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.

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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
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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 :)

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Interesting how everyone understands "proof without words" as "proof made of pictures". I read the title and immediately thought that every proof can be written without words, using just first order logic. I stopped there and thought that this is just another language, using different words - and I came to the conclusion that there can not be a mathematical proof without "words", because you have to get some information across! Sure, you can use different languages than English. But in the end, this boils down to the question, what is a word?

BTW: Unmentioned so far are category-theoretical proofs, which can sometimes be expressed very comprehensively as a sequence of diagrams. I am too lazy to look up a good example, because I already explained that I don't believe in the question.

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Formulas (first order or what not) are just words written in abbreviated form. That really does not count... –  Mariano Suárez-Alvarez Jul 6 '11 at 1:01
So are pictures. That's my point. –  Jesko Hüttenhain Aug 9 '11 at 17:35
I doubt there is any sense in which one can formalize the notion, but I think it is pretty clear that a proof written in the first order calculus, or any other calculus, is simply not a "proof without words". You do not believe in the question, you say, but I honestly cannot understand what that can possibly mean: there is certainly something that gets the name proof-without-words (there is even a section in the MAA Monthly dedicated exclusively to this, and it has run for decades!) and most people ---while probably not being able to explain exactly what they are--- recognize them. –  Mariano Suárez-Alvarez Sep 16 '11 at 17:33
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(I'd post this as a comment to Mariano Suárez-Alvarez, but I've not enough rep). From a ME thread.

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

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