# 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"... 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! Jul 9 '11 at 12:11
• My opinion is that almost every proof-without-words is improved by a few well-chosen words. 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. 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. Jan 23 '15 at 22:52

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

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).

• @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 Jun 29 '10 at 13:05
• @Johann: some days of the week, I am such a person; and from that point of view, the picture is a beautifully clear encoding of a certain bijection, and the formal construction of the bijection itself is a very beautiful proof. No beauty is destroyed!// I strongly believe that a proof with a clear intuition should also be clear as a formal proof. If not, either (usually) our formalism isn't as good as it could be, or (occasionally) our intuition really is overlooking some non-trivial subtleties. Jun 29 '10 at 14:37
• Am I the only one who doesn't understand this "proof" at all? Oct 17 '10 at 17:07
• @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. Nov 11 '10 at 13:40
• 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). 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:

An animation of the above is:

(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.

• I think it is just as easy to introduce some kind of logical gap in a written proof as in a graphical one. Mar 7 '10 at 23:41
• @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.) May 15 '10 at 20:22
• @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. Nov 29 '10 at 1:04
• +1 , Here is the wiki page for this. en.wikipedia.org/wiki/Missing_square_puzzle Oct 25 '14 at 7:17
• It might be noted that the success of the illusion partly depends on the fact this uses Fibonacci numbers (it is a coincidence I guess that the next newest answer is also about Fibonacci numbers!). Jan 23 '15 at 2:53

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

• I suppose this picture can also be adapted to obtain the stereographic projection proof that a sphere is a manifold? Dec 14 '09 at 23:47
• @Jason Dyer: What software did you use? Inkscape?
– user577
Dec 15 '09 at 1:48
• I usually use Inkscape for my vector-based needs, but this was just done with my Smartboard presentation software. Dec 15 '09 at 14:21
• This picture shows not only that they have same cardinality but that they are homeomorphic. Jan 23 '20 at 7:46

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:

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

• 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. May 3 '16 at 21:21
• 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. Oct 17 '16 at 16:37
• A complete result (guessed not shown) is for m or n odd : Any mxn board with 1 square removed has a neighborhood graph that has an hamiltonian cycle. Mar 30 '20 at 10:19
• I call this the "hungry snake" proof (imagine a game of Snake and you have to get the high score. The snake would have to assume some pose like this one). The exact same idea can be used to prove a lot of boards have the same property (for general n-dimensions). Nov 18 '20 at 3:46

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 putting 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

• fantastic ! Apr 17 '10 at 23:30
• Really exceptional! Jul 24 '10 at 6:49
• 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
• @Max: The inductive step is easy to figure out, since the rectangle above contains the rectangles from previous steps. Mar 16 '11 at 20:01

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)

• 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. Dec 14 '09 at 16:47
• 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. Dec 12 '11 at 13:31
• @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. Feb 15 '12 at 15:10
• 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 $n\times(n+1)\times(2n+1)$. Apr 10 '13 at 10:48
• @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:

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.

• Why would you resist? May 20 '10 at 17:41
• +1 for the "Conjecture & Proof" shout-out. Best, course, ever! Nov 10 '10 at 23:18
• Igen, nagyon jó. Feb 7 '11 at 5:08
• I really like this image and proof. The same idea works for other regular polygons, and I made images for pentagons, heptagons and so on, which you can find on my blog post at jdh.hamkins.org/no-regular-polygons-in-the-integer-lattice. (The code accept $n$ as input, and makes the image for an $n$-gon.) Dec 4 '16 at 16:24
• And here is my post on the hexagonal lattice: the only regular polygons to be found are triangles and hexagons. jdh.hamkins.org/no-regular-polygons-in-the-hexagonal-lattice Dec 9 '16 at 3:48

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

(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 https://www.maa.org/sites/default/files/Roger_B04151._Nelsen.pdf).

• @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! Feb 7 '11 at 2:55
• The same picture also gives an interesting formula for the integral of an inverse function! Jun 29 '11 at 0:57
• I guess this proof works only when $f$ and $g$ are both increasing? Nov 19 '15 at 19:16
• @GregMartin : The way this drawn assumes that $f(b)>f(a)$ and $g(b)>g(a)$, but that's all (and you can draw similar pictures for the other possibilities). If they're not increasing the whole way, then there will be stuff outside the shaded regions, but it will count both positively and negatively and so cancel. Thus, the shaded regions' areas do equal the stated integrals. Dec 7 '20 at 18:34

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.

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

• This proof is actually known to Archimedes and used in his Quadrature of the Parabola en.wikipedia.org/wiki/1/4_%2B_1/16_%2B_1/64_%2B_1/… Jul 18 '16 at 12:48
• Visual recursion. Awesome. Mar 6 '17 at 17:46
• Pierre Arnoux made a nice video about the geometric series, which has a version (with colours and animation) of this picture: youtu.be/6KQiTJLBwEw The video has words though! Sep 12 '18 at 8:52

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.

• Yes, this one is beautiful. May 15 '10 at 18:47
• 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? :) Jul 28 '13 at 18:35
• @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
• I've learned this and related proofs from Hilbert and Cohn-Vossen (but these proofs still originated mostly with Dadelin). Nov 9 '13 at 3:21
• I was confused by the perspective. In case anyone is having the same problem: the perspective is from a point that is (below the apex $S$ of the cone but) above the base of the cone (circle $k2$) and the bottom half-sphere ($G2$). I mistakenly thought we were looking up through $k2$ into the inside of the cone -- I think this is because in my browser, at least, the the circle $k2$ gets thicker when it passes behind the ellipse and should if anything get thinner. It's generally a nice drawing, though, and a nice proof. Mar 5 '14 at 22:49

## $$2 \pi > 6$$

   

• And similarly one proves that $\pi < 4$ by inscribing a circle in a square. Nov 16 '10 at 21:46
• 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. Mar 12 '11 at 22:07
• jstor.org/stable/10.5951/mathteacher.105.8.0632 Dec 2 '12 at 8:48
• Helpful to remember that pi is ratio of circumference of the circle to diameter. Dec 21 '16 at 15:54
• Another case for $\tau$. Jul 25 '18 at 16:53

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 (Wayback Machine)

• 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! 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. Feb 25 '13 at 22:56
• 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? Apr 19 '14 at 16:04
• For completeness, since a link is still missing: Mariano Suárez-Álvarez has given a beautiful improved version of the second image here: math.stackexchange.com/q/61483 Feb 24 '18 at 13:20
• I didn't hear about "Means inequalities" until today and I don't know what this is useful for, but I have re-created the first image in GeoGebra, so you can drag around the point E to get a better feel for the line lengths: geogebra.org/classic/ndrfsstq Feb 16 '20 at 2:26

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

• 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. Dec 14 '09 at 20:58
• 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. Dec 15 '09 at 3:23
• @HB: Um, Thomas Jefferson? 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... 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. 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.

• There's probably a nice three-dimensional rendition of this that doesn't require writing down all those numbers. Aug 16 '12 at 23:03
• +1 Superb. Is this original? If not, to whom is it attributed? Nov 22 '13 at 2:05
• @MichaelLugo the 3D rendition is elsewhere on this page mathoverflow.net/q/8851 (you commented on it), but I do prefer this version.
Jan 12 '16 at 16:21

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

and the reverse animation

• Mariano, thanks for fixing up my post! Dec 16 '09 at 13:56
• I... don't quite get it. I think I need a few more words: What's the dot representing in each picture? Dec 16 '09 at 15:01
• 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. Dec 16 '09 at 15:34

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

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.

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

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).

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

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 $n=3$:

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.)

• 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. Nov 11 '10 at 3:41

from Steven Strogatz's column: http://opinionator.blogs.nytimes.com/2010/04/04/take-it-to-the-limit/ (Wayback Machine)

• 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. Nov 8 '10 at 11:27
• 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... Nov 10 '10 at 6:21
• 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. Nov 10 '10 at 6:21
• Wikipedia attributes this proof to Leonardo da Vinci. You can make establish rigorous convergence by using triangles that inscribe and circumscribe the wedges. Nov 11 '10 at 3:04
• 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.

• It looks like your image is no longer available... Nov 16 '10 at 19:09
• 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. Jul 8 '11 at 21:00
• 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. Jul 10 '11 at 15:41
• @StevenGubkin If you still have it, could you put the picture back in? No one can see it. Oct 20 '15 at 16:05
• This post is very much related to this. Nov 17 '16 at 0:46

The sequence of pictures

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

• Thomas Harriot first proved this formula in 1603, apparently by a similar argument, though I have not seen his picture(s). 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 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. Apr 26 '10 at 20:57
• 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/… May 18 '10 at 19:04
• This same proof also appears at the very opening of this paper: arxiv.org/abs/1301.0352 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.

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

Sphere eversion

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

• As pretty as it is, that is nowhere understandable as a proof. More as an illustration. Mar 11 '10 at 16:44
• @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. Apr 6 '10 at 16:30
• I just watched the video, which was excellent, but it had a lot of words in it. Aug 19 '12 at 0:23
• The original picture has better image quality: andreghenriques.com/PDF/Eversion.pdf May 21 '21 at 23:00

Late to the party, but David Lehavi and Bob Palais both mentioned the proof that $\pi_1(SO(3))$ has an element of order 2. In fact it is the only nontrivial element, and so the double cover of $SO(3)$ is simply connected.

Here's an animated illustration of that fact, courtesy of Wikipedia (here):

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

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

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

• 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. Sep 14 '10 at 8:47
• 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. Sep 14 '10 at 9:22
• 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. Nov 16 '10 at 21:51
• 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. Apr 10 '11 at 20:18
• Contrary to gowers, I don't think it's clear that we are defining a sequence of continuous functions which converges uniformly. We aren't defining a sequence of functions at all, only a sequence of images of the interval under a function. If we choose the "wrong" sequence of parameterisations of this sequence of curves then we do not get a limit function. The sequence of parameterisations (which is not illustrated) is crucial to proving the existence of the limit object: without some indication of which parameterisations we must choose there is no proof. Jun 26 '14 at 7:45

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

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.

• It needed quite a long time for me to understand this. But, well, then it is amazing! Oct 28 '15 at 10:16
• 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. Nov 17 '16 at 2:57

This proof-without-words of the Pythagorean Theorem is far from a new one, but it's the first one I've ever seen 'in the wild' (this photo was snapped after finishing dinner at a Mongolian Grill restaurant):

• I imagine people from the other tables, watching somebody while taking a picture of an empty plate! Dec 4 '16 at 7:31

As you probably already know — there are lots 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...) Dec 14 '09 at 6:14

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

• Maybe it doesn't fit into the "non-trivial" category? 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 Mar 19 '14 at 19:02