# 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
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

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

The derivative of sine is cosine.

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It looks like your image is no longer available... – I. J. Kennedy 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. – Phil Isett Jul 8 '11 at 21:00
@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 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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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.)

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And what exactly is a proof about this? – darij grinberg Nov 10 '10 at 23:40
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

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

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

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

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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

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

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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-Alvarez Feb 7 '11 at 2:55
The same picture also gives an interesting formula for the integral of an inverse function! – Matt Noonan Jun 29 '11 at 0:57
I guess this proof works only when $f$ and $g$ are both increasing? – Greg Martin Nov 19 '15 at 19:16

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

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

n2 + 2 can:

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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.

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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
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
Where is the picture? – Patrick Da Silva 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. – KalEl Nov 7 '14 at 22:36

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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A 3D proof of a Fibonacci identity, that even includes a video:

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

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

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

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Here http://www.maa.org/sites/default/files/269122948517.pdf you can find Grace Lin's proof without Words that The Product of the Perimeter of a Triangle and Its Inradius Is Twice the Area of the Triangle (see the figure below)

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

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A nice proof for trigonometric equation:

$sin^2(x)+cos^2(x)=1$

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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

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)

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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
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
@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
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

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).

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

(Adapted from an entry I saw at Wolfram Demonstrations)

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Wow ! – Dinakar Muthiah Dec 19 '09 at 22:56
@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-Alvarez Jun 29 '10 at 13:05
Am I the only one who doesn't understand this "proof" at all? – mathreader 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. – Steven Gubkin 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). – Barry Cipra Oct 15 '11 at 2:17

Q: Can you tile with ?

 

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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

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.

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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
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
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
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
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

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

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

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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
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
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
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
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

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

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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## $$2 \pi > 6$$

   

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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
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

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

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

and the reverse animation

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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
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

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.

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Yes, this one is beautiful. – Andrés Caicedo May 15 '10 at 18:47
@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

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

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

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