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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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32 Answers

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95

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

alt text

(Adapted from an entry I saw at Wolfram Demonstrations)

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2 
Fantastic! I'm reminded of a question I wrote for a high-school math competition way back when, which hinged on a similar diagram that counts solutions to a + b + c = n in nonnegative integers. (Imagine a third, horizontal, line.) The problem was something like to count the solutions to a + b + c = 100 with a, b, c bounded above by some numbers close to 100; I still don't know an elegant way to solve the problem other than to see it from the triangle. – Harrison Brown Dec 14 at 19:12
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Wow ! – Dinakar Muthiah Dec 19 at 22:56
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33

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

Proof without words

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1 
I suppose this picture can also be adapted to obtain the stereographic projection proof that a sphere is a manifold? – Kevin Lin Dec 14 at 23:47
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I usually use Inkscape for my vector-based needs, but this was just done with my Smartboard presentation software. – Jason Dyer Dec 15 at 14:21
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30

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

Pythagorean Theorem, picture proof

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6 
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 at 20:58
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The 20th President of the US, James Garfield, independently discovered the proof obtained by halving the right-hand diagram along a diagonal of the square of side length c. It requires you to write down an equation, though. That's my favorite proof, but mostly because of the corollary that B. Obama isn't the first geeky POTUS. – Harrison Brown Dec 15 at 3:23
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@HB: Um, Thomas Jefferson? – Pete L. Clark Mar 6 at 3:23
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29

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!):

alt text

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.

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Page 340 of Hatcher's book: math.cornell.edu/~hatcher/AT/AT.pdf – sigfpe Dec 14 at 18:27
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This is sometimes called the Eckmann-Hilton argument: en.wikipedia.org/wiki/… – Kevin Lin Dec 14 at 20:46
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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 at 20:53
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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 Lin Dec 14 at 23:50
28

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.

alt text

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

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26

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

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

(Author: Man-Keung Siu)

alt text

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2 
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 at 16:47
24

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

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

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

Here is an image I found online

fibonacci_rectangle

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fantastic ! – Martin Brandenburg Apr 17 at 23:30
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19

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

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

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19

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

alt text

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I think it is just as easy to introduce some kind of logical gap in a written proof as in a graphical one. – Steven Gubkin Mar 7 at 23:41
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@Steven: I think there is some truth to your claim, but I don't agree fully. First, we may notice that most proofs rely much more on writing than on pictures, and so mathematicians have developed a better radar for "written gaps". Second, there is a very strong sense in which written proofs may be formalized and checked by computer. Picture proofs, unless they share quite a bit of the "discrete" character of written proofs, usually are not amenable to such treatment. (And the notions of discreteness I can think of pretty much ensure that the picture proof could be turned into words.) – Pietro KC May 15 at 20:22
17

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

http://img705.imageshack.us/img705/1970/pnorm1oo.gif

and the reverse animation

http://img689.imageshack.us/img689/6233/pnorminfoo.gif

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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 at 15:01
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The red line in xy-space satisfies the given equation. The dot gives the (a,b) coordinates of the same line in ab-space. The xy- and ab-spaces are linearly dual to each other. The resulting black and red shapes represent the unit balls in respective norms. – Igor Khavkine Dec 16 at 15:34
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16

The sequence of pictures

alt text alt text alt text alt text

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

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2 
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 at 22:31
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Haha, I'm happy to see these illustrations useful to someone! I created them some years ago, mainly to crystalize what I saw in my minds eye after finding some simple proofs of this identity online. The words accompanying these images can be found at planetmath.org/encyclopedia/… Also, original MetaPost source can be obtained from this unfortunately obscure link: images.planetmath.org:8080/cache/objects/5841/src/… – Igor Khavkine Apr 26 at 20:55
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There is an analogous proof using the fact that although the hyperbolic plane has infinite area, a triply asymptotic triangle has finite area, so once you pick one of the two triply asymptotic triangles containing your triangle, you're in business. The relevant picture's in my answer posted separately (I posted it before I had the reputation to leave comments): mathoverflow.net/questions/8846/… – Vaughn Climenhaga May 18 at 19:04
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15

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

Made in inkscape

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

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

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14

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

Proof for $n=3$:

(there should be a figure here...)

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

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

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13

Sphere eversion

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12

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.

alt text

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1 
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 at 23:06
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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 at 2:35
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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 at 3:03
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10

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

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9

In an attempt to push the bar towards the non-trivial, I'll mention the proof that the boundary complex of every polytope is shellable. The proof is virtually word-free but requires an actual movie rather than a still image: imagine yourself in a spaceship, taking off in a straight line from one of the facets, away from the polytope. Every once in a while a new facet is visible to you; under assumptions of general position, this provides a shelling of the complex (obviously, you need to fly off to projective infinity and come back on the other side).

This was assumed by Euler but first proved only in 1970 by Brugesser and Mani, who said that the idea came to him in a dream. More details here (search for "shellability") or here.

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11 
Why are there so many words and so few pictures in this answer? – David Eppstein Dec 14 at 23:07
5 
Because I couldn't a way to draw this, let alone animate, in a reasonable time. I trust that the description is helpful in imagining what the actual wordless proof is. – Alon Amit Dec 15 at 5:17
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I want a video! – Emil Jan 16 at 22:45
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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.

alt text

(That picture is slightly modified from p. 221 of this book, which has the whole proof in more detail.)

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7

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

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

Proof without words:

alt text

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

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

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5

  • 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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1 
For SO(3) has order 2 element: gregegan.customer.netspace.net.au/APPLETS/21/… – sigfpe Dec 14 at 15:29
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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 at 21:21
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Why are there so many words and so few pictures in this answer? – David Eppstein Dec 14 at 23:07
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@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 at 0:00
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It doesn't feature Feynman, but here's a video of a human doing the plate trick (just after 1 minute in): youtube.com/watch?v=CYBqIRM8GiY – Harrison Brown Dec 15 at 2:42
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Have a look at this document from an MIT-instructor: http://mit.edu/18.098/book/extract2009-01-21.pdf

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Means inequalities:

alt text (Can HM be worked in? I will add it to the picture if so)

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

alt text

The second image is due to Brian Sears

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2

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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There a proof of Erd˝os-Mordell Inequality 'without words' is an impressive one. Please follow the link http://forumgeom.fau.edu/FG2007volume7/FG200711.pdf

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2

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.

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1 
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 at 16:38
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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 - http://www.youtube.com/watch?v=e6JSTkwXg90

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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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0

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:

alt text

alt text

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