Proofs without words - MathOverflow [closed]

Proofs without words

2009-12-14T06:04:14Z

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)

Answer by Mariano Suárez-Alvarez for Proofs without words

2009-12-14T06:05:09Z

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

(Adapted from an entry I saw at Wolfram Demonstrations)

Answer by Mike for Proofs without words

2009-12-14T06:13:11Z

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

Answer by Mike for Proofs without words

2009-12-14T06:29:44Z

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)

Answer by David Lehavi for Proofs without words

2009-12-14T07:05:44Z

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

Answer by Alon Amit for Proofs without words

2009-12-14T07:53:45Z

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.

Answer by Alon Amit for Proofs without words

2009-12-14T08:05:58Z

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.

Answer by Steve Flammia for Proofs without words

2009-12-14T15:35:33Z

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

Answer by Jason Dyer for Proofs without words

2009-12-14T15:50:16Z

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

Answer by Harrison Brown for Proofs without words

2009-12-14T16:08:51Z

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.

Answer by Aaron Mazel-Gee for Proofs without words

2009-12-15T00:19:29Z

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.)

Answer by Steven Gubkin for Proofs without words

2009-12-15T01:31:55Z

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

The derivative of sine is cosine.

Answer by Agol for Proofs without words

2009-12-15T01:52:27Z

Sphere eversion

Answer by Darsh Ranjan for Proofs without words

2009-12-15T02:32:12Z

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.)

Answer by Igor Khavkine for Proofs without words

2009-12-16T11:32:32Z

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

and the reverse animation

Answer by Gil Kalai for Proofs without words

2009-12-19T17:45:58Z

Q: Can you tile

with ?

Answer by aorq for Proofs without words

2009-12-22T16:54:49Z

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.

Answer by vonjd for Proofs without words

2010-01-15T16:31:52Z

Have a look at this document from an MIT-instructor: http://mit.edu/18.098/book/extract2009-01-21.pdf

Answer by serargus for Proofs without words

2010-01-16T21:42:19Z

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

Answer by Marco Radeschi for Proofs without words

2010-02-22T15:25:59Z

The sequence of pictures

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

Answer by Sunni for Proofs without words

2010-03-05T16:59:16Z

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

Answer by Tobias Hagge for Proofs without words

2010-03-06T21:55:58Z

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:

Answer by shreevatsa for Proofs without words

2010-03-06T23:18:28Z

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.

Answer by Russell O'Connor for Proofs without words

2010-03-07T02:16:51Z

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

Answer by Kumar for Proofs without words

2010-03-07T11:01:10Z

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 -

Nyquist sampling theorem?
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 :)
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

Answer by Chris Conway for Proofs without words

2010-03-25T18:11:49Z

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.)

Answer by muad for Proofs without words

2010-03-25T19:00:35Z

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.

Answer by Vaughn Climenhaga for Proofs without words

2010-05-15T16:40:20Z

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.)

Answer by Dave Pritchard for Proofs without words

2010-05-15T22:32:52Z

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

Answer by Vaughn Climenhaga for Proofs without words

2010-05-20T01:27:35Z

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.

Answer by Bassam Abdul-Baki for Proofs without words

2010-06-22T02:02:17Z

http://home.comcast.net/~babdulbaki/MathProofs.html

http://home.comcast.net/~babdulbaki/PWW.html (See link at bottom.)

Answer by BlueRaja for Proofs without words

2010-06-25T02:45:08Z

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:

Answer by Franklin for Proofs without words

2010-07-11T15:11:05Z

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

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

Answer by Daniel Miller for Proofs without words

2010-07-14T22:15:56Z

There is a beautiful proof of the fact that a checkerboard with sides $2^{n}$, and one square removed can be tiled with $L$-shaped pieces formed by three squares. Given that a checkerboard of sides $2^{n-1}$ can be so tiled, then a square checkerboard of sides $2^{n}$ can be tiled by filling in the quarter in which the removed piece lies, and then placing an extra $L$-shaped tile with one square in each of the remaining three quarters.

Answer by Thierry Zell for Proofs without words

2010-08-07T03:37:33Z

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

Answer by muad for Proofs without words

2010-09-14T07:44:10Z

$$2 \pi > 6$$

Answer by Alexis Monnerot-Dumaine for Proofs without words

2010-09-14T08:24:03Z

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

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

Answer :

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

Answer by AndrewLMarshall for Proofs without words

2010-09-27T15:36:25Z

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

Answer by Kris Joanidis for Proofs without words

2010-12-28T02:42:47Z

The composition of two continuous mappings is continuous.

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

Answer by Daniel Parry for Proofs without words

2011-02-07T01:27:59Z

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 to http://www.math.ufl.edu/~mathguy/year/S10/int_by_parts.pdf)

Answer by Bob Palais for Proofs without words

2011-02-07T04:01:50Z

Here are some dynamic versions:

http://www.math.utah.edu/~palais/sums.html (two of the summation formulas mentioned above)

Several belt, plate, and tangle trick animations:

http://www.math.utah.edu/~palais/links.html

A visual derivation of complex multiplication:

http://www.math.utah.edu/~palais/newrot.swf

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

http://www.math.utah.edu/~palais/PythagorasIsosceles.html

and the general case:

http://www.math.utah.edu/~palais/Pythagoras.html

Answer by Chris Heunen for Proofs without words

2011-03-15T22:02:59Z

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.

Answer by thei for Proofs without words

2011-04-10T15:54:03Z

I suggest the videos of Viennot explaining the bijections between different families of objects counted by Catalan numbers:

http://web.mac.com/xgviennot/Cont_Science/vid%C3%A9os.html

Answer by Marco Golla for Proofs without words

2011-04-10T16:57:19Z

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).

Answer by leonbloy for Proofs without words

2011-06-28T14:42:08Z

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

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

Answer by Jesko Hüttenhain for Proofs without words

2011-06-28T16:46:28Z

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.

Answer by jkun for Proofs without words

2011-07-06T00:39:00Z

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 :)

Answer by jkun for Proofs without words

2011-07-07T23:25:37Z

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.

Answer by Phil Isett for Proofs without words

2011-07-08T21:49:24Z

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/

Answer by Ron Maimon for Proofs without words

2011-08-16T06:45:45Z

This proves the Minkowski version of the Pythagorean theorem:

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

Answer by Ben A. for Proofs without words

2011-08-16T12:23:01Z

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".

Answer by sclv for Proofs without words

2011-09-16T03:05:50Z

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

source: http://www.futilitycloset.com/2011/09/12/half-and-half/

Answer by euklid345 for Proofs without words

2011-09-16T03:19:31Z

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

Answer by isomorphismes for Proofs without words

2011-10-01T23:31:18Z

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

Answer by Martin Brandenburg for Proofs without words

2011-11-08T09:33:26Z

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.

Answer by Roberto Mizzoni for Proofs without words

2012-02-15T00:00:13Z

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: $$\binom{n}{k} = \frac{n}{(n-k)} \frac{(n-1)!}{(n-1-k)!\ k!} = \frac{n!}{(n-k)!\ k!}$$

Answer by Franklin for Proofs without words

2012-05-29T22:27:02Z

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

link text

Answer by Jon Cohen for Proofs without words

2012-08-16T22:01:48Z

The pathspace of any topological space is contractible.

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

Answer by Marc Chamberland for Proofs without words

2012-08-20T22:39:22Z

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).

Answer by Marc Chamberland for Proofs without words

2012-08-21T01:28:28Z

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 $~a^2 + b^2 + c^2 = ab + bc + ca$:

$$ $$

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