Question

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 1

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

(Adapted from an entry I saw at Wolfram Demonstrations)

Answer 2

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

Answer 3

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 4

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 5

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

Answer 6

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 7

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 8

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

Answer 9

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 10

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 11

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

and the reverse animation

Answer 12

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 13

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 14

$$2 \pi > 6$$

Answer 15

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). UPDATE: updated the URL for the picture.

Answer 16

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 17

The sequence of pictures

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

Answer 18

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 19

Sphere eversion

Answer 20

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 21

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

The derivative of sine is cosine.

Answer 22

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 23

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

Answer 24

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

Answer 25

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 26

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 27

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 28

Q: Can you tile

with ?

Answer 29

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

Answer 30

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.)