Proofs without words

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

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

Answer 2

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 hexagon 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 3

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
[email protected] [email protected]}

n² + 2 can:

Answer 4

(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 5

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

Proof with words: The [area of a] circle

Similar in concept to the above video:

Answer 6

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 7

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

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

Answer 8

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 9

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

Answer 10

Q: Can you tile with ?

$ $ $ $

Answer 11

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

Answer 12

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.

Answer 13

Given three mutually tangent circles in the plane, there exist exactly two circles tangent to all three.

Answer 14

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

(This is a draft of Chapter 4 of: Sanjoy Mahajan, Street-Fighting Mathematics, MIT Press 2010.)

Answer 15

Gluing two Mobius strips along their edges is a Klein bottle.

Answer 16

• 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 17

The pathspace of any topological space is contractible.

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

Answer 18

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 19

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 20

I was somehow challenged by the idea: do more abstract topics allow proofs without words? I came out with this example. Of course, it is disputable if it is really "without words", since some words of explanation should be given (here they are: vertical segments represent Banach spaces and subspaces; connecting segments between two of them represent a linear operator. Italic letters $a,b,c,d,e$ are the dimensions of the corresponding linear subspaces).

Answer 21

This proves the Minkowski version of the Pythagorean theorem:

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

Answer 22

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

Answer 23

.
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=$2(\cos \frac{3}{7}\pi+\cos \frac{5}{7}\pi+\cos \frac{1}{7}\pi).$

Answer 24

Posted this quite a while ago on math.SE as an answer to https://math.stackexchange.com/q/733754/214353; I think it has a place here too.

$\hskip7em\phantom{\ }$

To be rigorous, it maybe only proves that $f(z):=\displaystyle\lim_{n\to\infty}\left(1+\frac zn\right)^n$ is $1$ at integer multiples of $2\pi i$ (perhaps also that $f$ is $2\pi i$-periodic), but does not give any insight into why $f(z)=\text{something}^z$; still...

I've seen it much better rendered once somewhere on the web but could not find it again, so tried to reconstruct it myself

Answer 25

I've tried to read carefully through all the posts here, and I think no one has mentioned the excellent book by R.B. Nelsen which contains many more such examples. I give here an explicit citation:

Nelsen, Roger B. Proofs without words: Exercises in visual thinking. No. 1. MAA, 1993.‏

Here is an example which I have recently used:

Answer 26

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 (Wayback Machine)

Sorry for the link; new users may not use image tags.

Here's the image:

Answer 27

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 28

A crucial step in proving the triangle inequality for the $L^p$ norms is to prove Young's inequality: $ab \leq \frac{a^p}p + \frac{b^q}q$ when $a,b>0$ and $p,q > 1$ are conjugate ($\frac 1p + \frac 1q=1$). This can be proved by comparing the areas in the following picture:

An easy calculus exercise gives the blue area as $\frac{a^p}p$ and the green area as $\frac{b^q}q$, hence their sum dominates the area $ab$ of the rectangle outlined in red.

I learned this proof from Cohn's measure theory book, where it appears as an exercise (in Section 3.3).

Answer 29

Here 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.

Answer 30

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