# 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)

• 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 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
• @goblin, I am afraid that you have completely misunderstood the concept. The idea is pictures which have the rather amazing capability of immediately suggesting on the mind of the viewer the idea of a proof. How on earth you managed to get from the rather well-known idea involved in this question to «proofs without logic» is a mystery to me. – Mariano Suárez-Álvarez Jan 23 '15 at 3:55

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:

• Nelson's book was already mentioned eight years ago (but without any illustrations): mathoverflow.net/a/8849 (and was the second answer to question). – jeq Aug 3 '17 at 14:16

This image is a bijection between the puzzle rule and the semistandard tableau rule for Littlewood-Richardson coefficients. It is taken from this paper of Ravi Vakil, where it is attributed to Terry Tao.

The picture generalises to a bijection between rules in K-theoretic Schubert calculus, but I haven't seen a picture and don't currently have the patience to create one.

• too complicated.... – T.... Oct 24 '17 at 21:16

I wonder why Apostol's proof of the irrationality of $\sqrt{2}$ - which is as visual as a proof can be (in my opinion) - has not been mentioned: One can literally see at a glance that it proves what it's supposed to prove: the impossibility of a isosceles triangle with integer side length (by infinite descent):

Note that it's not a proof completely without words. It helps a lot to read the comments of the author:

Each line segment in the diagram has integer length, and the three segments with double tick marks have equal lengths. (Two of them are tangents to the circle from the same point.) Therefore the smaller isosceles right triangle with hypotenuse on the horizontal base also has integer sides.

But through own thinking one could come up with this by oneself (having in mind what's to be proved).

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.)

• I am guessing he did it by assembling four of the said right-triangles into a parallelogram. There is a path that bounces directly between the two longer sides. Mod out by the symmetry and you get a periodic path in the triangle. – Willie Wong Mar 11 '10 at 17:04

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.

• This proof without words has an awful lot of them! – I. J. Kennedy Apr 30 '11 at 17:39
• Indeed, there is a proof with only eleven words: rearrangement and arithmetic-geometric inequalities. (Details are left to the reader.) – dvitek May 4 '11 at 0:13
• I don't think drvitek's proof makes sense. – darij grinberg Jun 28 '11 at 14:55
• +1 It is quite a nice idea, if you actually do the drawing. – rem Jun 17 '14 at 10:23

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.

• Formulas (first order or what not) are just words written in abbreviated form. That really does not count... – Mariano Suárez-Álvarez Jul 6 '11 at 1:01
• So are pictures. That's my point. – Jesko Hüttenhain Aug 9 '11 at 17:35
• I doubt there is any sense in which one can formalize the notion, but I think it is pretty clear that a proof written in the first order calculus, or any other calculus, is simply not a "proof without words". You do not believe in the question, you say, but I honestly cannot understand what that can possibly mean: there is certainly something that gets the name proof-without-words (there is even a section in the MAA Monthly dedicated exclusively to this, and it has run for decades!) and most people ---while probably not being able to explain exactly what they are--- recognize them. – Mariano Suárez-Álvarez Sep 16 '11 at 17:33
• On the contrary, I would say that any picture that is a rigorous proof must first be formalised in some sense, and will then most probably be in words of some form. – Manuel Bärenz Mar 19 '14 at 12:36

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

• Link does not work. – ThiKu May 14 '15 at 8:57
• You are right, I'll delete the link until I hopefully find working one, thanks. @ThiKu – VividD May 14 '15 at 9:00
• This isn't a Fibonacci identity per se; for all $a$ and $b$, $(a+b)^3 = a^3+b^3+3ab(a+b)$. – Steven Stadnicki May 16 '15 at 0:40

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).

• You miss your graph.... – Hu xiyu Nov 29 '17 at 17:55
• That's one way of establishing $ab \leq \frac{a^p}{p} + \frac{b^q}{q}$, but I find this easier to grok via convexity of the exponential function. – Todd Trimble Nov 30 '17 at 0:32

Identity:

$$(a+b)^2=a^2+b^2+2ab$$

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:

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

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

• This is not quite in the spirit of the question... – Mariano Suárez-Álvarez Sep 16 '11 at 17:27
• +1: Thanks for this wonderful and beautiful link (be it in the spirit of the question or not). – Hans-Peter Stricker Sep 16 '11 at 18:03

The composition of two continuous mappings is continuous.

• Can you explain that? – M. Winter Feb 4 at 19:06

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/

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

• This answer has already been proposed, and after some discussion it was more or less agreed that this is not a proof-without-words in standard sense of the term. – Mariano Suárez-Álvarez Oct 1 '11 at 23:45
• It's not a proof without words but a visualization (i.e. statement without words) of the Yoneda Lemma. (But it depends: for the experts it may count as a proof.) – Hans-Peter Stricker Sep 5 '18 at 21:27

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

• For me the standard argument is just as visual, and clearer: I have n!/(n-k)! ordered lists of length k with no repetitions from an alphabet of n letters. If I group together all words using the same letters, there are k! members of each group, hence n!/(n-k)!(k!) groups. Each group corresponds to an unordered list. – Steven Gubkin Feb 15 '12 at 15:05
• I find it just as visual, and very clean, but even harder to depict. – Roberto Mizzoni Feb 15 '12 at 21:05

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

http://www.xavierviennot.org/contscience/videos.html

A nice proof for trigonometric equation:

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

• 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
• However, this could be a nice proof of $\int_{0}^{\pi/2}\sin^2 x\,\mathrm{d}x = \int_{0}^{\pi/2}\cos^2 x\,\mathrm{d}x = \frac{1}{2}\left(\frac{\pi}{2}\cdot 1\right)$ – Machinato Jul 18 '16 at 13:29
• Well this might be actually interpreted as a visualization of the fact that squaring a sine/cosine curve produces another (shifted up) sine/cosine curve, viz. $\cos^2(x)=\frac12(1+\cos(2x))$ and $\sin^2(x)=\frac12(1-\cos(2x))$; in this way it comes closer to be a proof of something. – მამუკა ჯიბლაძე Nov 12 '17 at 7:30

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



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