# 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"... 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! Jul 9 '11 at 12:11
• My opinion is that almost every proof-without-words is improved by a few well-chosen words. Feb 12 '12 at 0:47
• @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. Jan 23 '15 at 3:55
• If you cannot tell the difference between a proof-tree and a proof without words in the tradition of, say, the AMM Monthly, then that is clearly a limitation of yours. I would rather you start a meta thread, or a blog, instead of further polluting this thread with what is clearly rather orthogonal chatter. Jan 23 '15 at 22:52

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

• How is that "without words"? :-/ Mar 5 '10 at 17:23
• If you observe carefully on the graphs, you don't even need to write a word. Mar 5 '10 at 19:15

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?

• Are there more details on 1? 2 and 3 don't seem like proofs so much as examples, though maybe you are just putting them forth as challenges.
– j.c.
Mar 7 '10 at 11:54
• The link to the video seems to be dead. (OTOH, it was probably not that relevant for the topic here.) Dec 23 '20 at 8:41

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! 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.) May 4 '11 at 0:13
• I don't think drvitek's proof makes sense. Jun 28 '11 at 14:55
• +1 It is quite a nice idea, if you actually do the drawing.
– user22882
Jun 17 '14 at 10:23

Proof of the lantern relation (taken from the book: A Primer on Mapping Class Groups by Farb, B. and Margalit, D.)

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.... 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. Nov 30 '17 at 0:32

Identity:

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

• There's a similar figure for $(a+b)^2 = (a-b)^2 + 4ab$; but unfortunately there appears to be no similarly simple figure for $(a+b+c)^3 = (a+b-c)^3 + (a+c-b)^3 + (b+c-a)^3 + 24abc$, although there are some nice ideas for figures with some overlapping regions: mathoverflow.net/q/306394/88133. Dec 6 '19 at 15:33
• it would be simpler and more clear with all squares sharing a diagonal Nov 2 '20 at 14:47

This is apparently not was intended, but I think it qualifies. From Principia Mathematica: the proof of 1+1=2.

• Which image? The second link is dead and the first has no proof-like images. Nov 15 '16 at 23:26
• en.wikipedia.org/wiki/File:Principia_Mathematica_54-43.png Dec 4 '16 at 4:46
• I think you are referring to proof without English words... But I believe these statements are equivalent to English sentences. But still, it's a pleasure to read this. Dec 5 '19 at 11:53

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. Mar 11 '10 at 17:04

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... Jul 6 '11 at 1:01
• So are pictures. That's my point. 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. 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. Mar 19 '14 at 12:36

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

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

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... Sep 16 '11 at 17:27
• +1: Thanks for this wonderful and beautiful link (be it in the spirit of the question or not). Sep 16 '11 at 18:03

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

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. 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.) Sep 5 '18 at 21:27

The composition of two continuous mappings is continuous.

• Can you explain that? Feb 4 '19 at 19:06

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. Feb 15 '12 at 15:05
• I find it just as visual, and very clean, but even harder to depict. Feb 15 '12 at 21:05

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 $f(\pi/2)=1$. 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. Jun 25 '15 at 18:53
• However, this could be a nice proof of $\int_{0}^{\pi/2}\sin^2 x\,\mathrm{d}x = \int_{0}^{\pi/2}\cos^2 x\,\mathrm{d}x = \frac{1}{2}\left(\frac{\pi}{2}\cdot 1\right)$ 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. 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 $~a^2 + b^2 + c^2 = ab + bc + ca$:



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