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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)

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    $\begingroup$ I hope I am not alone in being (usually) unable to appreciate "proof by picture"... $\endgroup$
    – Suvrit
    Commented Jul 8, 2011 at 21:14
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    $\begingroup$ @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! $\endgroup$
    – Selene Routley
    Commented Jul 9, 2011 at 12:11
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    $\begingroup$ My opinion is that almost every proof-without-words is improved by a few well-chosen words. $\endgroup$
    – Joel David Hamkins
    Commented Feb 12, 2012 at 0:47
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    $\begingroup$ @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. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 23, 2015 at 3:55
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    $\begingroup$ 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. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 23, 2015 at 22:52

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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.

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    $\begingroup$ This proof without words has an awful lot of them! $\endgroup$
    – I. J. Kennedy
    Commented Apr 30, 2011 at 17:39
  • $\begingroup$ Indeed, there is a proof with only eleven words: rearrangement and arithmetic-geometric inequalities. (Details are left to the reader.) $\endgroup$
    – dvitek
    Commented May 4, 2011 at 0:13
  • $\begingroup$ I don't think drvitek's proof makes sense. $\endgroup$
    – darij grinberg
    Commented Jun 28, 2011 at 14:55
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    $\begingroup$ +1 It is quite a nice idea, if you actually do the drawing. $\endgroup$
    – user22882
    Commented Jun 17, 2014 at 10:23
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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.

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  • $\begingroup$ I first learned this in Dan Velleman's book "How to prove it." I'm not sure if he originated it or not. $\endgroup$
    – Jim Conant
    Commented Feb 7, 2011 at 2:36
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Proof of the lantern relation (taken from the book: A Primer on Mapping Class Groups by Farb, B. and Margalit, D.)

Proof of the lantern relation

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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.

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  • $\begingroup$ too complicated.... $\endgroup$
    – Turbo
    Commented Oct 24, 2017 at 21:16
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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

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  • $\begingroup$ How is that "without words"? :-/ $\endgroup$
    – Andrea Ferretti
    Commented Mar 5, 2010 at 17:23
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    $\begingroup$ If you observe carefully on the graphs, you don't even need to write a word. $\endgroup$
    – Sunni
    Commented Mar 5, 2010 at 19:15
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Identity:

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

enter image description here

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    $\begingroup$ 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. $\endgroup$
    – Zach Teitler
    Commented Dec 6, 2019 at 15:33
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    $\begingroup$ it would be simpler and more clear with all squares sharing a diagonal $\endgroup$
    – Pietro Majer
    Commented Nov 2, 2020 at 14:47
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This is apparently not was intended, but I think it qualifies. From Principia Mathematica: the proof of 1+1=2.

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    $\begingroup$ Which image? The second link is dead and the first has no proof-like images. $\endgroup$
    – Simply Beautiful Art
    Commented Nov 15, 2016 at 23:26
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    $\begingroup$ en.wikipedia.org/wiki/File:Principia_Mathematica_54-43.png $\endgroup$
    – Chris Conway
    Commented Dec 4, 2016 at 4:46
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    $\begingroup$ 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. $\endgroup$
    – justadzr
    Commented Dec 5, 2019 at 11:53
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I just scanned some old drawings, and found this one that fits right in.

It proves that the connected sum of two Klein bottles is homeomorphic to the connected sum of a Klein bottle and a Torus. enter image description here

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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.)

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    $\begingroup$ 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. $\endgroup$
    – Willie Wong
    Commented Mar 11, 2010 at 17:04
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There are $\binom{n+2}{4}$ equilateral triangles in the regular $n$-vertices-per-side triangular grid, by choosing $A<B<C<D$ from the set $\{1,...,n+2\}$: https://arxiv.org/abs/2211.00186

enter image description here

enter image description here

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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.

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    $\begingroup$ Formulas (first order or what not) are just words written in abbreviated form. That really does not count... $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jul 6, 2011 at 1:01
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    $\begingroup$ So are pictures. That's my point. $\endgroup$
    – Jesko Hüttenhain
    Commented Aug 9, 2011 at 17:35
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    $\begingroup$ 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. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Sep 16, 2011 at 17:33
  • $\begingroup$ 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. $\endgroup$
    – Manuel Bärenz
    Commented Mar 19, 2014 at 12:36
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A 3D proof of a Fibonacci identity, that even includes a video:

enter image description here

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  • $\begingroup$ Link does not work. $\endgroup$
    – ThiKu
    Commented May 14, 2015 at 8:57
  • $\begingroup$ You are right, I'll delete the link until I hopefully find working one, thanks. @ThiKu $\endgroup$
    – VividD
    Commented May 14, 2015 at 9:00
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    $\begingroup$ This isn't a Fibonacci identity per se; for all $a$ and $b$, $(a+b)^3 = a^3+b^3+3ab(a+b)$. $\endgroup$
    – Steven Stadnicki
    Commented May 16, 2015 at 0:40
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The picture below proves (without words) that it is possible to draw the classical Desargues configuration $D_{10}$ using points in the set $\{-3,-2,-1,0,1,2,3\}^2$. This picture was produced by Alex Ravsky as the answer to this MO-question.

enter image description here

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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

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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

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    $\begingroup$ This is not quite in the spirit of the question... $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Sep 16, 2011 at 17:27
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    $\begingroup$ +1: Thanks for this wonderful and beautiful link (be it in the spirit of the question or not). $\endgroup$
    – Hans-Peter Stricker
    Commented Sep 16, 2011 at 18:03
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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

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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/

enter image description here

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

alt text

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    $\begingroup$ 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. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Oct 1, 2011 at 23:45
  • $\begingroup$ 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.) $\endgroup$
    – Hans-Peter Stricker
    Commented Sep 5, 2018 at 21:27
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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:

alt text Exactly $n-k$ times:

alt text
By induction, a base case, and taking $k=n$ and $k=0$ for granted: $$\binom{n}{k} = \frac{n}{(n-k)} \frac{(n-1)!}{(n-1-k)!\ k!} = \frac{n!}{(n-k)!\ k!}$$

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    $\begingroup$ 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. $\endgroup$
    – Steven Gubkin
    Commented Feb 15, 2012 at 15:05
  • $\begingroup$ I find it just as visual, and very clean, but even harder to depict. $\endgroup$
    – Roberto Mizzoni
    Commented Feb 15, 2012 at 21:05
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Michael Goldberg, Constructions possible by ruler and compasses, Math. Mag. Vol. 51, No. 5, p. 283 (1 page) (Nov., 1978).

enter image description here

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alt text

The composition of two continuous mappings is continuous.

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    $\begingroup$ Can you explain that? $\endgroup$
    – M. Winter
    Commented Feb 4, 2019 at 19:06
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Here you have two issues created by me, all the points named have integer coordinates but because of practical difficulties I couldn't draw perfect figures.Can you see what mathematical propositions can be derived by extending these figures up to infinite number of circles? https://www.geogebra.org/m/j5yvst8s https://www.geogebra.org/m/vm4ffyvj [1]: https://i.sstatic.net/Be6SG.jpg

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  • $\begingroup$ If you are interested in getting a clear picture in full details you can search it on GeoGebra. I have drawn figures for two of such cases. $\endgroup$
    – Janaka Rodrigo
    Commented Aug 23, 2023 at 7:54
  • $\begingroup$ We have lots of applications related to the convergent geometric series but none was available for convergent series of angles. By carefully examining the provided figures with geometry theorems in mind it's not big challenge to realize figures can be used to evaluate the sums of infinite series of angles $$arccot(3) + arccot(7) + arccot(13)+.... $$ and $$arccot(2)+ arccot(8) + arccot(18) + .....$$ and show that the sums equal to$ π/2$. $\endgroup$
    – Janaka Rodrigo
    Commented Aug 23, 2023 at 9:59
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Countable union of countable sets is one of my all-time favorites (and surely one of the all-time best):

here

It's a picture of Cantor's pairing function.

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  • $\begingroup$ my only concern is that the link might rot $\endgroup$
    – no upstairs
    Commented Dec 7 at 12:16
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Trig proof

A nice proof for trigonometric equation:

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

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    $\begingroup$ 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 $f(\pi/2)=1$. $\endgroup$
    – Steven Stadnicki
    Commented May 16, 2015 at 0:45
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    $\begingroup$ @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. $\endgroup$
    – Mario Carneiro
    Commented Jun 25, 2015 at 18:53
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    $\begingroup$ However, this could be a nice proof of $\int_{0}^{\pi/2}\sin^2 x\,\mathrm{d}x = \int_{0}^{\pi/2}\cos^2 x\,\mathrm{d}x = \frac{1}{2}\left(\frac{\pi}{2}\cdot 1\right)$ $\endgroup$
    – Machinato
    Commented Jul 18, 2016 at 13:29
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    $\begingroup$ 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. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Nov 12, 2017 at 7:30
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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 $~a^2 + b^2 + c^2 = ab + bc + ca$:

$$ $$

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

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  • $\begingroup$ I think when you say proofs without words you need to use diagrams and all simplifications should be explained by figures. Actually the best way is to use only one figure. $\endgroup$
    – Janaka Rodrigo
    Commented Aug 23, 2023 at 10:05
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