# 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)

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where possible could people also either note the image source or explain/provide a link to a "how to" for constructing the associated diagram? I think that such would also be helpful for folks` – Carter Tazio Schonwald Dec 14 '09 at 23:57
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 aka Rod Vance 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

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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Formulas (first order or what not) are just words written in abbreviated form. That really does not count... – Mariano Suárez-Alvarez 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-Alvarez 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. – Turion Mar 19 '14 at 12:36

Here http://www.maa.org/sites/default/files/269122948517.pdf 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.

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

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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

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

http://web.mac.com/xgviennot/Cont_Science/vid%C3%A9os.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/

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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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This is not quite in the spirit of the question... – Mariano Suárez-Alvarez 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 Stricker Sep 16 '11 at 18:03

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

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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

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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## protected by Scott Morrison♦Oct 11 '13 at 0:51

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