## Proofs without words [closed]

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 2009 at 23:57
I hope I am not alone in being (usually) unable to appreciate "proof by picture"... – S. Sra Jul 8 2011 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 2011 at 12:11
I am actually quite fond of this question, David! I tend to make comments on answers that are not relevant, and they have a tendency to get deleted after that. – Mariano Suárez-Alvarez Sep 16 2011 at 17:34
My opinion is that almost every proof-without-words is improved by a few well-chosen words. – Joel David Hamkins Feb 12 2012 at 0:47

## closed as no longer relevant by S. Sra, Mark Meckes, Mark Sapir, Felipe Voloch, Mariano Suárez-AlvarezAug 21 at 1:54

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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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).

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The pathspace of any topological space is contractible.

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

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

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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:

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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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.

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

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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-Alvarez Oct 1 2011 at 23:45

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 2011 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 2011 at 18:03

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

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Of cause, this is not intuitive and it isn't elementary at all, but when I was asked to give the shortest proof of $\pi_1(S^1)=\mathbb{Z}$ I could imagine I answered $S^1\cong\mathbb{R}/\mathbb{Z}$.

I have to apologize if this becomes an example of misunderstanding the question, but I wanted to state that in my opinion "Proof without words" doesn't need to mean "Proof with picture".

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This proves the Minkowski version of the Pythagorean theorem:

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

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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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Another proof of the sum of the first $n$ squares, relying on the knowledge of the formula for the sum of the first $n$ numbers:

$1^2 + 2^2 + \dots + n^2 = n(n+1)(2n+1)/6$

This one has a similar flavor to the fabled proof by Gauss of the sum of the first $n$ numbers. It's a good follow up for students after Gauss's proof.

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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 :)

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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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Formulas (first order or what not) are just words written in abbreviated form. That really does not count... – Mariano Suárez-Alvarez Jul 6 2011 at 1:01
So are pictures. That's my point. – Jesko Hüttenhain Aug 9 2011 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 2011 at 17:33

(I'd post this as a comment to Mariano Suárez-Alvarez, but I've not enough rep). From a ME thread.

$$\sum_{k=1}^n (-1)^{n-k} k^2 = {n+1 \choose 2} = \sum_{k=1}^n \; k = \frac{(n+1) \; n}{2}$$

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I'm quite surprised no-one pointed out this one yet:

Theorem. The trefoil knot is knotted.

Proof.

$\square$

Some comments: a 3-colouring of a knot diagram D is a choice of one of three colours for each arc D, such that at each crossing one sees either all three colours or one single colour. Every diagram admits at least three colourings, i.e. the constant ones. We'll call nontrivial every 3-colouring in which at least two colours (and therefore all three) actually show up. It's easy to see (one theorem, more pictures!) that Reidemeister moves preserve the property of having a nontrivial 3-colouring, and that the unknot doesn't have any nontrivial colouring.

The picture shows a (nontrivial) 3-colouring of the trefoil.

EDIT: I've made explicit what "nontrivial" meant -- see comments below. Since I'm here, let me also point out that the number of 3-colourings is independent of the diagram, and is itself a knot invariant. It also happens to be a power of 3, and is related to the fundamental group of the knot complement (see Justin Robert's Knot knotes if you're interested).

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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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Algebraic manipulations in monoidal categories can also be performed in a graphical calculus. And the best part is that this is completely rigorous: a statement holds in the graphical language if and only if it holds (in the algebraic formulation). See for example Peter Selinger's "A survey of graphical languages for monoidal categories". There are many instances, for example in knot theory studied via braided categories. The following specific example comes from Joachim Kock's book "Frobenius Algebras and 2D Topological Quantum Field Theories", and proves that the comultiplication of a Frobenius algebra is cocommutative if and only if the multiplication is commutative.

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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:

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 might be trivial but integration by parts has a nice proof without words:

(Got from: Roger B. Nelsen, Proof without Words: Integration by Parts, Mathematics Magazine, Vol. 64, No. 2 (Apr., 1991), p. 130; the original link is to http://www.math.ufl.edu/~mathguy/year/S10/int_by_parts.pdf)

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@Daniel, I've turned the PDF into a PNG, and inserted the relevant part. I did keep the URL to the PDF for reference. Thanks, by the way! – Mariano Suárez-Alvarez Feb 7 2011 at 2:55
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The composition of two continuous mappings is continuous.

Bloody thing won't let me embed the image...

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from Steven Strogatz's column: http://opinionator.blogs.nytimes.com/2010/04/04/take-it-to-the-limit/

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Nice, but that reminds me of the "proof" of $2=\pi$ by approximating a straight line of length 2 by starting with a circle with this line as diameter, then two circles with one half of the line as diameter each, then for circles with on quarter of the line as diameter, ... One still has to find an argument that a geometric process converges at all and converges to the desired result. Both cannot be deduced purely from looking at a picture. – Johannes Hahn Nov 8 2010 at 11:27
Wikipedia attributes this proof to Leonardo da Vinci. You can make establish rigorous convergence by using triangles that inscribe and circumscribe the wedges. – S. Carnahan Nov 11 2010 at 3:04

A classic one, from the late 19th century, that surprized Peano's contemporaries.

Question : "A curve that fills a plane ? You must be kidding"

Well, of course a formal proof was necessary, but it is still one of my favorites.

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How can you be sure that you're eventually covering all the points with irrational or transcendental coordinates? And giving a sequence of curves which fill more and more of the plane isn't the same as giving a single curve that does it all at once - it's not clear that such a limiting curve exists just looking at the pictures. – Michael Burge Sep 14 2010 at 8:47
Existence of the limits object is something that is very often forgotten. For example most Introductions to fractals give geometric descriptions of Koch's snowflake etc. via such an iteration but don't prove that there exists a limit of this iteration. – Johannes Hahn Sep 14 2010 at 9:22
Project: Fill the square one pixel at a time by following (an approximation to) this curve; then find some suitable baroque music accompaniment; then upload it to youtube. – Michael Hardy Nov 16 2010 at 21:51
If you look at the picture in detail you can see that you are defining a sequence of continuous functions that converge uniformly. It's also clear from the picture that the image is dense. Therefore the limiting function exists and its image (being dense and compact) is the whole square. Of course, this proof isn't 100% visual but the non-visual part -- the basic facts about uniform convergence and compactness -- can be regarded as background knowledge. So I think it's a nice example. – gowers Apr 10 2011 at 20:18
Remarkably, no picture nor mention to it was made in Peano's article, the construction being completely based on ternary expansions. The picture of a sequence converging to a square-filling curve appeared one year later in the paper by Hilbert. – Pietro Majer Nov 17 2011 at 14:14
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## $$2 \pi > 6$$

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And similarly one proves that $\pi < 4$ by inscribing a circle in a square. – Michael Hardy Nov 16 2010 at 21:46
At first I was thrown off by this, because I was looking at area and not circumference. The area of an inscribed regular 12-sided polygon in the unit circle is also 3. – Todd Trimble Mar 12 2011 at 22:07
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In the movie category, I'm surprised that no-one has yet posted a link to Moebius Transformations Revealed.

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But what does that movie prove? – Mariano Suárez-Alvarez Nov 8 2010 at 3:50
@Mariano: it doesn't prove anything, but then again neither do any proofs without words. They merely give us insight into the proof, and in that respect, any movie has even more potential than a simple image. I think we will soon see very innovative approaches in movie-proofs. – Thierry Zell Nov 8 2010 at 3:59
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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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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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This proof without words has an awful lot of them! – I. J. Kennedy Apr 30 2011 at 17:39

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
conway@math.princeton.edu asoifer@princeton.edu

n2 + 2 can:

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