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What are some of the more beautiful proofs you know? I am measuring beauty in two dimensions -- first, how conceptually elegant is it and second, how aesthetically pleasing is it.

Context: I work at a econ consulting firm. We're mostly math majors or very quantitative econ majors. A buddy and I are trying to decide what to write on the glass door to the office we share. Currently it has a graph of quality of Brad Pitt's movies against how frequently he was shirtless in that movie. Time to upgrade that...

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closed as primarily opinion-based by Ayman Hourieh, Jack M, Jonas Meyer, Siméon, anorton Aug 27 '14 at 21:48

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise.If this question can be reworded to fit the rules in the help center, please edit the question.

Proof of Pythagoras theorem by dividing a right triangle by an altitude using similar triangles? –  Assaultous2 Aug 27 '14 at 15:46
I think perhaps you meant "aesthetically" and not "ascetically." –  Bey Aug 27 '14 at 16:03
I think you should post a picture of the Brad Pitt door so we know what we're working with. =) –  Vincent Aug 27 '14 at 16:09
Remember to leave a tiny margin near the edge of the door for future proofs of significant number theory theorems. –  fixedp Aug 27 '14 at 16:16
While this Question has garnered an enthusiastic response, it seems appropriate for Community Wiki treatment (since multiple "right" answers are invited). –  hardmath Aug 27 '14 at 21:21

16 Answers 16

up vote 13 down vote accepted

Barak beat me to my #1 choice. This would be second:

enter image description here

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Well this one's actually much prettier with all the colors and everything (would fit perfectly on that door next to Brad Pitt). –  barak manos Aug 27 '14 at 16:01
It's also slightly more accessible for those less mathematically inclined than Euler's Identity (though I certainly like that better). –  Vincent Aug 27 '14 at 16:04
Well check this out: math.stackexchange.com/a/728239/131263... It's an answer to a similar question (well, more or less identical as a matter of fact). Back then I answered with the answer given here by @paw88789. –  barak manos Aug 27 '14 at 16:09
I actually like this proof better than the other. –  Quang Hoang Aug 27 '14 at 16:10


$\qquad\qquad\qquad\qquad\quad$ Geometric Explanation of the Binomial Theorem


$\qquad\qquad\qquad\qquad\qquad\qquad$ Proof that $~\displaystyle\sum_{k=1}^n(2k-1)=n^2$

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Wow this is amazing! –  zerosofthezeta Aug 27 '14 at 17:29
Now try drawing the proof for $(a+b)^4$ :-) –  Nate Eldredge Aug 27 '14 at 18:59
This is very good. If I need to explain the binomial theorem to an alien, I shall use this! :) –  Rusan Kax Aug 27 '14 at 21:18

Cosines and Sines Around the Unit Circle

Unit Circle Angles

Trigonometric Angle Sum and Difference

Trig Angle Sum and Difference

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I'm fond of Euclid's proof of the infinitude of primes: For any finite set $S=\{p_1, p_2,\dots, p_k\}$ of prime numbers, let $N=p_1\cdot p_2\cdot\cdots\cdot p_k+1$. Then $N$ isn't divisible by any prime in $S$. Hence it is divisible by some other prime. Hence the set $S$ does not include all primes. Thus there must be infinitely many primes.

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For an esoteric accompaniment to that, one could write down Euler's product formula of the Riemann zeta function nearby: $$\sum_{n=1}^\infty \frac{1}{n^s} = \prod_{p\in\,\text{primes}} \frac{1}{1-p^{-s}}.$$ (If there were finitely many primes, what would be true of the harmonic series?) –  Semiclassical Aug 27 '14 at 16:04
For clarity's sake, it's also possibly that N itself is prime... but then it's also not in S, hence etc. etc. –  Dancrumb Aug 27 '14 at 16:53
@Dancrumb: Even if $N$ is prime, it is divisible by itself, and so I think the above is still ok even in that case. –  paw88789 Aug 27 '14 at 17:35

For me, it's Conway's inverse proof of the Morley equilateral triangle:

enter image description here

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I tried to find problems from different areas. My five suggestions are.

Sophomore' dream. The formula for the problem is: $$\begin{align}\int_0^1 x^{-x}\,dx &= \sum_{n=1}^\infty n^{-n}\end{align}$$ You can find facts about the problem and the proof of it at Sophomore's dream wikipedia article.

Bretschneider's formula. This is an expression for the area of a general convex quadrilateral. $$K = \sqrt {(s-a)(s-b)(s-c)(s-d) - abcd \cdot \cos^2 \left(\frac{\alpha + \gamma}{2}\right)}$$ It is the generalization of Brahmagupta theorem and Henon's formula. You can find the proof at Bretschneider's formula wiki article.

Feuerbach's circle. It is a circle that can be constructed for any given triangle.

enter image description here

It is also named nine-point circle because it passes through nine significant concyclic points defined from the triangle. Find more at Nine-point circle wiki article.

Taxicab numbers. If you want a funny story and numbers on the door.

I remember once going to see him when he was lying ill at Putney. I had ridden in taxi-cab No. 1729, and remarked that the number seemed to be rather a dull one, and that I hoped it was not an unfavourable omen. "No", he replied, "it is a very interesting number; it is the smallest number expressible as the sum of two [positive] cubes in two different ways.

$$1729 = 1^3 + 12^3 = 9^3 + 10^3$$ More details about the story and list of numbers at Taxicab number and 1729 wiki articles. Also Ramanujan's wikipedia page could be interesting.

Monty Hall problem. Or a door-within-doors.

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

enter image description here

More details at Monty Hall problem wikipage.

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Proof of Euler's Identity: $$e^{\pi{i}}+1=0$$

BTW, your question is more or less a copy of "Simple" beautiful math proof, so you might wanna check it out too. There's some great colorful stuff there, my answer being somewhere in the middle.

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The proof that $\frac{22}{7} > \pi$.

$$ \begin {align*} 0 &< \displaystyle\int_0^1 \frac {x^4 \left( 1 - x \right)^4}{1 + x^2} \, \mathrm{d}x \\&= \displaystyle\int_0^1 \frac {x^4 - 4x^5 + 6x^6 - 4x^7 + x^8}{1 + x^2} \, \mathrm{d}x \\&= \frac {22}{7} - \pi. \end {align*} $$


The Pythagorean Theorem.



Proof that $ \displaystyle\sum_{k=1}^{n} k^3 = \left( \displaystyle\sum_{k=1}^{n} k \right)^2 $: a proof without words.

Proof that the sum of the cubes is the square of the sum

Number Theory

  1. Deriving Binet's Formula.
  2. Finding two irrationals $x,y$ such that $x^y$ is rational. If $x=y=\sqrt2$ is an example, then we are done; otherwise $\sqrt2^{\sqrt2}$ is irrational, in which case taking $x=\sqrt2^{\sqrt2}$ and $y=\sqrt2$ gives us: $$\left(\sqrt2^{\sqrt2}\right)^{\sqrt2}=\sqrt2^{\sqrt2\sqrt2}=\sqrt2^2=2.\qquad\square$$


Binomial coefficients equal alternating sum of squares $-$ see leonbloy's answer.

Visual proof

On the left, you have the alternating sum as an inclusion-exclusion of squares: the total sum is the number of coloured cells.

On the right, you have those L shaped shapes rearranged in the top left of a 6x6 grid. If you think of each cell as a coordinate $(x_1,x_2)$ that gives two elements chosen from the set $\{1, 2 \cdots 6\}$, it's seen that the elements are choosen with $ x_2 > x_1$, what corresponds to a combination (no repetition, and no order).

The others are well known, but, just for the sake of completeness...

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

As a side note, this link is excellent if you want to find your own and decide if proofs you see are actually nice.

Also, if you want to see a list of awesome proofs without words, see here.

Visual proof 2

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Proof that $22/7$ exceeds $\pi$ is really a nice one. –  user153012 Aug 27 '14 at 23:10

The rudimentary differential equation proof of Euler's formula in the complex plane $e^{i \pi}=-1$, where $i=\sqrt{-1}$. First, via $\frac{d}{d\theta}$,

$$e^{i\theta}=f(\theta)+ig(\theta) \implies ie^{i\theta}=f^{\prime}(\theta)+ig^{\prime}(\theta)=if(\theta)-g(\theta).$$

Comparing real and imaginary parts, $f(\theta)=g^{\prime}(\theta)$ and $f^{\prime}(\theta)=-g(\theta)$ which implies

$$f^{\prime \prime}(\theta)+f(\theta)=0 \implies f(\theta)=\cos(\theta),\: g(\theta)=\sin(\theta).$$

Evaluating at $\theta=\pi$, gives $e^{i\pi}=-1$.

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If you are going to make the assumption that $\frac{d}{d\theta} e^{i\theta} = ie^{i\theta}$ then Euler's result follows immediately from the fact that multiplication by $i$ is a quarter angle rotation in the complex plane, and motion at a quarter angle to position is a circle. –  DanielV Aug 28 '14 at 5:05

The proof for the irrationality of $\sqrt{2}$ is pretty simple and satisfying, I think. It's a very easy result to achieve, but the proof is very elegant and has some nice symmetry.

Assume $\sqrt{2} = \frac{p}{q}$ with p and q relatively prime (totally simplified).

$2q^2 = p^2$

$p^2$ is even

the square of an odd number is odd, so $p$ must be even. Let $p=2r$



$q^2$ is even

the square of an odd number is odd, so $q$ must be even

contradiction: $p$ and $q$ are both even, so they are not relatively prime. $\sqrt{2}$ must be irrational.

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Here are a few visual proofs that

$$\text{arctan}(1) + \text{arctan}(2) + \text{arctan}(3) = \pi$$

One by user KennyTM:

Visual Proof 1

More by user dldarek:

Visual Proof 2

I think the lattice nature of the proofs would look nice on a door.

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Don't forget to follow the links to their posts, the original posts should have priority for votes. –  DanielV Aug 27 '14 at 20:59

Some suggestions:

$1$. The proof for the Gaussian integral

$$\int_{-\infty}^{\infty}e^{-x^2} \mathrm dx=\sqrt{\pi}$$

$2$. The proof for Euler's solution to the Basel problem

$$\frac {\ \ \pi^2}6=\sum_{n=1}^{\infty}\frac 1{n^2}=\frac 1{1^2}+\frac 1{2^2}+\frac 1{3^2}+\frac 1{4^2}+\cdots+\frac 1{n^2}+\cdots$$

$3$. The proof for Wallis' product $$\frac \pi 2=\frac 21 \cdot \frac 23\cdot \frac43\cdot\frac45\cdot\frac65\cdot\frac67\cdots $$

From the above it is interesting to note how $\pi^{\frac 12}$, $\pi$ and $\pi^2$ can be computed using an integral, an infinite product, and an infinite sum respectively.

Perhaps something more relevant for a glass door would the equations written on the glass window by John Nash (Russell Crowe) in the movie "A Beautiful Mind"!

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Euler's identity in matrix form (link for proof):

$$ \color{#10a}{\large{e^{i \, \mathbf{\Pi}} + \mathbf{I} = \mathbf{0}} }$$


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Already suggested about 3 hours ago (math.stackexchange.com/a/911068/131263) –  barak manos Aug 27 '14 at 18:58
Not in matrix form I'm afraid @barakmanos. –  Dmoreno Aug 27 '14 at 19:01
Oh... got it :) –  barak manos Aug 27 '14 at 19:04
But the matrix form is just a special case of continuous functional calculus... –  Freeze_S Aug 27 '14 at 20:48
And? This is a door. –  user153012 Aug 27 '14 at 23:16

The classification of finite simple groups -- so there would finally be a single reference that could be given for this important result. ;)

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A still image from the top-voted entry at http://mathoverflow.net/questions/8846/proofs-without-words along with the equation it proves, $1+2+\cdots+(n-1)={n\choose2}$, could be good. (Note: the entry there was originally just a still. Personally I find the animation a little unpleasant, but that may just be me.)

Added later: The original version of this proof without words, which appeared in "A Discrete Look at $1+2+\cdots+n$" by Loren Larson, can be found at http://www.matem.unam.mx/~rod/teaching/mac/larson-discrete_look_gauss_series.pdf (see Figure 7 there).

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The proof of the interpolation theorem three steps which seems redundant yields an amazing result Or the proof for the gamma function at 1/2 gives pi otherwise known as (1/2)!=π

EDIT:as noted in the comments square root of pi is actually the value of of the gammq function at 1/2 which is defined for (n-1)!

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You are wrong. $\Gamma(1/2) = \sqrt{\pi}$, on the other hand $\Gamma(n)=(n-1)!$ for positive integers, so "$(1/2)!$" could be $\Gamma(3/2)=\sqrt{\pi}/2$, but we dont use factorial notation for nonintegral numbers. That is one of the reason why we introduce Gamma function! Bytheway the idea is good, it is really an amazing result! –  user153012 Aug 27 '14 at 17:40
Oops sorry you are right I will add an edit right away –  Avrham Aton Aug 27 '14 at 18:03

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