Beautiful, simple proofs worthy of writing on this beautiful glass door [closed]

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...

• Proof of Pythagoras theorem by dividing a right triangle by an altitude using similar triangles? Commented Aug 27, 2014 at 15:46
• I think perhaps you meant "aesthetically" and not "ascetically."
– Bey
Commented Aug 27, 2014 at 16:03
• I think you should post a picture of the Brad Pitt door so we know what we're working with. =) Commented Aug 27, 2014 at 16:09
• Remember to leave a tiny margin near the edge of the door for future proofs of significant number theory theorems. Commented Aug 27, 2014 at 16:16
• While this Question has garnered an enthusiastic response, it seems appropriate for Community Wiki treatment (since multiple "right" answers are invited). Commented Aug 27, 2014 at 21:21

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

• Well this one's actually much prettier with all the colors and everything (would fit perfectly on that door next to Brad Pitt). Commented Aug 27, 2014 at 16:01
• It's also slightly more accessible for those less mathematically inclined than Euler's Identity (though I certainly like that better). Commented Aug 27, 2014 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. Commented Aug 27, 2014 at 16:09
• I actually like this proof better than the other. Commented Aug 27, 2014 at 16:10

$\qquad\qquad\qquad\qquad$

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

$\qquad\qquad\qquad\qquad\qquad\qquad\quad$

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

• Wow this is amazing! Commented Aug 27, 2014 at 17:29
• Now try drawing the proof for $(a+b)^4$ :-) Commented Aug 27, 2014 at 18:59
• This is very good. If I need to explain the binomial theorem to an alien, I shall use this! :) Commented Aug 27, 2014 at 21:18

Cosines and Sines Around the Unit Circle

Trigonometric Angle Sum and Difference

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

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.

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?

More details at Monty Hall problem wikipage.

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.

• 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?) Commented Aug 27, 2014 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. Commented Aug 27, 2014 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. Commented Aug 27, 2014 at 17:35

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.

Here are a few visual proofs that

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

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

• Don't forget to follow the links to their posts, the original posts should have priority for votes. Commented Aug 27, 2014 at 20:59

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 $e^{i\pi}=-1$.

• 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. Commented Aug 28, 2014 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 $p=2r$

$2q^2=4r^2$

$q^2=2r^2$

$q^2$ 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.

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

Calculus

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*}

Geometry

The Pythagorean Theorem.

Algebra

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

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: $$\left(\sqrt2^{\sqrt2}\right)^{\sqrt2}=\sqrt2^{\sqrt2\sqrt2}=\sqrt2^2=2.\qquad\square$$

Combinatorics

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

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.

• Proof that $22/7$ exceeds $\pi$ is really a nice one. Commented Aug 27, 2014 at 23:10

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"!

A still image from the top-voted entry at https://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).

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

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

Cheers!

• Already suggested about 3 hours ago (math.stackexchange.com/a/911068/131263) Commented Aug 27, 2014 at 18:58
• Not in matrix form I'm afraid @barakmanos. Commented Aug 27, 2014 at 19:01
• Oh... got it :) Commented Aug 27, 2014 at 19:04
• But the matrix form is just a special case of continuous functional calculus... Commented Aug 27, 2014 at 20:48
• And? This is a door. Commented Aug 27, 2014 at 23:16

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)!

• 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! Commented Aug 27, 2014 at 17:40
• Oops sorry you are right I will add an edit right away Commented Aug 27, 2014 at 18:03