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

Question

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...

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

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

Answer 5

Cosines and Sines Around the Unit Circle

Trigonometric Angle Sum and Difference

Answer 6

$\qquad\qquad\qquad\qquad$

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

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

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

Answer 7

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"!

Answer 8

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)!

Answer 9

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

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

Cheers!

Answer 10

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.

Answer 11

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).

Answer 12

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

Answer 13

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.

Answer 14

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

Answer 15

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...

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

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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.

Answer 16

Here are a few visual proofs that

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

One by user KennyTM:

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More by user dldarek:

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