# Lesser known derivations of well-known formulas and theorems

What are some lesser known derivations of well-known formulas and theorems?

I ask because I recently found a new way to derive the quadratic formula which didn't involve completing the square as is commonly taught. Doing so I was wondering what other proofs and derivations for other formulas that have remained unknown to most people? Whether it be because the proof is too complex or less pretty, I still find it insightful to see different ways to solve a problem. To me it makes me understand proofs better, and thus also giving a better comprehension of it.

\begin{align} &\text{Given a quadratic function } f:\\[0.1em] f &=ax^2+bx+c = a(x-r_1)(x-r_2) = ax^2-a(r_1+r_2)+ar_1r_2\\[0.1em] a &= a,\enspace \frac{b}{a} = -(r_1+r_2),\enspace \frac{c}{a} = r_1r_2\\[1em] f' &= 2ax+b, \enspace f'(x) = 0 \Rightarrow x = -\frac{b}{2a}\\[0.2em] \text{This is an}& \text{ extremum of } f \text{, and is equidistant from each root } r_1, \enspace r_2 \text{ as shown:}\\[0.4em] \frac{b}{a} &= -(r_1+r_2) \iff -\frac{b}{2a} = \frac{r_1+r_2}{2} \\[1em] \Rightarrow \enspace &\text{The roots are of the form } r= -\frac{b}{2a}\pm d\\[1em] \frac{c}{a} = r_1r_2 &= (-\frac{b}{2a}+d)(-\frac{b}{2a}-d) = \frac{b^2}{4a^2}-d^2\\[0.2em] \Rightarrow\enspace& d^2 = \frac{b^2}{4a^2}-\frac{c}{a} = \frac{b^2}{4a^2}-\frac{4ac}{4a^2} = \frac{b^2-4ac}{4a^2}\\[0.2em] \Rightarrow\enspace&d = \pm\frac{\sqrt{b^2-4ac}}{2a}\\[1em] \text{Which yields }& r = -\frac{b}{2a}\pm d = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\enspace\square \end{align}

• I really like this derivation personally. Its very nicely geometrically motivated Nov 11, 2014 at 22:43
• You may find these 100 proofs of Pythagorean Theorem interesting
– user170039
Nov 14, 2014 at 12:20
• I love this derivation, but it's worth noting that it's basically completing the square in disguise: shifting your extremum to zero is exactly equivalent to putting the quadratic in the form $s^2-t$, where $s=x+(b/2a)$. Nov 15, 2014 at 0:22

The part of Wilson's theorem which states that $(p-1)!\equiv -1 {\mod p}$ for any prime $p$ is normally proved by grouping elements in the product $(p-1)!$ with their inverses, but it also admits a proof using Sylow theory by first showing that there are $(p-2)!$ Sylow $p$-subgroups of $S_p$. To do this, observe first that a Sylow $p$-subgroup of $S_p$ is generated by a permutation of $p$ objects, of order $p$. There are $(p-1)!$ of these, and each subgroup is isomorphic to a cyclic group with $p$ elements, which has $p-1$ generators.

By the third Sylow theorem, we then have $(p-2)!\equiv 1 {\mod p}$ and the result follows.

In 1955, Hillel Furstenberg gave a topological proof for the fact that there are infinitely many prime numbers which derives a contradiction by observing that if there were only finitely many, a certain set would be closed that from topological arguments is known not to be closed.

The full proof can be found here.

• I wouldn't say that's a topological proof; there isn't any actual topology used besides the definition of a topology on a space (and the definition is nothing more than convenient nomenclature or bookkeeping). It's still a cool proof, though. Nov 15, 2014 at 0:17
• Here is a version of Fuerstenberg's proof: We are arguing about periodic sets of integers. The set $N_p$ of all integers prime to $p$ is periodic, and the intersection of two periodic sets is periodic. If there were only finitely many primes the set $\{1,-1\}$ would be periodic. Nov 21, 2014 at 12:37

Here's a proof due to Zagier that every prime $p = 4n + 1$ is the sum of two squares: On the finite set $A = \{(x, y, z)\in \mathbb{Z}^{\geq 0}:\, p = x^2 + 4yz\}$, define $$f(x, y, z) = \begin{cases} (x + 2z, z, y - x - z) & \text{if x < y - z}; \\ (2y - x, y, x - y + z) & \text{if y - z < x < 2y}; \\ (x - 2y, x -y + z, y) & \text{if x > 2y}. \end{cases}$$ The function $f$ has exactly one fixed point: $(1, 1, n)$. Any involution $g$ of $A$ (i.e., function with $g = g^{-1}$) must have a number of fixed points $F_g$ equal to $\#A \pmod{2}$, since $\frac{1}{2}\left(F_g + \#A\right)$ is an integer by Burnside's theorem. (Specifically, it's the number of orbits of $A$ under $g$). Thus the involution $(x, y, z) \to (x, z, y)$ must have also have an odd number of fixed points; in particular, there must exist at least one fixed point $(a, b, b)\in A$. Thus $p = a^2 + (2b)^2$, as required.

• This is Zagier's famous "one-sentence proof."
– Mike
Nov 15, 2014 at 1:29
• I elaborated on some of the details that would ordinarily be omitted in a research paper, but there is something to be said for a one-sentence proof of a nontrivial result. Nov 15, 2014 at 1:31

The Cantor-Bernstein theorem is most commonly proved using an argument that involves the integers. In fact, there is a simple proof based on an elementary fixed point lemma.

The uniform boundedness principle states that the pointwise boundedness of a collection of continuous linear operators on a Banach space implies uniform boundedness (with respect to the operator norm).

While the most preferred modern proof—relying on Baire's category theorem—is simple and elegant, there is actually an older proof based on a more transparent, more elementary, but also more tedious inductive argument.

Namely, if a collection of continuous linear operators fails to be bounded with respect to the operator norm, then one can find a point at which pointwise boundedness fails, too. The proof involves constructing a sequence of operators that exhibit ever larger values at certain points, and then “pushing” these outliers to infinity to obtain a point at which pointwise boundedness breaks down. Accordingly, this older proof is coined the technique of the “gliding hump.” See Megginson (1998, p. 49) for more details.

The proof in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563 that $(1+1/n)^n \to e$

which uses the arithmetic-geometric mean inequality (AGMI), which we will use in the form $((v_1+v_2+...v_n)/n)^n > v_1v_2...v_n$ (all $v_i$ positive) with equality if and only if all the $v_i$ are equal.

Consider $n$ values of $1+1/n$ and $1$ value of $1$. By the AGMI, $((n+2)/(n+1))^{n+1} > (1+1/n)^n$, or $(1+1/(n+1))^{n+1} > (1+1/n)^n$.

Consider $n$ values of $1-1/n$ and $1$ value of $1$. By the AGMI, $(n/(n+1))^{n+1} > (1-1/n)^n$ or $(1+1/n)^{n+1} < (1+1/(n-1))^n$.

Therefore $(1+1/n)^n$ is increasing and $(1+1/n)^{n+1}$ is decreasing.

Since $0 < (1+1/n)^{n+1}-(1+1/n)^n = (1+1/n)^{n}/n <4/n$, the two sequences converges to a common limit.

To show that $(1+1/n)^{n} < 4$, note that $(1+1/n)^{n} <(1+1/n)^{n+1} \le 2^2$ (with $n=1) =4$,

Pythagorean Theorem

When i was in school, I struggled hard to understand and remember the proof of Pythagorean Theorem. But when i came across the below proof I was surprised a lot. ( How simple it is!!!)

Let's say a triangle has three sides a, b, c. Take four such triangles.

Three of these have been rotated 90°, 180°, and 270°, respectively. Each has area ab/2. Let's put them together without additional rotations so that they form a square with side c.

The square has a square hole with the side (a - b). Summing up its area (a - b)² and 2ab, the area of the four triangles (4·ab/2), we get

c² = (a - b)² + 4.ab/2
= a² - 2ab + b² + 2ab
= a² + b²

Here is an animation of the same proof from wikipedia:

Recall the Quadratic Reciprocity Law which states that for $p,q \in \mathbb{Z}$ prime with $p \neq q$: $$\left(\frac{p}{q}\right) \left(\frac{q}{p}\right) = (-1)^{\frac{p-1}{2} \frac{q-1}{2}}.$$ Where $\left(\frac{p}{q}\right)$ is the Legendre symbol of $p$ with respect to $q$.

When I was first shown this theorem it was proved with Eisentein's method. But this law has a lot of proofs. Here is a list of 246 proofs of the Quadratic Reciprocity Law, hosted by the University of Heidelburg. Notably, both Gauss and Kronecker each have 8 different proofs.

Brahmagupta's Formula which finds the area for a cyclic quadrilateral based on side lengths.

The formula is shown below where K is the area and $$a, b, c$$, and $$d$$ are the side lengths.

$$K=\frac{1}{4} \sqrt{(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)}$$

Here is a proof of the formula.