What's your favorite proof accessible to a general audience? [closed]

Question

What math statement with proof do you find most beautiful and elegant, where such is accessible to a general audience, meaning you could state, prove, and explain it to a general audience in roughly $5 \pm\epsilon$ minutes. Let's define 'general audience' as approximately an average adult with education and experience comparable to someone holding a bachelor's degree in any non science major (e.g. history) from an average North American university.

Answer 1

I really like the proof of $$\sum_{i=1}^n i = \dfrac{n(n+1)}{2}$$ in which $1 + 2 + \cdots + (n-1) + n$ is written forwards then backwards and summed. It is claimed that Gauss had come up with this when he was just a child, although contested.

The proof

Let $$s = 1 + 2 + \cdots + (n-1) + n.$$ Clearly, $$ s = n + (n-1) + \cdots + 2 + 1.$$

Sum to get $$2s = \underbrace{(n+1) + (n+1) + \cdots + (n+1) + (n+1)}_{n \text{ times}}.$$

Hence, $$2s = n(n+1),$$ and $$s = \dfrac{n(n+1)}{2}.$$

Answer 2

I'm a bit reluctant to throw another answer on the pile, especially because I think there are other lists on this website which serve a pretty similar purpose. However, I think an excellent 5 minute blurb could be given to a general audience on the trick, attributed to von Neumann, for performing a fair coin toss when only a biased coin with unknown bias is available. There is a wikipedia entry on this. Here is an informal description:

Suppose you have a biased coin. The chance that the coin comes up heads or tails (assume both are actually possible) are unknown to you, but do not change from toss to toss. You and your friend Jane wish to use this coin to decide (fairly) which of you gets the top bunk at math camp.

Jane makes the following observation. Suppose the coin is flipped twice in a row. The possible outcomes are: \begin{align*} HH && TT && HT && TH \end{align*} Now, we do not know how likely each of these outcomes is, but one thing is certain: $$ \text{ The outcomes $HT$ and $TH$ are equally likely.}$$ Because of this observation, you agree on the following fair way to settle the dispute. You "call" the outcome $HT$ while Jane "calls" the outcome $TH$, and then proceed to flip the coin twice. If either $HH$ or $TT$ occurs, a mistrial is declared and you start over, flipping the coin another two times. Eventually, either $HT$ or $TH$ will occur, and the dispute is settled.

I think this works well because:

• It is simple -- simple enough to fit into 5 minutes.
• It requires no special knowledge from the audience. People generally have pretty reasonable built-in intuition for probability.
• It is a beautiful, but also practical, idea -- making it effective as "Math P.R."
• You can also actually demonstrate the procedure, to make sure it is understood, without any special equipment. Just take something such as a thimble which can land either of two states, but for which it is not clear that either outcome is equally likely.

Answer 3

The harmonic series diverges because otherwise there exists a finite number \begin{align*} S &= 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\dotsb \\ &= \left(1+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}\right)+\dotsb \\ &> \left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{6}+\frac{1}{6}\right)+\dotsb \\ &= 1+\frac{1}{2}+\frac{1}{3}+\dotsb \\ &= S \end{align*}

Answer 4

I'd probably go for Euclid's beautiful proof of the infinitude of primes, as it doesn't require much knowledge beyond elementary school.

Edit: Another possibility might be the pictorial proof that the derivative of $x^2$ is $2x$ by a square whose side length, $x$, increases and what that means to the rate of change of the area

Answer 5

Existence of Eulerian walks and the whole 'seven bridges of Königsberg' story. It's a cliché, but it's not about numbers (which is a plus when talking to a general audience), and it's something people can find at least mildly amusing. The argument is simple and everyone can follow it, and the best thing is, it's not just a clever solution to what looks like a random puzzle, which I feel is the impression people can get when shown other simple proofs. Starting from the seven bridges perspective and replacing shores and islands with vertices and bridges with lines, this proof is the best toy example I know of of the power of mathematical abstraction.

Answer 6

There exist irrational $x$ and $y$ such that $x^y$ is rational.

proof: if $\sqrt{2}^\sqrt{2}$ is rational we're done, otherwise we consider $(\sqrt{2}^\sqrt{2})^\sqrt{2}$, which evaluates to $2$.

edit: simpler to fit any audience, but somewhat related, the proof without words of the irrationality of $\sqrt{2}$

later edit: also Euclid's proof of the infinity of the primes is a good candidate

Answer 7

The fact that $$1 = \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ ...$$ by drawing the following picture of a square with a side length of $1$:

Answer 8

If you would say ten minutes, I'd go with $\sqrt2$ is irrational. Because that's a bit time consuming, since it can be a bit confusing for people less familiar with these things.

Instead, let's prove that $2+2$ is $4$.

Begin with axioms, mathematical proofs begin with axioms:

1. $0$ is a natural number, and $1$ is the successor of $0$.
2. For every natural number $n$, there is a successor $n+1$.
3. For every natural number $n$, $n+0=n$ and $n+(k+1)=(n+k)+1$.
4. $2=1+1$, $3=2+1$, $4=3+1$.

Theorem. $2+2=4$.

Proof. $2+2=2+(1+1)=(2+1)+1=3+1=4$.

The proof is important, since this shows both that proofs begin with axioms, and that we have to prove things which may seem obvious to us. But also, it will reassure them that $2+2=4$.

Answer 9

The Mutilated chessboard problem:

Suppose a standard 8x8 chessboard has two diagonally opposite corners removed, leaving 62 squares. Is it possible to place 31 dominoes of size 2x1 so as to cover all of these squares?

has an easy parity-based solution:

The puzzle is impossible to complete. A domino placed on the chessboard will always cover one white square and one black square. Therefore a collection of dominoes placed on the board will cover an equal numbers of squares of each colour. If the two white corners are removed from the board then 30 white squares and 32 black squares remain to be covered by dominoes, so this is impossible. If the two black corners are removed instead, then 32 white squares and 30 black squares remain, so it is again impossible.

Answer 10

I am reading some great responses, thank you all, keep them coming please, I love these. I'd also contribute with another of my favorite ones. Not so much a proof but a 'reasonable argument" for the area of a circle. Most 'general audience' members know $$a=\pi r^2$$ but I am often surprised at how few of them have ever seen a proof for this, similarly but not to same extent for Pythagorean theorem, everyone has seen/used these theorems but few have every seen a proof or even a sketch of a proof,

Answer 11

Plenty of people are dumbfounded when you suggest to them that $0.\overline{999}$ might be equal to $1$. Regardless of whether or not the following is rigorous, I've found it to be a wonderful way of demonstrating the property that only requires an understanding of simple algebra.

\begin{align} \textbf{let } x &= 0.\overline{999}\\ 10 \cdot x &= 10 \cdot 0.\overline{999}\\ 10x &= 9.\overline{999}\\ 10x - x &= 9.\overline{999} - x = 9.\overline{999} - 0.\overline{999}\\ 9x &= 9\\ x &= 1\\ 0.\overline{999} &= 1 \end{align}

Answer 12

I'm fond of the proof of Fermat's little theorem by 'counting necklaces'. It is a nice application of combinatorics that most people can follow. You start by stating the theorem as $p \mid a^p - a$ to avoid explaining congruences. It goes like this:

Take an alphabet with $a$ letters, and form all possible strings $S$ of length $p$. There are $a^p$ of them. We remove all the strings consisisting of a single character, and look at the remaining $a^p-a$ strings. We want to split them into groups of size $p$. Consider the strings as necklaces: connect the first and last letter. Imagine that two necklaces are 'friends' if you can obtain one from the other by sliding letters around, so for example the string $ABABA$ would be friends with $BABAA$. We claim that if a string $S$ cannot be broken into several copies of a shorter string $T$, so $S \neq TT\dots$ for some $T$, then $S$ has exactly $|S|$ friends. This is easy to argue informally, because it can't have more, and if it had fewer, then it would repeat itself. But $p$ is a prime, so the strings can't possibly be broken down to pieces. Hence every group of friends has exactly $p$ members. (If $\varepsilon$ is big, and you have the time, you can argue that being friends in this case is an equivalence relation). Since every friend group has $p$ members, we have divided $a^p - a$ strings into $p$ groups, so $p \mid a^p-a$.

Answer 13

Problem: An old car has to travel a 2-mile route, uphill and down. Because it is so old, the car can climb the first mile, the ascent, no faster than an average speed of 15 mi/hr. How fast does the car have to travel the second mile, on the descent it can go faster, of course, in order to achieve an average speed of 30 mi/hr for the trip?

Solution/Proof: It's impossible! Let $r$ be the rate of descent. We use the formula $$ \mathrm{time} = \frac{\mathrm{distance}}{\mathrm{rate}}; $$ the ascent takes $\frac{1}{15}$hr, the descent takes $\frac{1}{r}$hr, and the total trip should take $\frac{2}{30}=\frac{1}{15}$hr. Thus we have $$ \frac{1}{15}+\frac{1}{r}=\frac{1}{15}\Longleftrightarrow \frac{1}{r}=0, $$ which is impossible. So the car cannot go fast enough to average 30 mi/hr for the 2 mile trip.

Historical note: This question was actually sent to Albert Einstein by his friend Wertheimer. Einstein, and his friend Bucky, enjoyed the problem and had the following to say in a response letter to Wertheimer:

Your letter gave us a lot of amusement. The intelligence test fooled us both (Bucky and me). Only on working it out did I notice that no time is available for the downhill run!

More detail about this exchange between Einstein and Wertheimer may be found here.

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I think this problem is excellent for a general audience because the problem statement is crystal clear, and the knee-jerk reaction (i.e., 45 mi/hr) is completely off-base. It also does not take too much time to explain that the uphill journey takes $\frac{1}{15}$hr while, in order to average 30 mi/hr for a 2 mile trip, the total trip must take a total of $\frac{2}{30}=\frac{1}{15}$hr. Thus, uh oh! No time for the downhill run.

Answer 14

Here's a variation on $$\sum_{n=1}^{R} n =\frac{R(R+1)}{2}$$ I stumbled onto this a while back.

shows $$1+2+3+4+5+6+7 = \frac{1}{2}\cdot b\cdot h + \frac{1}{2}\cdot b$$ $$1+2+3+4+5+6+7 = \frac{1}{2}\cdot 7\cdot 7 + \frac{1}{2}\cdot 7$$ then $$1+2+3+4+5+6+7 = \frac{1}{2}\left( 7\cdot 7 + 7\right)$$ thus $$1+2+3+4+5+6+7 = \frac{7(7+1)}{2}$$

Answer 15

Insolvability of the $15$-puzzle.

Not directly accessible to everyone but I found that with a lot of effort most people will eventually understand it.

(Picture from Wikimedia commons.)

Theorem. Slide the blocks until 14 and 15 have swapped positions and all other blocks have returned to their initial positions. You won't succeed: it's impossible.

Sketch of Proof. The required number of moves is odd, because it corresponds to the permutation (14 15). (The general audience may learn this fact by drawing pictures that illustrate that the parity of the number of orbits changes with every move, or because they learned to comute determinants at some points in their lives.) On the other hand, the required number of moves is even because at each step the 'hole' moves from a black to a white square when colored as a chessboard, so since the hole returns to its original position it is impossible to swap 14 and 15.

Answer 16

It might be too short, but my favourite mathematical proof of a not-too-mathematical concept is the ancient Chinese proof (actually1, not apocryphally) of Pythagoras' theorem. First you can explain what Pythagoras' theorem means. Then, you introduce this image:

Now you're ready to tell them the proof.

1. Behold!
     (QED)

Answer 17

Every rational number has a repeating decimal.

Basically, once you know how to do long division to generate digits, all you need is a very naive pigeonhole principle.

When doing the long division for $\frac{p}{q}$, the remainder is always in $0,1,\dots,q-1$. And thus, you must eventually get the same remainder to the right of the decimal, at which point, the digits start to repeat.

This is a sentimental favorite of mine. I came up with this proof in the sixth grade, when our teacher taught us that $\pi$ didn't repeat and was exactly $\frac{22}{7}$. I recall explaining the proof to a friend in the schoolyard. It was my first proof.

(One point would have to be clarified - that a "terminating" decimal is a repeating decimal with the "0" repeating...)

Answer 18

Let me make it clear that I don't count this example as a proof, but I'm assuming you're using proof as a synonym of plausibility argument.

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The Principle of Mathematical Induction: Let $P$ be a property about natural numbers which is true of $1$ and whenever it is true of a number, it is true of its successor. Then all natural numbers satisfy $P$.

Proof: It's true for $1$, therefore, by hypothesis, it's true for $2$. Therefore, applying the hypothesis to $2$, it is true for $3$. Therefore, applying the hypothesis to $3$, it is true for $4$. And so on and so forth.

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Accompanying this presentation, I'd make a parallel between the principle of mathematical induction with its proof and an infinite row of sufficiently close domino tiles, analyzing the cases where the first one falls and where the first one doesn't fall.

Answer 19

I like the proof of the handshaking lemma in graph theory. I've taught it to a few students from ages 12-18 and they all seem to understand.

Lemma: There is an even number of vertices in a (finite) graph of odd degree.

The proof revolves around the fact that odd$\times$odd$=$odd, and even$+$odd$=$odd and similar facts like that.

It involves another simple lemma as well, even easier than the handshaking lemma.

Lemma 2: $\sum_v $deg$(v)=2*$the number of edges.

I like these proofs because it takes five minutes to show some legit pure mathematics to any one. Quick consequences of this are Eulers formula $V-E+R=2$ and Euler circuits etc.

Answer 20

For an audience with a bit of geometrical intuition about the sphere (earth), and with a markable sphere at hand to draw on, I like the proof that on a sphere, the area $\Delta$ of a spherical triangle that has angles $A$, $B$, and $C$ has area $\dfrac{(A+B+C-180°)}{720°}\odot$, where $\odot$ is the area of the sphere. (The audience doesn’t even need to know that the area of a sphere with radius $r$ is $4\pi r^2$).

First, the audience must understand that the sides of a spherical triangle must be shortest paths between the vertices, which are arcs of great circles. Some string and an appeal to airplane flights can make this point.

Then one draws the complete great circle containing each side of the triangle. These three circumferences divide the sphere’s surface into three “double lunes,” which are like the peel of an orange slice together with its antipodal peel. Each double lune contains the original spherical triangle, and each also contains the antipodal triangle. Elsewhere, the double lunes don’t overlap, and together, they cover the entire sphere.

Therefore the total area of the double lunes equals the area of the sphere plus four times the area of the triangle, which is the excess coverage beyond just covering the sphere once. They covered the triangle and antipodal triangle three times each, whereas only two of the six covers are needed to complete the coverage of the entire sphere.

The double lune with corners at $A$ and $A$’s antipode covers $\dfrac{2A}{360°}$ of the sphere, so $\frac{(2A+2B+2C)}{360°}\odot=\frac{(A+B+C)}{180°}\odot$ exceeds the sphere by four times the triangle’s area. Four times the triangle’s area is then $\frac{(A+B+C)}{180°}\odot-\odot=\frac{(A+B+C-180°)}{180°}\odot$, and the area of the triangle is $\frac{(A+B+C-180°)}{4\cdot180°}\odot$.

Thus the area $\Delta$ equals $\dfrac{A+B+C-180°}{720°}\odot$.

Answer 21

The Stable Marriage Theorem of Gale & Shapley. There is hardly anything in the statement or proof that even looks like mathematics to a "general audience". I haven't tried telling it to a general audience myself, but I'm sure a skilled expositor could get it across in $5+\varepsilon$ minutes, for some sufficiently large value of $\varepsilon$.

Reference: D. Gale and L. S. Shapley, College admissions and the stability of marriage, Amer. Math. Monthly 91 (1962), 9-15.

Answer 22

I would choose the proof that the set of reals is too big to form a sequence from it. However I'm not sure if 5min would be enough time ;)

Answer 23

Geometric series.

\begin{align} S&=\quad\;\,1+\frac1x+\frac1{x^2}+\frac1{x^3}+\dotsb\\ Sx&=x+1+\frac1x+\frac1{x^2}+\frac1{x^3}+\dotsb\\ Sx-S&=x\\ S(x-1)&=x\\ S&=\frac{x}{x-1} \end{align}

And, if time permits, a variant:

\begin{align} S&=\quad\;\,\quad\;\,\frac1x+\frac2{x^2}+\frac3{x^3}+\dotsb\\ Sx&=\quad\;\,1+\frac2x+\frac3{x^2}+\frac4{x^3}+\dotsb\\ Sx^2&=x+2+\frac3x+\frac4{x^2}+\frac5{x^3}+\dotsb \end{align} Rearranging, then summing: \begin{align} Sx^2&=x+2+\frac3x+\frac4{x^2}+\frac5{x^3}+\dotsb&Sx&=1+\frac2x+\frac3{x^2}+\frac4{x^3}+\dotsb\\ S&=\quad\;\,\quad\;\,\frac1x+\frac2{x^2}+\frac3{x^3}+\dotsb&Sx&=1+\frac2x+\frac3{x^2}+\frac4{x^3}+\dotsb\\ \hline Sx^2+S&=x+2+\frac4x+\frac6{x^2}+\frac8{x^3}+\dotsb&2Sx&=2+\frac4x+\frac6{x^2}+\frac8{x^3}+\dotsb \end{align} Thus, we have: \begin{align} Sx^2+S&=x+2Sx\\ Sx^2-2Sx+S&=x\\ S(x^2-2x+1)&=x\\ S&=\frac x{x^2-2x+1} \end{align}

As I finish typing this, I realize that perhaps the variant isn't as "accessible to a general audience" as the first one.

Answer 24

There's one statement with proof I really like that seems accessible to a wide audience rather quickly.

Statement: For any natural number $n$, there exist $n$ consecutive natural numbers that do not contain a prime number.

Proof: Consider the sequence $(n+1)!+2, (n+1)!+3, \ldots ,(n+1)!+(n+1)$. These numbers are divisible by $2, 3, \ldots, n+1$, respectively. In a five-minute version to a non-science Ba, one shouldn't mention the word 'factorial' (just say something like 'multiply the first ... numbers'). The main part required is that $2$ divides both $2\cdot a$ and $2$, hence $2$ divides $2+2\cdot a$, and similarly for other numbers than $2$.

If there's time left or if one has another $5\pm\epsilon$ minutes, one could explain that there are infinitely many primes (via Euclid's argument). When this seems contradictory (which might very well be the case, when one's talking to, say, a historian), it is a nice idea to talk about how large these factorials actually become.

Answer 25

A group of people stranded on a desert island (that aren't all the same sex) can produce at most a number of generations equal to one less than the population size while avoiding incest.

Proof: Take any descendent of the initial population, call them Alice.

Then one of Alice's parents is of the previous generation to Alice herself (i.e. if Alice is of generation 5, then at least one of her parents is of generation 4).

Also, Alice has more ancestors than that parent (since Alice's ancestors are the ancestors of both of her parents, who can't have any ancestors in common since we're not allowing incest).

So now let's climb up the genealogical tree to this parent of Alice's. In so doing the number of ancestors in the initial population "above us" goes down by at least one, and the generation goes down by exactly one.

Keep doing this until you climb back up to the initial population. You'll make a number of jumps equal to Alice's generation (if we call people in the initial population generation zero), and at each jump the number of ancestors goes down by at least one. Since we're now back up at the initial population, we have 1 ancestor (ourselves). So we lowered the number of ancestors a number of times equal to Alice's generation, and ended up with 1. So, Alice's generation can't be any more than one less than the number of ancestors she has (if she has 6 ancestors, she can't be of generation 6 or more because then we would have subtracted something from 6 six times and ended up with 1). Now, in the best case scenario Alice has everyone as an ancestor, so her generation can't possibly be more than one less than the initial population size. $\blacksquare$

Depending on how badly algebra class has mentally scarred your friend, you may find it easier to use names for the various numbers - $n$ and $G$ and whatnot. I was trying to avoid that because some people really do start to get nervous as soon as they hear you say "Let $x$...".

Be sure to illustrate with a drawing of an example genealogical tree as you explain. Even I'm not sure I'd be able to follow the above if the explanation was purely verbal.

Answer 26

In high school one learns that $\left(-a\right)\left(-b\right)=ab$, but in my experience the proof is never given, or a clear reason. The proof is easy using the number systems that people in America are comfortable with, i.e $\mathbb{R}$, and of course the notion that $a\left(0\right)=0$.

Proof: $\left(-a\right)\left(0\right)=0$

$\implies \left(-a\right)\left(b+\left(-b\right)\right)=0$

$\implies\left(-a\right)\left(b\right)+\left(-a\right)\left(-b\right)=0$

$\implies \left(a\right)\left(b\right)+ \left(-a\right)\left(b\right)+\left(-a\right)\left(-b\right)=\left(a\right)\left(b\right)$

$\implies 0+\left(-a\right)\left(-b\right)=\left(a\right)\left(b\right)$

$\therefore \left(-a\right)\left(-b\right)=\left(a\right)\left(b\right)$

Answer 27

Problem: A red ribbon is tied tightly around the earth at the equator (assume the earth is a perfect sphere). How much more ribbon would you need if you raised the ribbon 1 ft above the equator everywhere?

Answer: Only a tad bit more than 6 ft!

Solution: Let $r$ be the radius of the earth in feet. Then the circumference (length of the ribbon) is $2\pi r$. When we increase the radius by $1$ foot, the new radius is $r+1$, so the new circumference is $2\pi(r+1)$. Thus, you need $$ 2\pi(r+1)-2\pi r = 2\pi \approx 6.28 $$ extra feet of ribbon.

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I think this is a very non-intuitive answer (and one you can certainly explain in a very brief time). Firstly, you are not given the radius of the earth at the beginning, and that seems a bit weird. Secondly, only $6.28$ feet!? That seems quite bizarre that you could raise the ribbon by $1$ foot everywhere around such a huge object as a planet and only need a little more than $6$ feet of ribbon to do it.

If you have more time on your hands, then you could even discuss the many generalizations of this kind of argument where you are not dealing with circular objects. For example, see Ravi Vakil's paper The Mathematics of Doodling in the American Mathematical Monthly.

Answer 28

The empty set is contained in any other set. Suppose there exists a set $A$ such that $\varnothing \not\subset A$. Then exists $x \in \varnothing$ such that $x \not\in A$. Oh, wait!

Answer 29

I would explain the Pigeon Hole principle, or one of its many guises.

In fact, I remember explaining the PHP to a non math student while playing bridge.

I told him that one gets at least $4$ cards of some suit- which is really the PHP.

You can come up with many other interesting "real life" examples, and it's fun.

Answer 30

How about the proof that if $p=2^{n}-1$ is prime, then $k=(2^{n}-1)\cdot 2^{n-1}$ is a perfect number?

[The factors of $k$ are $1, 2,4,\cdots,2^{n-1}, p, 2p, 4p,\cdots, 2^{n-1}p$ .]