122
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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.

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  • 39
    $\begingroup$ "If there are $\omega$ supercompact cardinals, then it is consistent that $\aleph_{\omega+1}$ has the tree property." :-) $\endgroup$ – Asaf Karagila Jan 17 '15 at 23:24
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    $\begingroup$ I would recommend watching numberphile on youtube, he has a lot of videos that would be appropriate answers to this question. $\endgroup$ – DanielV Jan 17 '15 at 23:38
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    $\begingroup$ If you wanted an inaccessible proof of an accessible fact to a general audience, I'd suggest the proof of the fact that the shortest path between two points is a straight line... $\endgroup$ – Mehrdad Jan 23 '15 at 23:35
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    $\begingroup$ @AsafKaragila, what does that even mean? $\endgroup$ – dfeuer Jan 26 '15 at 2:42
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    $\begingroup$ It is unfortunate that this question is now attracting delete votes. $\endgroup$ – user 170039 Jun 14 at 17:46

39 Answers 39

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

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  • 4
    $\begingroup$ Actually, is was Gauss who came up with this :) $\endgroup$ – Ivo Terek Jan 17 '15 at 23:25
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    $\begingroup$ Edited to say "when he was just a child", this should cover most claims :) $\endgroup$ – MathMajor Jan 17 '15 at 23:34
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    $\begingroup$ As a motivation note that this formula can also be used to calculate the number of clinks of glasses in a group of n+1 people. $\endgroup$ – Roman Reiner Jan 18 '15 at 1:23
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    $\begingroup$ @GabrielH It was known long before Gauss. And the story about Gauss is possibly apocryphal. $\endgroup$ – Thomas Andrews Jan 19 '15 at 4:26
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    $\begingroup$ For what? Fermat was proving theorems about triangular numbers a century before Gauss was born. As for the story about Gauss being apocryphal, the real question is, what source do you have to support it? Bell's book contains a lot of bunk, and it is a source of a lot of random myths about mathematicians. @GabrielH $\endgroup$ – Thomas Andrews Jan 19 '15 at 4:51
75
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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.
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  • 2
    $\begingroup$ Excellent suggestion! This is a very simple and very creative solution to a problem that is very easily stated. Thank you! $\endgroup$ – userX Jan 19 '15 at 6:46
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    $\begingroup$ Small comment: we are assuming that the coin's bias does not affect the independence of each coin flip (i.e. each coin flip is independent of the others). $\endgroup$ – Ryan Jan 19 '15 at 15:57
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    $\begingroup$ I upvoted this terrific answer, but the statement "People generally have pretty reasonable built-in intuition for probability" is probably the most demonstrably false statement on this entire SE spinoff. $\endgroup$ – dotancohen Jan 20 '15 at 6:44
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    $\begingroup$ @sweeneyrod: No the chance of you choosing either side is not 50% simply because you chose it. Whether you knew the probabilities does not make a difference because for your method to work you have to assume that you randomly choose between two choices perfectly unbiasedly! $\endgroup$ – user21820 Jan 22 '15 at 11:51
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    $\begingroup$ @sweeneyrod If you don't know whether the coin's biased towards H or T, then it's Mathematically valid to call H or T and toss the coin once. However, this requires John to trust your claim that you don't know. This is unacceptable in many situations where coin tosses are used; eg. if John loses the toss, he may feel like you cheated. By using HT/TH combinations, you can both verify that it's fair, without any need for trust. $\endgroup$ – Warbo Jan 24 '15 at 17:16
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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*}

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64
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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

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  • 36
    $\begingroup$ Yes! But never forget to state what happens if $p_1p_2...p_n+1$ isn't prime, or the audience ends up thinking it's always prime (like it happened to me). $\endgroup$ – user2103480 Jan 18 '15 at 0:11
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    $\begingroup$ @Amjad If $p_1p_2\cdots p_n$ are all the primes, then $p_1p_2\cdots p_n+1$ is prime. And it is not prime. And it is zero. That is the nature of a proof by contradiction. :) $\endgroup$ – Thomas Andrews Jan 19 '15 at 14:29
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    $\begingroup$ @ThomasAndrews I think Amjad is referring to the fact that the proof buys you more than the infinitude of primes. If $p_1,p_2,...,p_n$ are the first $n$ primes, then all of the prime factors of $N=p_1p_2...p_n+1$ are larger than $p_n$. So you can always produce a larger prime using this method, but you might have to factor $N$ to find it. $\endgroup$ – Steven Gubkin Jan 19 '15 at 19:44
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    $\begingroup$ The derivative shows a relationship between area and sidelength - you draw a square with sidelengths "x" and then extend it by "h". For an "input" of a small h to the sidelength you get an "output" of 2xh + h^2 in terms of area. As the derivative measures, in loose terms, output over input, you get a rate of change of 2x + h for the area, with h getting neglectably small. Nothing too special about this rather informal proof, but I kinda like it. $\endgroup$ – user2103480 Jan 21 '15 at 18:27
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    $\begingroup$ I was just being irritatingly pedantic - if you do the proof as "by contradiction," then we are assume $p_1,\dots,p_n$ are all the primes, and the contradiction thus allows you deduce anything, including that $p_1p_2\dots p_n+1$ is prime or even that $1=2$. @Trengot $\endgroup$ – Thomas Andrews Jan 23 '15 at 17:46
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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.

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    $\begingroup$ i have to agree this is one of the best examples fitting all the criteria for general audience etc.. thank you! $\endgroup$ – userX Jan 18 '15 at 1:48
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    $\begingroup$ The vast majority of the general audience seem to think that "mathematics is about numbers". I like this example because it has almost nothing to do with numbers; it demonstrates that maths is far more than just opaque equations and stuff. $\endgroup$ – MathematicalOrchid Jan 19 '15 at 9:54
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    $\begingroup$ @MathematicalOrchid: Mathematics had mostly not been about opaque equations and stuff until recently... $\endgroup$ – user21820 Mar 21 '15 at 5:44
62
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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

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  • 33
    $\begingroup$ Keep in mind the definition of "general audience". Will a general audience understand exponent rules with irrational exponents? Will they even appreciate what an "irrational number" is? $\endgroup$ – Jack M Jan 17 '15 at 23:34
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    $\begingroup$ I see jack's point but I have to say this is undeniably beautiful ! $\endgroup$ – userX Jan 17 '15 at 23:38
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    $\begingroup$ @1110101001: that expression evaluates to $2$ so I'm pretty sure it isn't irrational $\endgroup$ – Alessandro Codenotti Jan 18 '15 at 1:16
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    $\begingroup$ A better proof is $(\sqrt{2})^{\log_2{9}} = 3$. $\endgroup$ – user203787 Jan 18 '15 at 23:23
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    $\begingroup$ @Alessandro It is better because it is constructive. To show that $\log_2{9}$ is irrational suppose it were equal to $m/n$. Then $2^m = 9^n$ which is impossible since $2$ does not divide the right hand side. $\endgroup$ – user203787 Jan 20 '15 at 14:36
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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 $1$:

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  • $\begingroup$ There's a nice proof in there that $\sum_{n\geq1}2^{1-2n}=\frac23$ if you shade the rectangles too. $\endgroup$ – samerivertwice Jun 15 at 12:24
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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 $2+2=4$.

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    $\begingroup$ You know what, you can even go for broke, and write $n+1$ for the successor. $\endgroup$ – Asaf Karagila Jan 17 '15 at 23:43
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    $\begingroup$ I like the 2+2=4 but I am wondering if it would not strenghten the prejudice that mathematicians are pettifoggers. Have you ever shown that proof to a general audience? I would be interested in hearing about the audience's reaction. $\endgroup$ – Taladris Jan 18 '15 at 22:48
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    $\begingroup$ I do not think that a "proof of 2 + 2 = 4" will leave any positive impression about mathematics on an average educated person. $\endgroup$ – user203787 Jan 18 '15 at 23:20
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    $\begingroup$ @Taladris As a teacher, this is precisely the point I am most interested in. If we assume that, "the beauty of math is not accessible to the general public", a corollary of that may be "math can only be enjoyed by some elite club of gifted thinkers". I don't believe this to be true, since none of us were born into this elite club. We all entered somehow into this set of people who enjoy and see beauty in math. I think statements such as ones found on this page are catalyst to open the doors wide so that more people belong to this set (see youtu.be/Yexc19j3TjE ) $\endgroup$ – userX Jan 19 '15 at 1:57
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    $\begingroup$ A somewhat similar experience: during a lesson on rational exponents in high school, my colleague show that if $\frac{a}{b}=\frac{c}{d}$, we have $x^\frac{a}{b}=x^\frac{c}{d}$ for any $x>0$. He really enjoy this proof; students (that I think to be not far from the "general audience") don't see the point. PS: I am not a downvoter. $\endgroup$ – Taladris Jan 19 '15 at 10:14
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The Mutilated chessboard problem:

enter image description here

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.
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  • $\begingroup$ The picture is inconsistent - On the checkerboard the Black corners are missing, but there is a picture of two white squares and you talk about missing white corners ? $\endgroup$ – Falco Jan 20 '15 at 9:57
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    $\begingroup$ @Falco, the 2x1 piece is simply a domino and the proof is good in both cases. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 20 '15 at 11:00
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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,

enter image description here

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    $\begingroup$ @Ian: π is usually defined in terms of half the circumference of the unit circle, in which case there is nothing much to prove to derive the circumference of an arbitrary circle. The difficult part if mathematical rigour is desired is actually in defining circumference and area themselves... $\endgroup$ – user21820 Jan 18 '15 at 6:01
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    $\begingroup$ @userX Seeing that you really get a triangle when you "unroll" the circle is nontrivial. $\endgroup$ – Zubin Mukerjee Jan 19 '15 at 4:30
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    $\begingroup$ This picture annoys me because it stops one step too early. There should be one more step that takes the right-hand half of the triangle, turns it upside down and places it atop the left-hand half of the triangle to obtain an $r\times \pi r$ rectangle, eliminating the need to invoke $A=\frac12bh$. $\endgroup$ – MJD Jan 19 '15 at 4:50
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    $\begingroup$ I feel that this explanation is more intuitive. $\endgroup$ – user26486 Jan 19 '15 at 5:16
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    $\begingroup$ "Reasonable arguments" in pictorial format aren't always reasonable. $\endgroup$ – 200_success Jan 19 '15 at 12:59
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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}

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    $\begingroup$ I don't like these proofs at all, since it always looks like mathematical trickery, if you add/multiply over several lines, because there are also some fake proofs for 1=0 with the same process. - I prefer this proof: What is 1/3 ? 0.333... ok and what is 0.3333 times 3 ? 0.99999 -> 1/3 times 3 is one. So it's the same. $\endgroup$ – Falco Jan 19 '15 at 9:47
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    $\begingroup$ @Falco I think that argument is less valid. If somebody believes $0.\overline{999}<1$, why not also believe that $0.\overline{333}<\frac{1}{3}$? $\endgroup$ – Peter Olson Jan 20 '15 at 3:18
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    $\begingroup$ @PeterOlson Then I would ask them: Ok, please write 1/3 for me in decimal notation... exactly 1/3 not more or less. And if they accept you cannot write 1/3 exactly with a finite number of decimals, they will truly understand why we need periodical numbers - double win ;-) $\endgroup$ – Falco Jan 20 '15 at 9:38
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    $\begingroup$ @Falco What if they accept that you can't write 1/3 exactly with a finite number of digits, but also say you can't do it with an infinite number of digits either? Your request doesn't seem that different from Please write 1 for me in decimal notation starting with 0 followed by a decimal point. Exactly 1, not more or less. to which the response would likely be that you can't do that, with either a finite or infinite number of decimals. $\endgroup$ – Peter Olson Jan 20 '15 at 14:34
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    $\begingroup$ We could debate endlessly what the "average person's" mathematical intuition is I suppose. But personally, I don't think it's any more obvious that 1/3=.333... than that 1=.999... But saying that .999... x 10 = 9.999..., I think most people would accept. You've said it's an infinite number of 9s, so "pushing" one in front of the decimal doesn't take away any at the end. $\endgroup$ – Jay Jan 21 '15 at 14:40
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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 $p \mid a^p-a$.

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  • $\begingroup$ I remember enjoying proofs of this based on group theory or combinatorics.. I don't remember a 'counting necklaces' proof but I will look that up. Thank you $\endgroup$ – userX Jan 18 '15 at 2:06
  • $\begingroup$ @userX I just added the proof here :) $\endgroup$ – Johanna Jan 18 '15 at 2:10
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    $\begingroup$ I see it, very nice! I can see why you like it, it is a very clever argument, and no group theory whatsoever, pure combinatorics and almost no combinatorics, amazing. thank you for sharing this. $\endgroup$ – userX Jan 18 '15 at 2:39
  • $\begingroup$ @userX I'm glad you like it too :) It's one of my favorite proofs. $\endgroup$ – Johanna Jan 18 '15 at 2:41
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    $\begingroup$ A good way to spice this proof up* is to change necklaces to pizzas and beads to pizza toppings. (*pun not intended, but then reconsidered and intended after all.) $\endgroup$ – Yoni Rozenshein Jan 19 '15 at 0:31
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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.


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.

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  • 1
    $\begingroup$ Very nice and surprising! I enjoyed this very much and I think the historical anecdote adds to the appeal $\endgroup$ – userX Jan 18 '15 at 18:18
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    $\begingroup$ Here's another way to think about this without getting into calculations. Imagine a phantom car that leaves at the same time and travels at the desired speed of 30 mi/hr. We need to arrive at the finish line at the same time as the phantom car—but after the first mile, the phantom car has just arrived at the end. $\endgroup$ – Théophile Jan 19 '15 at 5:04
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Here's a variation on $$\sum_{n=1}^{R} n =\frac{R(R+1)}{2}$$ I stumbled onto this a while back.

enter image description here

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 $$1+2+3+4+5+6+7 = \frac{7(7+1)}{2}$$

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    $\begingroup$ I've never seen this before. Nice! $\endgroup$ – Brian Fitzpatrick Jan 20 '15 at 19:54
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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.

15 puzzle

(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.

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    $\begingroup$ There is a nice proof where you don't need to talk about permutations. Notice that when you move left or right then the number of disorders (a disorder is a pair of pieces with numbers such that $x<y$ and $y$ is before $y$ while going line by line horizontally) do not change but when you move up or down then number of disorders change by an odd number and number of the row where the empty place is changes by one (again odd number). The starting position has even number of disorders and empty piece in row number 1. If you swap 1 and 2 you have 1 disorder and empty piece in row 1. Contradiction. $\endgroup$ – pw1822 Jan 18 '15 at 23:06
  • $\begingroup$ I love these puzzles, and can solve them quite quickly (although I guess it's not hard once you know how), and this little problem is one of my favorites. $\endgroup$ – Alfred Yerger Jan 19 '15 at 4:21
  • $\begingroup$ It wasn't clear to me at first that "Slide the blocks until 14 and 15 have swapped positions" meant that all the other numbers must be returned to their starting locations. I only guessed that was a requirement since without it, it's easy to swap 14 and 15. $\endgroup$ – Todd Wilcox Jan 20 '15 at 14:12
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    $\begingroup$ @ToddWilcox Good point I've noticed before that this tends to confuse people. I have added it, hope it's clear now! $\endgroup$ – Myself Jan 20 '15 at 14:14
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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:

Diagram showing how square on hypotenuse plus four identical right angle triangles is equal to the squares on the other two sites, plus four of the same right angle triangles

Now you're ready to tell them the proof.

1. Behold!
     (QED)
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    $\begingroup$ I agree, I would even say that familiarity with at least one beautiful proof of the Pythagoras theorem should be a legal requirement to graduate from high school. $\endgroup$ – userX Jan 19 '15 at 2:36
  • $\begingroup$ I'd support the proof of the theorem with the following gif: i.imgur.com/W8VJp.gif $\endgroup$ – András Hummer Jan 19 '15 at 15:33
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    $\begingroup$ @AndrásHummer It could be that it only works for that specific triangle or that the water doesn't entirely fill the square up by a few plank lenghts for all we know. It is not a proof by any means, while the above picture is. $\endgroup$ – user26486 Jan 24 '15 at 0:31
  • $\begingroup$ I prefer this diagram, originally devised by Thabit ibn Qurra around 900 AD. See here for a couple of animated versions. $\endgroup$ – PM 2Ring Jan 26 '15 at 10:12
  • $\begingroup$ Nice example, but the Behold! is from Bhaskara. $\endgroup$ – André Nicolas Mar 21 '15 at 5:30
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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...)

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  • $\begingroup$ So, what did your teacher say when s/he saw your proof that his/her assertion was a contradiction? $\endgroup$ – WetSavannaAnimal Jan 19 '15 at 5:22
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    $\begingroup$ @WetSavannaAnimalakaRodVance I actually never showed her the proof. I argued with her for a long time, because I did the division for $\frac{22}7$ and saw it repeated, and argued with her a bit, possibly creating a scene. :) She said, "Compute some more digits, it will stop repeating." So I gave up arguing, but came up with the general argument. $\endgroup$ – Thomas Andrews Jan 19 '15 at 5:25
  • $\begingroup$ Could we use this as a 2-for-one? Meaning first we introduce Pigeon Hole Principle, anticipating a laugh at the silly name followed by the typical 'what is that good for?' Seems to be one could explain both the PHP and this classic result in a very reasonable amount of time. As a bonus, prove $\pi \neq \frac{22}{7}$, wait 3-for-one! very nice! $\endgroup$ – userX Jan 19 '15 at 16:56
  • $\begingroup$ Well, we didn't "prove" $\pi\neq \frac{22}7$, we just proved that if $\pi$ does not repeat, then $\pi\neq \frac{22}7$ :) @userX $\endgroup$ – Thomas Andrews Jan 19 '15 at 18:40
  • $\begingroup$ Can you expand upon this? I don't see why the digits should start to repeat the moment you get the same remainder. Forgive me for asking such an elementary question. $\endgroup$ – dalastboss Jan 19 '15 at 19:47
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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.


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.


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.

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  • $\begingroup$ yes, thank you. I mean "plausibility argument" $\endgroup$ – userX Jan 18 '15 at 1:35
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    $\begingroup$ I like the dominoes picture. The picture I first learned was similar: "Suppose you're climbing a ladder. If you know you can reach the first rung, and you know that if you can reach any rung then you can reach the next one up, then in fact you can climb the whole ladder." $\endgroup$ – Jesse Madnick Jan 18 '15 at 3:20
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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.

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  • $\begingroup$ Thank you Jack, I like this one too. I also vaguely remember a variation of this problem. I think the question is roughly "if 15 people work in an office, is it possible that each gets along with exactly 5 other people?" using your Lemma 2 and counting degrees of vertices leads to a simple and clean proof concluding 'not possible'. $\endgroup$ – userX Jan 19 '15 at 6:07
  • $\begingroup$ Of course, permission to use this example? :) $\endgroup$ – Jack Davies Jan 19 '15 at 12:53
  • $\begingroup$ Yes, by all means use. $\endgroup$ – userX Jan 19 '15 at 21:47
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    $\begingroup$ On a similar vein, I've found the proof that any graph has at least two vertices of the same degree to be an easily accessible proof. One of the nicest bits about it is that the proof calls on the pigeonhole principle, which, if you confine your discussion to finite graphs, does a great job of illustrating the difference between common sense and a formal proof. $\endgroup$ – sju Jan 25 '15 at 16:41
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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 $\dfrac{A+B+C-180°}{720°}\odot$.

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    $\begingroup$ I found the 'markable sphere at hand to draw on' very important. With your explanation and the diagrams here web.cortland.edu/jubrani/375ch5.pdf I was able to follow it. I think this is excellent. Once understood, I think the general audience would feel like they learned something non-trivial, and the requirements to understand this are very minimal. I had never seen this before, thank you! $\endgroup$ – userX Jan 19 '15 at 7:11
  • $\begingroup$ I remember learning this in graduate school. I thought it was beautiful and was shocked that I’d never known about it. It’s also a nice starting point for further ideas about non-Euclidean geometry. Question: If the five angles of a regular pentagon each measure 90°, where are you? $\endgroup$ – Steve Kass Jan 19 '15 at 15:48
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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.

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  • $\begingroup$ very nice, it seems to me in a classroom setting one could take this further and actually carry out the algorithm live. If one could pull this off it would be a very nice example of beautiful mathematics that may not feel like math to participants, I do agree on the +$\epsilon$ point.. might be a large epsilon :-) thank you! $\endgroup$ – userX Jan 18 '15 at 1:46
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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 ;)

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    $\begingroup$ Even explaining (properly) what a real number is would take about 20 minutes at least. $\endgroup$ – Jack M Jan 17 '15 at 23:33
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    $\begingroup$ I would not say exactly what real number is. Everyone from that audience should have some intuition and knowledge about real numbers from school and that would be enough in my opinion to understand the proof. $\endgroup$ – pw1822 Jan 17 '15 at 23:37
  • $\begingroup$ Just to be clear, Are you talking about cantors diagonalization argument ? $\endgroup$ – userX Jan 17 '15 at 23:48
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    $\begingroup$ Yes I mean this proof. $\endgroup$ – pw1822 Jan 17 '15 at 23:53
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    $\begingroup$ yes, I very much like that one there is a very nice presentation here youtu.be/elvOZm0d4H0 on such proof at 3:35 mark, it is very elegant. thank you for suggesting this one.. and they do it in a tone appropriate for general audiences.. very well done! $\endgroup$ – userX Jan 18 '15 at 2:03
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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.

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    $\begingroup$ It might be easier to demonstrate the variant by multiplying the terms in $S(S-1)$ (using the $S$ from the original problem). Thus $$\frac1x+\frac2{x^2}+\frac3{x^3}+\dotsb = \big(1+\frac1x+\frac1{x^2}+\dotsb \big)\big(\frac1x+\frac1{x^2}+\frac1{x^3}+\dotsb \big) = \big(\frac{x}{x-1}\big) \big(\frac{x}{x-1} - 1\big) = \frac{x}{(x-1)^2}.$$ $\endgroup$ – Théophile Jan 19 '15 at 5:21
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    $\begingroup$ Is there a reason you use $\frac{1}{x}$ for the common ration, rather than $x$? It makes it so much harder to read. $\endgroup$ – Thomas Andrews Jan 19 '15 at 14:25
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    $\begingroup$ @ThomasAndrews I was wondering that too. $\endgroup$ – Théophile Jan 19 '15 at 15:47
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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.

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    $\begingroup$ yes.. i like this one too.. I concur, I think this can be shared with a general audience and I think is a very nice question and very nice solution, thank you! $\endgroup$ – userX Jan 18 '15 at 2:18
  • $\begingroup$ That is, of course, $n+1$ consecutive numbers... $\endgroup$ – Thomas Andrews Jan 19 '15 at 5:53
  • $\begingroup$ Somehow, I once started getting confused if $1$ or $2$ should be used and, for some reason, got it mixed up by now. Thanks for pointing it out - I just edited accordingly. $\endgroup$ – HSN Jan 19 '15 at 15:22
  • $\begingroup$ There's nothing wrong with introducing a term like factorial if the concept is simple, which it is in this case. $\endgroup$ – Théophile Jan 19 '15 at 15:54
  • $\begingroup$ +1. Also you can prove this result using Primorials. $\endgroup$ – Bumblebee Apr 27 '15 at 5:45
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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.

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  • $\begingroup$ You can probably give a much better bound than $n-1$. Somewhere in the vicinity of $\log_2 n$. $\endgroup$ – Asaf Karagila Jan 18 '15 at 0:04
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    $\begingroup$ Are half-brothers can have sex in your definition of incest? $\endgroup$ – Asaf Karagila Jan 18 '15 at 0:22
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    $\begingroup$ How do you assign a "generation" to an individual? How is "incest" defined here? Is having offspring with one's seventh cousin twice removed "incest" here? Since that term is non-constant among cultures it needs a definition. $\endgroup$ – Jeppe Stig Nielsen Jan 18 '15 at 0:28
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    $\begingroup$ Incest means sex with someone with whom you have a common ancestor in the initial population. Generation is zero for the initial population, and otherwise is one more than the maximum of your parents' generations. $\endgroup$ – Jack M Jan 18 '15 at 1:45
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    $\begingroup$ Incest is not fully inductive. $\endgroup$ – Joshua Jan 19 '15 at 22:49
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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: $\left(-a\right)\left(0\right)=0$

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

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

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

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

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

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    $\begingroup$ Accessible, perhaps, but I think it's a challenge for a general audience to appreciate. They could easily be confused about why we need a bunch of formal rules just to prove another formal rule. $\endgroup$ – Erick Wong Jan 19 '15 at 3:11
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    $\begingroup$ I agree with Krampus that this should be much better-known, but I also agree with Erick that it might not fit the topic of this question. (I cover this topic when I teach future mathematics teachers; I think every teacher of mathematics should be able to demonstrate why “negative times negative is positive” as rigorously as this, if asked to, say by a curious student, but I’m not sure the proof needs to be taught in the elementary or high school curriculum.) $\endgroup$ – Steve Kass Jan 19 '15 at 4:29
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    $\begingroup$ I've actually been teaching basic algebra this way for more than a decade, except we learn the names of the field axioms and every step needs to be justified by such, or a theorem or a definition. So the above is nearly identical to what I have been doing in class, except we would include the justifications on each step such as commutativity, associativity, cancellation law of addition, distributive law etc. Seems ridiculous to me that students go thru 12+ years of math and never take a week or two to learn what they can and can't do [field axioms]. $\endgroup$ – userX Jan 19 '15 at 6:23
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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.


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.

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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!

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    $\begingroup$ Is this really such a great example? I mean, for one thing I don't see why the statement "for all sets $A$, $\varnothing \subset A$" would be of any interest whatsoever to the average adult -- it's barely of interest to me! Besides which, no one will actually get anything out of this unless they first understand (1) "what is a set?", (2) "when is one set contained in another?" (precisely), and (3) "what is the empty set?" (and why would we bother defining such an object at all?). I don't thing someone can be exposed to these ideas in 5 minutes, except very superficially. $\endgroup$ – Mike F Jan 18 '15 at 0:55
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    $\begingroup$ @MikeF I agree, "for all sets $A$, $\varnothing \subset A$" seems like a hard sell to a general audience, yet I find the proof so short and elegant and potent... thats why I enjoyed it.. your point is valid, how much of this can we share with a general audience in 5 minutes.. the trouble with doing this superficially is that i don't think they would enjoy it as much.. $\endgroup$ – userX Jan 18 '15 at 2:13
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    $\begingroup$ Strictly speaking, $\varnothing \subset A$ does not hold for $A = \varnothing$. You probably meant $\varnothing \subseteq A$. $\endgroup$ – wchargin Jan 18 '15 at 4:21
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    $\begingroup$ @WChargin: Plenty of people use "$\subset$" the way that you use "$\subseteq$", and use notation like "$\subsetneq$" or similar when strict containment is meant. $\endgroup$ – Mike F Jan 18 '15 at 5:49
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    $\begingroup$ I think that's rather trivial. $\endgroup$ – Elliot Gorokhovsky Jan 20 '15 at 22:59
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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.

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    $\begingroup$ A bridge squeeze, I suppose, is a sort of pigeonhole principle... $\endgroup$ – Thomas Andrews Jan 19 '15 at 18:42
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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 $1, 2,4,\cdots,2^{n-1}, p, 2p, 4p,\cdots, 2^{n-1}p$ .]

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