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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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    $\begingroup$ "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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One of my favorite proofs given constraints such as these is:

Theorem. $n^2 - n$ is even for all natural numbers $n$.

The proof can be carried out in many different ways depending on your "general" audience.

I have written up a sketch of the entire talk on MO and re-mentioned it on MESE.

See also the couple of different generalizations around the identity $(n-1)n = n^2 - n$.

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    $\begingroup$ So many approaches to this, very nice! This showcases, to some extent, the role creativity plays in math and in problem solving in general. I think it's excellent. Thank you! by the way, did you actually give this talk? and if so, how did it go? was the audience of general background? $\endgroup$ – userX Jan 19 '15 at 19:18
  • $\begingroup$ @userX Yes, I have given this talk to pre-service and in-service secondary school teachers (who were graduate students at Teachers College Columbia University). I gave it as a "model" lecture for students attending a teaching seminar; I think it went pretty well, since the attendees looked to be quite engaged (and I was invited back to give another talk the next year - that one ended up being about the multiplication table). $\endgroup$ – Benjamin Dickman Jan 19 '15 at 19:41
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    $\begingroup$ very nice! This sort of creativity party is what I enjoy most. Thank you! $\endgroup$ – userX Jan 19 '15 at 19:51
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I really like proofs using the pigeonhole principle I give two examples I think most people should know should know.


Example 1:

In a party with $n$ persons there are always two persons who have shaken hands with the same number of people. Proof: clearly in parties people don't shake hands with the same person twice (for sufficiently low alcohol levels). Intitially a person can shake hands any number of times from $0$ to $n-1$ ($n$ possible values). But notice if someone shook $0$ times then no one shook $n-1$ times, hence there are at most $n-1$ possible distinct values for the number of hands each person shook.

Since there are $n-1$ values and $n$ persons at least two must have shook the same amount of hands

Example 2:

In a party of $6$ or more there is a group of three persons who all know each other or a group of three persons such that none of them know each other.

Proof: Choose a random person, call him Bob. Then the remaining people (who number at least $5$) can be classified into two groups: those that know Bob and those that don't. The largest of these groups has at least three persons. Suppose the group of people who do know Bob is larger than $3$ (the other case is analogous). Now suppose there are two persons in that group that know each other. Pick Bob and those two persons and you have a group of three which all know each other. So the statement is true, the other case is when none of the people in the group know each other, pick three of the people in the group and then those three people don't know each other. So the statement is again true.

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  • $\begingroup$ That seems to be a really good mix of clever and relatable to regular people. I like it. $\endgroup$ – Fred Hamilton Jan 25 '15 at 23:21
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Taking $\epsilon = 5$ minutes here, we presented a lecture to psychologists, engineers, philosophers and also mathematicians over Game Theory, and how to take better decisions with examples, and different problems.

So we took about $10$ minutes to explain Nash Equilibrium to them, using the most famous Prisoner's Dilemma.

I believe everyone on the lecture understood the concept and went out of there knowing what this important idea has to do with economy, games and also mathematics.

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While possibly a bit silly, I find that the (utterly trivial) proof of the uniqueness of identity elements very nicely illustrates how "abstract" mathematical proofs "work" and how, at least not totally trivial, questions can get very simple answers if posed correctly.

While the proof itself obviously does not require 5 minutes to present one would probably have to use a few minutes to introduce the concept of an identity element (and why its of any interest) and possibly give some comments on what one means by mathematical "structure" more generally.

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I love the proof that a finite group of even order has an element that is its own inverse. I realize that groups are not nearly as well understood as they should be to a general audience, but that is the fault of our education priorities, as groups are everywhere in math and nature.

Anyway, the proof. Let $G$ be a group of finite even order. We want to show there is an element $h\in G$ such that $h\ne e$ and $h^2=e$, where $e$ is the group identity. We pick an element $g\in G$. If $g^2\ne e$, then we remove the elements $g, g^{-1}$, which must be unique. Removing 2 elements from an even set leaves an even set. We can continue to remove pairs of elements and their inverses until we find an element which is its own inverse (and we are done), or else we finally get to the last 2 elements of our group (the smallest even set), which must be the identity $e$ and some other element $h$ (since our procedure will never remove $e$). These two elements must include the inverse of $h$ by the inverse property of groups, and since $e$ cannot be the inverse of $h$, $h$ must be its own inverse. QED

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    $\begingroup$ I actually love this proof too, but I think explaining it to first semesters abstract algebra students would take a good 20-30 minutes, if they already know what a group is. I am not sure this is a 5 minute one for the general public. $\endgroup$ – userX Jan 19 '15 at 5:59
  • $\begingroup$ Yes, you are probably and unfortunately correct. We need to teach groups much earlier and connect them with the concept of symmetry, which I think would resonate with more people. $\endgroup$ – user452 Jan 21 '15 at 0:54
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Proof that $\sqrt 2$ is irrational: Any non-integer fraction multiplied by itself cannot be an integer.

(So a full length proof along these lines would first have to show that the prime factorization of integers is unique, and this turns out to be rather hard. But any young kid who has learned about prime factorization will accept this without proof.)

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There's no way to tune a piano in perfect harmony.

There are twelve half-steps in the chromatic scale, twelve notes in each octave of the keyboard. Start at middle "C", and ascend a perfect fifth to "G". That's seven half steps up, with a frequency ratio of 3/2. Drop an octave to the lower "g" -- that's twelve half steps down, and a frequency ratio of 1/2. Continuing around the "circle of fifths" twelve times, and dropping an octave seven times, brings you back to middle "C", a frequency ratio of 1.

So, $1 = (\frac{3}{2})^{12} \times (\frac{1}{2})^7$, or $3^{12}=2^{19}$.

Ask your piano tuner next time about those fifths.

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    $\begingroup$ For a non-musician the term "perfect harmony" might need to be explained (or defined) first. $\endgroup$ – Paŭlo Ebermann Jan 24 '15 at 9:34
  • $\begingroup$ The term is used here to be more suggestive than precise. A more precise statement might be, "There's no way to tune a piano to just intonation simultaneously in all keys." $\endgroup$ – Steve Mitchell Jan 27 '15 at 20:10
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A proof that needs five seconds rather than five minutes, but I find instructive nonetheless: Suppose you have a line in $\mathbb{R}^2$ together with two points $A$ and $B$ on the same side of the line. Determine the point where the distance travelled by an object moving from A to B being reflected at the line is minimal.

Too lazy to draw up a diagram. Show people the problem, then reflect one of the points on the line and they will suddenly see...

A great example to show the beauty of mathematics without involving anything abstract or scary numbers. Also a good opportunity to explain some basics.

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One of the first proofs I was shown was in a discrete structures class. I don't think this would survive the rigors of formal proof, but I really enjoyed it.

Definitions:

$N$ is the set of numbers starting with 0, 1, 2, $\dots $

A number in our set $N$ is odd if and only if there exists another number in our set $N$ that when multiplied by two and added to one yields the number in question. Or more formally

$\forall a \in N$, a is odd if $\exists b \in N \ni a = 2b + 1$

To Prove:

If $n$ is odd then $n^2$ is odd

Proof: Direct

Let us assume that we take an "odd" number, $n$. Then "n is odd" is true. Let us prove that $n^2$ is true.

By our assumptions $n$ is odd

This is the critical part of the proof, that might need some convincing

Since $n$ is odd we can use the definition of an odd number above and state that, $ \exists x \in N \ni n=2x+1$.

Now we can substitute our number $n$ with $2x+1$ and state $n=2x+1$

Further to prove $n^2$ is odd, we need only show that $(2x+1)^2$ is odd

This also might need some examples to convince

By distributive property, $(2x+1)^2=4x^2+4x+1$

Now we can refactor and rewrite $4x^2+4x+1=2(2x^2+2x)+1$

We know from rules of arithimetic that $2x^2+2x$ is a number which we can call $z$ and $z=2x^2+2x$

By substitution $2(2x^2+2x)+1=$2z+1$

$\therefore$ by definition of odd number $2(2x^2+2x)+1$ is odd

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