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The background of my question comes from an observation that what we teach in schools does not always reflect what we practice. Beauty is part of what drives mathematicians, but we rarely talk about beauty in the teaching of school mathematics.

I'm trying to collect examples of good, accessible proofs that could be used in middle school or high school. Here are two that I have come across thus far:

(1) Pick's Theorem: The area, A, of a lattice polygon, with boundary points B and interior points I is A = I + B/2 - 1.

I'm actually not so interested in verifying the theorem (sometimes given as a middle school task) but in actually proving it. There are a few nice proofs floating around, like one given in "Proofs from the Book" which uses a clever application of Euler's formula. A very different, but also clever proof, which Bjorn Poonen was kind enough to show to me, uses a double counting of angle measures, around each vertex and also around the boundary. Both of these proofs involve math that doesn't go much beyond the high school level, and they feel like real mathematics.

(2) Menelaus Theorem: If a line meets the sides BC, CA, and AB of a triangle in the points D, E, and F then (AE/EC) (CD/DB) (BF/FA) = 1. (converse also true) See: http://www.cut-the-knot.org/Generalization/Menelaus.shtml, also for the related Ceva's Theorem.

Again, I'm not interested in the proof for verification purposes, but for a beautiful, enlightening proof. I came across such a proof by Grunbaum and Shepard in Mathematics Magazine. They use what they call the Area Principle, which compares the areas of triangles that share the same base (I would like to insert a figure here, but I do not know how. -- given triangles ABC and DBC and the point P that lies at the intersection of AD and BC, AP/PD = Area (ABC)/Area(DBC).) This principle is great-- with it, you can knock out Menelaus, Ceva's, and a similar theorem involving pentagons. And it is not hard-- I think that an average high school student could follow it; and a clever student might be able to discover this principle themselves.

Anyway, I'd be grateful for any more examples like these. I'd also be interested in people's judgements about what makes these proofs beautiful (if indeed they are-- is there a difference between a beautiful proof and a clever one?) but I don't know if that kind of discussion is appropriate for this forum.

Edit: I just want to be clear that in my question I'm really asking about proofs you'd consider to be beautiful, not just ones that are neat or accessible at the high school level. (not that the distinction is always so easy to make...)

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Surely you know the books ''Math! Encounters with Undergraduates'', ''Math Talks for Undergraduates'' and ''The beauty of doing Mathematics'' by Serge Lang. They were written with your same intention. –  Giuseppe Tortorella Sep 8 '11 at 11:15
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I remember Lang giving a talk at Caltech supposedly aimed at undergraduates, but far more advanced. Lang called on a postdoc in the front row, apparently thinking he was a student, and the postdoc couldn't answer. At one point, he wrote out a complicated formula, and challenged Prof. Ramakrishnan, "Do you teach your students this?" "No." "So you see, Caltech is not better than anywhere [sic] else." After a bit more, he went back to the formula and corrected a sign. Ramakrishnan called out, "That, we teach." –  Douglas Zare Sep 14 '11 at 0:38
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Even the notion of proof may not be accessible at high school level –  Fernando Muro Sep 15 '11 at 20:33
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In the US, most high school mathematics classes are not based on proofs. Theorems are introduced without reference to proof. Only in a few limited situations are students asked to prove anything, such as in some geometry classes and some calculus classes. Instead, high school math classes concentrate on introducing objects, their properties, and how to manipulate those objects. Finding neat accessible proofs to show them is reasonable, but this is very different from finding neat accessible math to show them. E.g., I think the Chaos Game is accessible, but few students can handle the proofs. –  Douglas Zare Sep 17 '11 at 0:39
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As far as I remember, the thing I liked the most in high school maths (age of 14) was the so called Ruffini's rule: $(x-a)$ divides a polynomial $P(x)$ if and only if $P(a)=0$. It looked to me so incredibly easy and so full of consequences. I hope they still learn it with a proof. –  Pietro Majer Sep 18 '11 at 20:12

44 Answers 44

Cantor's diagonal argument.

The warm-up could be an equally beautiful proof, namely that the rationals are countable.

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I am a bit torn on this one. At least I believe before one would have to also prove, say, that the cardinality of the rationals is equal to that of the naturals (and/or related things). Because if not the 'insight' that there are more reals than naturals might seem too 'obvious' to make a proof enlightening. –  quid Sep 8 '11 at 10:51
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But the enumeration proving that the rationals are countable is also almost as beautiful as the diagonal argument! Perhaps they could be combined. –  J.C. Ottem Sep 8 '11 at 11:11
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Yes, I agree; combined it could be very interesting. I should have made my agreement clearer in my first comment; sorry about that. –  quid Sep 8 '11 at 11:19
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The counting-rationals argument loses some appeal when you have to deal properly with skipping the non-reduced fractions. The visual proof of course establishes that $\mathbb{N} \times \mathbb{N} \cong \mathbb{N}$ as sets, and the countability of $\mathbb{Q}$ follows by applying an unrelated lemma on cardinality of surjective images. Glossing over a lemma like that seems exactly like what a beautiful proof (especially an exemplary proof) should avoid. –  Ryan Reich Sep 9 '11 at 21:09
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There is a simple bijection between the positive rationals and positive integers shown by the fusc function. Let $f(n)$ be the number of ways of representing $n$ as a sum of powers of $2$ with at most $2$ copies of each power. $f(4)=3$ since $4=4,2+2,2+1+1$. The sequence $\{f(n)\}_{n=0} = \{1,1,2,1,3,2,3,1,4,...\}$. Over natural numbers $n$, $g(n) = f(n-1)/f(n)$ hits each positive rational precisely once. See also the Calkin-Wilf tree, how Euclid's algorithm reduces relatively prime ordered pairs. The sequences of numerators and denominators in order are offset copies of $\{f(n)\}$. –  Douglas Zare Sep 10 '11 at 21:33

The formula $1 + 2 + ... + n = n(n+1)/2$ can be proved on middle school level: Assume first n is even. Then there are n/2 pairs (1,n), (2,n-1), ..., (n/2,n/2+1) those sum is always n+1. Thus the overall sum is n/2*(n+1). The case when n is odd can be treated in the same manner.

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Actually this proof would be more approriate for primary school (age 10 or so). Smarter kids (like Gauss) figure it out themselves. –  GH from MO Sep 8 '11 at 12:04
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I prefer the "visual proof" because you don't have to reason, you just "see" it: put 1 little square in the first raw, two in the second,... n in the last raw, and you've obtained a figure of which you can easily compute the area as: (AreaOfBigSquare-AreaOfSquaresOnTheDiagonal)/2+ AreaOfSquaresOnTheDiagonal=$(n^2-n)/2+n=n(n+1)/2$. Of course to pass from $(n^2-n)/2+n$ to $n(n+1)/2$ the kid must have learned some "algebra". –  Qfwfq Sep 8 '11 at 13:48
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No, the best proof is the visual one where you draw a triangle with 1, 2, ..., n circles in each row, and another row below the last with n + 1 circles. Anything in the smaller triangle can be specified by two "coordinates" in the last row, obtained by dropping down parallel to the sides. Thus, $\binom{n + 1}{2}$ using the definition of "n + 1 choose 2"; to get a formula for that number you could use the argument in Dan's comment. This one is nice because it is both visual and a bijective proof, rather than computational. –  Ryan Reich Sep 8 '11 at 15:31
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How about this: write the numbers from $1$ to $n$ in row, then write them again in a row below, but backwards. Add up the columns, to get $n+1$ each time, so $n(n+1)$ in total, because there are $n$ columns. Divide by $2$. –  euklid345 Sep 9 '11 at 4:22

There is a very elegant proof that there exists no continuous injection from the plane into the real line. The proof can basically be given by drawing a picture on the blackboard.

Suppose there is such an injection $f$. Let $x$ and $y$ be distinct points in the plane and let $g_1$ and $g_2$ be paths from $x$ to $y$ such that $g_1(r_1)\neq g_2(r_2)$ for $r_1,r_2\in (0,1)$. Now this implies that $f\circ g_1((0,1))\cap f\circ g_2((0,1))=\emptyset$, contradicting the intermediate value theorem.

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But what is the point of proving a theorem as intuitively obvious as this in the eyes of a high-schooler? –  darij grinberg Sep 8 '11 at 10:18
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Also, continuous functions in more than 1 dimension are usually not defined in high school... –  darij grinberg Sep 8 '11 at 10:19

Extending on Ralph's answer, there is a similar very neat proof for the formula for $Q_n:=1^2+2^2+\dots+n^2$. Write down numbers in an equilateral triangle as follows:

    1
   2 2    
  3 3 3
 4 4 4 4

Now, clearly the sum of the numbers in the triangle is $Q_n$. On the other hand, if you superimpose three such triangles rotated by $120^\circ$ each, then the sum of the numbers in each position equals $2n+1$. Therefore, you can double-count $3Q_n=\frac{n(n+1)}{2}(2n+1)$. $\square$

(I first heard this proof from János Pataki).

How to prove formally that all positions sum to $2n+1$? Easy induction ("moving down-left or down-right from the topmost number does not alter the sum, since one of the three summand increases and one decreases"). This is a discrete analogue of the Euclidean geometry theorem "given a point $P$ in an equilateral triangle $ABC$, the sum of its three distances from the sides is constant" (proof: sum the areas of $APB,BPC,CPA$), which you can mention as well.

How to generalize to sum of cubes? Same trick on a tetrahedron. EDIT: there's some way to generalize it to higher dimensions, but unfortunately it's more complicated than this. See the comments below.

If you wish to tell them something about "what is the fourth dimension (for a mathematician)", this is an excellent start.

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@Frederico: Careful. The sum of cubes does not correspond to a tetrahedron, but rather a pyramid. (Which does not have this nice symmetry) However the slightly modified sum $$Q_N^'=\sum_{n=1}^N n\frac{n(n+1)}{2}=\frac{1}{2}\sum_{n=1}^N n^3+\frac{1}{2}\sum_{n=1}^N n^2$$ will correspond to a tetrahedron. By using this symmetry we find $$Q_N^' = \frac{1}{4} (3N+1)\left(\frac{N(N+1)(N+2)}{6}\right)$$ since each entry will be $3N+1$ and the tetrahedral numbers are $\binom{N+3}{3}=\frac{N(N+1)(N+2)}{6}$. From here we can deduce that $$\sum_{n=1}^N n^3=\frac{(N^2+N)^2}{4}.$$ –  Eric Naslund Sep 16 '11 at 23:31
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In short, you won't have a nice generalization of the solution for $n=2$ to higher dimensions. Such a generalization is not expected either since Faulhaber's Formula is not so simple. –  Eric Naslund Sep 16 '11 at 23:39
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@Eric Oh, you're right. :( What a pity, it would've been an even neater generalization. –  Federico Poloni Sep 17 '11 at 9:57
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Cool. Both the example and the follow up comments. –  Manya Sep 19 '11 at 8:30

Morley's Theorem.

Wikipedia's proof is completely elementary and only involves trigonometric identities and euclidean plane geometry.

There is also a proof by Alain Connes, based on affine geometry techniques. Of course it is a bit more technical, but again it involves math that doesn't go much beyond the high school level, and could be appreciated by the most gifted students

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Thanks. Actually this example was given by Rota in a famous paper about the phenomenology of mathematical beauty as an example of a theorem that is surprising but not beautiful (in response to Hardy's claim that beauty arises from a feeling of surprise.) Of course, it is open to debate... Thanks for the Connes reference, I didn't know about that. –  Manya Sep 8 '11 at 11:40
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See also Conway's proof (linked at the Wikipedia article). –  Todd Trimble Sep 8 '11 at 12:05
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@Manya: Well, I guess it's a matter of taste. Personally, I do find this theorem beautiful. Thank you for the remark –  Francesco Polizzi Sep 8 '11 at 12:40
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But the question was asking for a beautiful proof, not a beautiful theorem. –  euklid345 Sep 9 '11 at 4:51
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P.S. (@ eucklid) Good point. Rota actually claimed that neither the theorem nor any of the proofs of it (thus far?) are beautiful. –  Manya Sep 19 '11 at 8:16

Seeing the struggle of many students with standard trigonometry, I especially like the rational parametrization of $x^2+y^2=1$ (which is equivalent to listing all Pythagorean triples) by starting from $\sin^2\phi+\cos^2\phi=1$ and then using $$ \sin\phi=\frac{2t}{1+t^2}, \quad \cos\phi=\frac{1-t^2}{1+t^2}, \qquad t=\tan\frac{\phi}2. $$ Note that the formulas are usually used in the context of integration of rational expressions in sine and cosine.

At the same time, a more general "geometric" argument (applicable to general quadratics), due to Bachet (1620), is still at school level. Namely, fix a single rational point on the curve, $(x _ 0,y _ 0)$ say, and consider the intersection points of the curve and straight lines $y-y_0=t(x-x_0)$ with rational slope $t$ passing through the point.

A beauty here is because of variety of different geometric and analytic methods for solving a classical arithmetic problem.

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Since coordinate geometry was invented after 1620 I wonder how Bachet could have done that. –  Franz Lemmermeyer Sep 8 '11 at 13:23
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Franz, I will be definitely happier if you are a little bit more constructive in your critisism: you seem to be the right person to explain why the method is usually attributed to Bachet! –  Wadim Zudilin Sep 10 '11 at 6:12

I recommend Kelly's proof of the Sylvester-Gallai theorem (the original proof of Gallai was about 30 pages long, this one takes a few lines). The theorem and the proof can be read here.

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@Emil: There are at least 3 points on $l$, at least two on one side of the perpendicular. Call the closer of these two $B$ and the farther of these two $C$. Now let $m$ denote the line $PC$, then the pair $(B,m)$ has a smaller distance than $(P,l)$. Contradiction. –  GH from MO Sep 8 '11 at 12:08
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Thanks, now I got it. I misread it first. –  Emil Jeřábek Sep 8 '11 at 12:36

The Gale-Sharpley stable marriage theorem, http://en.wikipedia.org/wiki/Stable_marriage_problem .

The algorithm and its proof are very much accessible to school students. Despite its innocuous look, the algorithm is not easy at all to invent.

On a similar note, Hall's theorem: http://en.wikipedia.org/wiki/Hall%27s_marriage_theorem#Graph_theory . This looks like a recreational puzzle but actually is closer to university mathematics than everything done in high school.

Here is another combinatorial exercise which, properly presented, does not even look like mathematics: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=279550#p279550 . The thing I don't like about it is that the standard "gotcha" proof (explained in the usual, informal way) requires a bit too much concentration to understand - some students might fail at it and take it as an example that mathematical proofs are something one either believes or not, rather than something one can check. Of course, one can formalize the proof, but this requires quite an amount of time in a high school class.

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I’ll point out that the “combinatorial exercise” is actually useful. Assume that we are given a sequence $s_1,\dots,s_n$ of symbols of arities $a_1,\dots,a_n\ge0$, respectively. It is easy to see that if we can arrange them into a well-formed term, then $\sum_ia_i=n-1$. The exercise says that, conversely, if this identity holds, then there exists a (unique) cyclic permutation of the string $s_1\dots s_n$ which is a valid term in Polish (prefix) notation. Among other things, this allows you to count the well-formed terms consisting of $n$ symbols. –  Emil Jeřábek Sep 8 '11 at 12:29

Two of my favorites: (1) The parameterization of all primitive Pythagorean triples; (2) The formula for the $n$th Fibonacci number in terms of the golden ratio $\phi = \frac{1+\sqrt{5}}{2}$, with the corollary that $\displaystyle \lim_{n \longrightarrow \infty} F_n/F_{n-1} = \phi$.

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1) Many elementary binomial identities or identities with Fibonacci numbers have beautiful proofs. Let me only mention the matrix representation of Fibonacci numbers whose determinant gives Cassini's identity.

2) Another elementary problem is the following: Is it possible to cover a checkerboard from which one white and one black square have been removed with dominoes? To show that this is possible run through the board in a cyclical way. Observe that on this path between a white and a black square are an even number of squares. Since I don't know how to make figures I indicate such a path for a 4x4-board: ((1,1),(1,2),(1,3),(1,4),(2,4),(2,3),(2,2),(3,2),(3,3),(3,4),(4,4),(4,3),(4,2),(4,1),(3,1),(2,1)).

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I would also add the (checkerboard) proof that if you remove opposite corners from the 8x8 board, the resulting figure can't be tiled with dominoes; if presented without the board coloration it's not immediately obvious, and its proof provides a wonderful a-ha moment that should be easily accessible for high school students (if not earlier). These (along with the various graph-theory problems) also have the advantage of showing students that mathematics is about more than just numbers. –  Steven Stadnicki Sep 8 '11 at 21:31

The theorem of "friends and strangers": the Ramsey number $R(3,3)=6$. Not only can the proof be understood by high-school students, a proof can be discovered by students at that level via something akin to the Socratic method. First students can establish the bound $R(3,3) > 5$ by 2-coloring the edges of $K_5$:
           R33
Then they can reason through that a 2-coloring of the edges of $K_6$ must contain a monochromatic triangle, and so $R(3,3)=6$: in every group of six, three must be friends or three must be strangers.

After this exercise, an inductive proof of the 2-color version of Ramsey's theorem is in reach.

An added bonus here is that one quickly reaches the frontiers of mathematics: $R(5,5)$ is unknown! It can be a revelation to students that there is a frontier of mathematics. And then one can tell the Erdős story about $R(6,6)$, as recounted here. :-)

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R(3,3)=6 sticks in my memory as the one time I managed to explain mathematics to a non-mathematical friend in the pub –  Yemon Choi Sep 8 '11 at 21:57
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This might be apocryphal, but I've read a story about a sociologist who was surprised to discover such patterns of friendship or non-friendship among his subjects, and pondering the deep psychological origins. If true, a wonderful argument for the need for math literacy. –  Thierry Zell Sep 9 '11 at 1:11
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From Jacob Fox's lecture notes: "In the 1950’s, a Hungarian sociologist S. Szalai studied friendship relationships between children. He observed that in any group of around 20 children, he was able to find four children who were mutual friends, or four children such that no two of them were friends. Before drawing any sociological conclusions, Szalai consulted three eminent mathematicians in Hungary at that time: Erdős, Turán and Sós. A brief discussion revealed that indeed this is a mathematical phenomenon rather than a sociological one." –  Joseph O'Rourke Sep 9 '11 at 12:07
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Re the Szalai story: $R(4,4) = 18$. –  Joseph O'Rourke Sep 9 '11 at 13:36
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I finally tracked down where I must have read about Szalai's story for the first time: N. Alon and M. Krivelevich's article on extremal combinatorics in the Princeton's companion. Also available at: cs.tau.ac.il/~krivelev/papers.html –  Thierry Zell Sep 29 '11 at 17:30

The proof that $\sqrt{2}$ is irrational is a nice example of proof by contradiction.

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@Manya My conception of beauty has changed since I first saw that proof. At that time, I did indeed consider it quite beautiful. Now, however, it's so embedded in me that it's hard to appreciate its beauty. –  Quinn Culver Sep 8 '11 at 13:58

Going by the parameters of the question, I don't see why the proof would necessarily need to be of a sophisticated theorem. I think Euclid's proof of the infinitude of primes is beautiful and definitely accessible to a high school audience. Having given the proof, one might reflect on some of its features that generalize to many other contexts, like proof by contradiction or the ability to use a clever construction to avoid infinite enumeration.

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No! Do not use this proof as an occasion to talk about proof by contradiction. Euclid's proof of this proposition was not by contradiction. The conventional practice of rearranging it into a proof by contradiction not only adds an extra complication that serves no purpose, but also leads to confusions such as the belief that if you multiply the first n primes and add 1, the result is always prime. See the paper by me and Catherine Woodgold on this: "Prime Simplicity", Mathematical Intelligencer, autumn 2009, pages 44--52. –  Michael Hardy Sep 15 '11 at 20:03
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It IS by contradiction, just not the kind of contradiction we have in mind. –  Mohamed Alaa El Behairy Sep 28 '11 at 1:35
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To complete Michael Hardy's comment, Euclid's original proof proves the following statement: given any finite list of primes, we can extend the list by finding a prime not in the list. (Proof: multiply the primes in the list and add one; any prime factor of this new number is a prime not in the original list.) So it's a constructive way of taking a list of primes and producing another prime; we don't have to assume that the original list was "all the primes" (as in the contradiction proof) or that they were the first n primes. –  shreevatsa Dec 23 '11 at 4:42

Euler's Bridges of Konigsberg problem. You can give it to students for five minutes to play with, watch them get annoyed, and then offer them the classical simple and beautiful impossibility proof. I think a lot of high school students, and even bright middle school students, would be totally convinced.

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One of the keys to making a proof accessible to high school students (or just non-mathematicians) is to make the answer relevant. This gives a dual responsibility, to ensure that the theorem is motivated and that the proof is accessible. The proof of the infinity of the primes has been mentioned already and is a fantastic example. You can lead students in to it using the (almost trivial) proof that there is no largest integer.

Another example is the classification of the regular polyhedra. With good students and models you can even lead them to the proof there there are at most 6 regular polytopes in 4d (actually showing they all exist is a little harder).

Keeping with polyhedra, the Euler characteristic is also powerful. Start with balloons and get the students to draw lines freely so you get a tiling. Then get them to count faces, vertices and edges. David Eppstein collected 19 proofs to choose from, several of which would be perfect for non-mathematicians: http://www.ics.uci.edu/~eppstein/junkyard/euler/

As a final example (and to show that it does not have to be deep mathematics to motivate) you can consider the question of blocking a square on a chess board and filling the remainder with tromioes. It starts with a puzzle, you can get people to play with, and leads to a lovely induction proof: http://www.cut-the-knot.org/Curriculum/Games/TriggTromino.shtml

Actually polyominoes are a fantastic source of many other fun, non-trivial but accessible proofs.

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The Halting Problem. The first time I saw this was my senior year in high school and it completely blew me away. All you need is a notion of what an algorithm is and very basic logic (enough to recognize that assuming $A$ and deriving $\neg A$ is a contradiction)

In a similar vein, Russell's Paradox. The problem here is that you need some basic set theory, so this is more for advanced high school students.

The first beautiful proof I saw in high school (it was beautiful at the time, but now seems too trivial) was the fact that for a geometric series $a, ax, ax^2, \dots$ the sum of the first $n$ terms is $a \cdot \frac{1-x^n}{1-x}$. I thought this was cool because of all the cancellation that seems to come out of nowhere. The treatment here is nice.

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If we're going for "beautiful" rather than just "neat", I'd vote for the Eckmann-Hilton argument. Although it's reasonably abstract for a high-schooler, it should still be quite accessible, and has a lot of "beautiful" symmetry, especially if you look at the nice circle of proofiness.

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Do you really think middle school or high school students will appreciate proving that a monoid is commutative, or the applications to higher homotopy theory? Perhaps I went to the wrong middle school. –  Douglas Zare Sep 8 '11 at 20:44
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Depends on the high school students. The assertion is certainly a nice first-round contest-level problem. And it is probably more interesting than many things done in school mathematics (then again, everything is...). –  darij grinberg Sep 8 '11 at 23:29
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How compelling do you think this symbol manipulation would be for a wide audience? Say, those who would not understand what you are asking if you ask them to show that a left identity equals a right identity. –  Douglas Zare Sep 9 '11 at 0:26

You should certainly look at the two books by Ross Honsberger, "Mathematical Gems I" and "Mathematical Gems II". A favourite example of mine the proof due to Conway that there are configurations of checkers below the half plane on an infinite board that allow you to move a checker 4 rows into the upper half plane, but not five rows.

The negative result is an ingenious argument using nothing more than the quadratic formula, but provides a great example of to apply mathematics in unexpected contexts.

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In the general game "Poset Chomp" the first player always has a winning strategy. The proof is a one-line strategy stealing argument, hence nonconstructive. In fact, a winning strategy is unknown in most cases, which makes the result interesting and mysterious. For a good quick account see here.

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@GH: I'm aware of the history of chomp and of simple posets and the strategy-stealing argument. Have you worked with middle school students who were not selected to be competitors in a math competition? What percentage do you think will understand and be impressed by a nonconstructive existence result about an abstract game they have not seen before? When I tell members of the general public about mathematics, I try to relate it to concrete objects and situations I think they know beforehand. –  Douglas Zare Sep 9 '11 at 0:08

I suggest Euler's polyeder formula with application to the Platonic solids. One first observes that instead of polyeders one can consider graphs in the plane, counting the unbounded region as a face. One observes next that one can just as well allow the edges to be pieces of curves. Then one observes that the formula $F - E + V = 2$ is preserved if one edge or one vertex is added, thus the formula is proved by induction. Then use the formula to prove that there are no regular polyeders except the five Platonic solids as follows: faces can only be triangles, squares or pentagons, edges are always common to exactly two faces, and for the number of faces that meet in a vertex there are a number of possibilities that one checks. For each case, $E$ and $V$ can be expressed in terms of $F$, and the formula gives the possible values of $F$.

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The proof, by counting inversions, that you can't interchange the 14 and 15 in the 15 puzzle, just by sliding, is accessible to high-school students, introduces important ideas, and might be found beautiful.

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i suggest the proof archimedes wanted on his tombstone and its relatives. i.e. since two solids with the same horizontal slice area at every height have the same volume, hence by pythagoras, the volume of a cylinder equals the sum of the volumes of an inscribed cone and an inscribed solid hemisphere.

to generalize this, the volume generated by revolving a solid hemisphere around a planar axis in 4 space equal that generated by revolving a cylinder minus that generated by revolving a cone. Using the fact that the center of gravity of a cone is 1/4 the way up from the base, one obtains the volume of a 4-sphere, as pi^2/2 R^4.

a generalization of the first computation is that of the volume of a bicylinder (intersection of two perpendicular cylinders), since it is the difference of the volumes of a cube containing the bicylinder and a square based double pyramid also inscribed in the cube.

I find these beautiful, but of course that is subjective.

I also like euclid's argument for pythagoras, and for constructing a regular pentagon, but they are hard to reproduce here briefly.

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Sperner's lemma (in dimension 2 to keep it visual). The proof in Francis Su's Monthly paper, Rental harmony: Sperner's lemma in fair division is especially easy to visualize. Theris a non-empty content, you can have students ponder the role of the hypotheses. And fair division applications allow to motivate it via concrete applications.

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There's a bizarre application of Sperner's lemma that's hard for high-school students (but I did give a talk to a group of them about it): one can't dissect a square into an odd number of non-overlapping triangles, all of the same area. –  paul Monsky Sep 9 '11 at 4:53

Why not some elementary theorems of Euclidan geometry? As I recall, the more general and fundamental theorems were just taken as given in my schooling, but I think many of them can be given accessible and beautiful proofs. Here are some good ones:

1) The Pythagorean theorem. (many lovely proofs)

2) Parallelograms having congruent bases and heights have the same area. (Euclid's proof is pretty.)

3) Use 2 to derive that similar triangles have corresponding sides in common proportion.

4) Two distinct circles have at most 2 points of intersection.

5) Prove the formula for volume of a pyramid without using calculus.

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What I absolutely dislike about 4) is how it cements the common misconception that mathematics is about giving painstakingly difficult proofs to intuitively obvious statements. Part 3) is only slightly better in this aspect. The rest are pretty good. –  darij grinberg Sep 9 '11 at 11:50
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The arguments are not so painful. #4 merely involves some observations relating the center of a circle, isoceles triangles, and the fact that two distinct lines intersect at most once. In any case I disagree with the sentiment. Part of the mathematical way of thinking is resisting the urge to accept things just because they seem obvious at first, and always demand that your knowledge but put on a firmer footing. I believe this is the essential "life lesson" students should take from mathematics. Sadly it is not being imparted in today's secondary schools much. –  Monroe Eskew Sep 9 '11 at 19:00
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I disagree. In school, mathematical proofs are like castles built on sand - not only do most students never realize what they are for, but they often tend to be sloppy right up to flawed (not "flawed" in the sense of "informal", but flawed in the sense of arguments that wouldn't be accepted as a correct proof even in a published paper), and the idea that proofs can be interesting is totally missing (at best they are considered a necessary evil by students and teachers alike). Adding to this a "revelation" that mathematicians prove trivial things in complicated (for students, at least) ... –  darij grinberg Sep 9 '11 at 21:04
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... ways doesn't help. In reality, mathematics is maybe 1% about proving things that are intuitively obvious (even topology), and 99% about proving things that are either surprising or seem to be useful in proving surprising things. Skepticism is a good life lesson, but it is better taught by providing examples of false intuitively obvious assertions with counterexamples than by providing examples of correct intuitively obvious assertions with their seemingly redundant proofs. –  darij grinberg Sep 9 '11 at 21:07

I think the fundamentals of the theory of quadratic residues modulo a prime qualify.

It is easy to explain what residue classes modulo a prime $p$ are, and to formulate statements of this kind:

1. The product of two quadratic residues modulo $p$ is a quadratic residue.

2. The product of a quadratic residue and a quadratic nonresidue modulo $p$ is a quadratic nonresidue.

3. The product of two quadratic nonresidues modulo $p$ is a quadratic residue.

(Note that I am not counting $0$ as a quadratic residue, nor as a quadratic nonresidue.)

Now 1 and 2 are very easy to show. 3 is not. What do we do?

First, it is easy to see that every quadratic residue is the square of exactly $2$ distinct residues modulo $p$. Thus there are exactly $\frac{p-1}{2}$ quadratic residues modulo $p$. Hence, there are exactly $\left(p-1\right)-\frac{p-1}{2}=\frac{p-1}{2}$ quadratic nonresidues modulo $p$. Now let $a$ and $b$ be two quadratic nonresidues. If $ab$ is a quadratic nonresidue, then there are at least $\frac{p-1}{2}+1$ different residues $x$ modulo $p$ for which $ax$ is a quadratic nonresidue (namely, each of the $\frac{p-1}{2}$ quadratic residues qualifies as such $x$ (by statement 2), but the quadratic nonresidue $b$ also qualifies), which leads to at least $\frac{p-1}{2}+1$ different quadratic nonresidues (since distinct $x$'es lead to distinct $ax$'es), contradicting the fact that there are only $\frac{p-1}{2}$ quadratic nonresidues modulo $p$. Thus, $ab$ must be a quadratic residue, and 3 is proven.

This indirect argument is, I believe, understandable to high school students. The only two theorems we used are:

A. Every quadratic residue is the square of exactly $2$ distinct residues modulo $p$.

B. If $a$ is a nonzero residue modulo $p$, then distinct $x$'es lead to distinct $ax$'es.

Both of these theorems can be derived from the following well-known fact:

F. If a prime divides a product of two integers, then it divides one of these two integers.

Proof of A: Assume that $a^2 \equiv b^2 \equiv c^2 \mod p$ for three integers $a$, $b$, $c$ pairwise incongruent modulo $p$. Then, $a^2 \equiv b^2 \mod p$ rewrites as $p\mid \left(a+b\right)\left(a-b\right)$. Hence (by fact F), at least one of $a+b$ and $a-b$ is divisible by $p$. Since $a$ and $b$ are incongruent modulo $p$, this can only mean that $a+b$ is divisible by $p$. Similarly, $b+c$ and $c+a$ are divisible by $p$. But therefore $2a=\left(a+b\right)+\left(c+a\right)-\left(b+c\right)$ must also be divisible by $p$. Since $p$ cannot be $2$ (as there are no three integers pairwise incongruent modulo $2$), this yields that $a$ is divisible by $p$. Similarly, $b$ and $c$ are divisible by $p$, which contradicts with their being incongruent. This proves A.

The proof of B is much simpler. The fact F is also used in one possible proof of statement 2. (However we can also prove 2 using 1 by the same trick as we used to prove 3 using 2.)

We have thus used the fact F a lot of times, but other than that, we didn't apply anything nontrivial - not even the theorem that a nonzero polynomial over a field cannot have more roots than its degree (this fact is often used in university-level treatises of quadratic residues).

The hard part is to tell students what is interesting about quadratic residues. Maybe cryptography?

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After two concrete answers, let me give a third, rambling answer:

A few years ago I decided to completely quit elementary geometry because it was still taking up half of my time even as I was already studying in university. Given that I have been doing it for most of my schoolyears, this should give an impression of how much there is to be done there - and all of it is comprehensible to a good school student.

I will not go into details, but this is a community wiki post ;)

First, there are so many lines in a triangle meeting at one point that one can reasonably ask whether there exist three symmetrically-defined lines not doing so. (There are, of course: e. g., the reflections of the medians in the corresponding altitudes.) This does not change the fact that each concurrence theorem is still a nontrivial result asking for a proof. Some of the basic cases can be handled with Ceva; harder results can take a dozen of pages to prove. The points where these lines concur often have several equivalent characterizations and collinearity properties (like the Euler line); this led Clark Kimberling to make an encyclopedia of such points similar to Sloane's On-Line Encyclopedia of Integer Sequences. An easy example is the concurrence of the lines $AX$, $BY$, $CZ$, where $X$, $Y$, $Z$ are the points where the incircle of triangle $ABC$ touches the sides $BC$, $CA$, $AB$. (The point where they concur is known as the Gergonne point of triangle $ABC$, known as $X_7$ in 1.) A not-so-easy example: If $O$ is the circumcenter of triangle $ABC$, then the lines connecting $A$, $B$, $C$ with the circumcenters of triangles $OBC$, $OCA$, $OAB$ concur. (This is the Kosnita point, aka $X_{54}$ in 1.) Then there are more complicated things, like: Consider the points where the excircle of triangle $ABC$ opposite to $A$ touches the extended sides $AB$ and $AC$. Let $M_a$ be the midpoint between these two points. Let $M_b$ and $M_c$ be defined similarly. Let the incircle of triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at $X$, $Y$, $Z$. Then, the lines $M_aX$, $M_bY$, $M_cZ$ concur (at a point which is $X_{1122}$ in 1; this is something I have discovered back in schooltime when playing around with dynamic geometry software).

Of course, concurrent lines aren't even half of the fun. There are theorems like Feuerbach's, stating that the nine-point circle touches the incircle and the excircles. This is some centuries old. Here is one found in 2000 by Floor van Lamoen: The medians of a triangle subdivide it in six little triangles (of equal area, by the way); the circumcenters of these triangles lie on one circle! Or here is another tangency property: If the incircle of triangle $ABC$ touches the side $BC$ at a point $X$, then the incircles of triangles $ABX$ and $ACX$ touch each other.

All theorems I mentioned can be proven in a synthetic way, i. e., using merely the part of elementary geometry studied in school, without coordinates or overly long computations. ("Overly long" is subjective and I am well aware of this.) This makes the field completely accessible to students. This is not to say that advanced mathematics doesn't shed some new light on it. For instance, one could try generalizing the above-mentioned Kosnita point by looking for all the points $P$ such that the lines connecting $A$, $B$, $C$ with the circumcenters of triangles $PBC$, $PCA$, $PAB$ concur. The answer turns out to be that the set of such points $P$ is a cubic curve known as the Neuberg cubic of triangle $ABC$. The sheer amount of interesting points on it allow for some neat applications of the group law on cubics to elementary geometric theorems.

I have been rather sparse with sources, as I haven't been keeping track of them for years. Some can be found on my links page, but it has not been updated since 2008 or so. Nowadays Jean-Louis Ayme's blog is the best place to find more synthetic proofs than one could ever read. There are also some good books. (Sorry for linking to my page this often; it is the one place on the internet I am most familiar with...)

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The trefoil knot is non-trivial.

Proof: It has a tricoloring: alt text And the existence of a tricoloring is preserved by Reidermeister moves. QED

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+1 because it is so obviously technically uncomprehensible to a high school student (at myn hiogh school) yet plausibly beautiful, and beautifully plausible. –  roy smith Sep 16 '11 at 5:05
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@Roy: I disagree with you. I don't think that this is incomprehensible to high school students. The notion of knot is pretty intuitive, and it's very easy to explain what a tricoloring is. Also, the Reidemeister moves are pretty easy things to explain (ok -- I'm not talking about the actual proof that they generate everything). The most difficult aspect of the argument is probably to convince a high school student that there is need for proof. Namely, isn't the trefoil obviously non-trivial!? For that, it might be good to show them some monster unknots first... –  André Henriques Sep 16 '11 at 5:17
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When I was shown the Reidemeister moves in school, several of my classmates and I made the objection, in essence, that it wasn't clear that they generated everything. Worse, since we didn't have topology to work with, we didn't really have a "real" definition to compare it with, so it felt to us that the real issues were being swept under the rug. –  Daniel McLaury Jun 15 '12 at 18:49

The following post from the "Everything Seminar" blog would make an excellent lesson in my opinion (link is below). It starts from a simple, but clever "hats puzzle" and then presents an infinite version of the puzzle which is solved (quite amazingly and beautifully) using the axiom of choice. It exemplifies the gap between what we expect to happen and what actually happens which is encountered in mathematics from time to time. Also, you don't really need to introduce any complicated concept, only describe what an equivalence relation is. The mere definition of an equivalence relation is beautiful mathematics in my opinion and the way in which such a simple concept can be used to solve a difficult (impossible?!) problem as above shows how interesting and intriguing mathematics can be.

The relevant post is here: http://cornellmath.wordpress.com/2007/09/13/the-axiom-of-choice-is-wrong/

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I disagree with, "You don't need to introduce any complicated concept." Do you think the students are already familiar with the Axiom of Choice? They aren't even comfortable with infinite sets. Presenting counterintuitive results which they can't otherwise verify and which are not connected to anything they have seen may be damaging to students new to mathematics. They may well reject mathematics (and at least the AOC) as not sensible, and they may be right unless you are extremely careful in the presentation, much more so than the blog you link, which was not aimed at high school students. –  Douglas Zare Sep 16 '11 at 23:25
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The positive integers are not the only infinite set involved in this example. If it were, you wouldn't need to use the Axiom of Choice. You are using the Axiom of Choice on equivalence classes of subsets of the natural numbers. I really think this example would be damaging. Normally, we start proving things we think are obvious, then prove things we believe, then use proofs to establish the truths of things which are uncertain. These students might or might not have hit the first stage and you want to jump to proving paradoxes with questionable axioms. Students will reject math. –  Douglas Zare Sep 17 '11 at 0:18
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Also, I disagree that the Axiom of Choice is intuitive. I think it's only intuitive if you overlook what it is really saying, that there exist oracles for situations where we don't have oracles. (The reals can be well-ordered? Ok, tell me how to compare them, or the least element of this set.) You are asking students to accept the step of memorizing the output of an oracle they don't have, but which somehow "exists." This would have turned me off from mathematics. There are much better things to show middle school and high school students than paradoxes which follow from questionable axioms. –  Douglas Zare Sep 17 '11 at 0:26
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I agree with Douglas Zare that this topic would not be appropriate for the intended audience. But I disagree with him abut the axiom of choice "really saying, that there exist oracles for situations where we don't have oracles." The axiom of choice is about sets, not oracles (unless you take "oracle" to mean simply set), and it certainly isn't about our "having" anything. After all, we don't even "have" all the natural numbers. It seems to me that part of the problem with the hats puzzle is that the solution pretends that sets (obtained by AC) can be used as oracles. –  Andreas Blass Sep 17 '11 at 15:55

This is maybe ambitious, for the details are obviously not completely accessible to the high school level; but the beauty of the ideas is, and this video is really a superb example of divulgation. Smale's theorem on the eversion of the 2-sphere and Thurston's construction.

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Minkowsky's Theorem (every convex region in the plane of area greater than 4 that's symmetric about the origin contains a lattice point other than (0,0)) is not at all obvious (are you sure you can't squeeze a sufficiently large "blob of irrational slope" in there?) but has a beautiful, simple, and surprising geometric proof.

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Do you want to give a hint about the proof? –  Manya Nov 19 '11 at 12:22
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Draw the region $R$ in the plane. Cut the plane into 2x2 squares by cutting along the lines $x=2i$ and $y=2j$ for all integers $i,j$; each square contains some part of $R$ (possible none.) Stack the squares on top of each other. Since $R$ has area greater than 4, there exist two squares whose parts of $R$ overlap. Write down what this means algebraically, apply symmetry and convexity, and construct the nontrivial lattice point. –  user6542 Nov 20 '11 at 11:14

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