An example of a beautiful proof that would be accessible at the high school level?

Question

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 Grünbaum 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 $BC,$ $AP/PD = \operatorname{Area} (ABC) / \operatorname{Area}(DBC).$) 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...)

Answer 1

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.

Answer 2

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 $K_5$:
          
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. :-)

Answer 3

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.

Answer 4

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.

Answer 5

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 $\displaystyle \lim_{n \longrightarrow \infty} F_n/F_{n-1} = \phi$.

Answer 6

The trefoil knot is non-trivial.

Proof: It has a tricoloring:

^(source)

And the existence of a tricoloring is preserved by Reidermeister moves. QED

Answer 7

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.

Answer 8

Cantor's diagonal argument.

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

Answer 9

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.

Answer 10

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: the round track puzzle. 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. [Update, a decade later: I have written an expanded version of the solution up as the solution to Exercise 5.2.4 in my Fall 2020 Math 235 notes.]

Answer 11

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

Answer 12

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.

Answer 13

This topological proof of the fundamental theorem of algebra is accessible to high school students, particularly those at the precalculus level.

     ^(source)

There are two major problems with this. First, while the winding number is intuitive, it takes effort to define it rigorously. Second, you also want to establish the basic property that the winding number doesn't change as you deform a curve without going over the origin, which again is difficult to establish rigorously without topology. Without these details, you might call this a hand-waving argument instead of a proof. It's good to give references to where these results will be established rigorously, and to give arguments for other results which are more complete.

Nevertheless, I like presenting this proof for several reasons. I think it's beautiful. Geometrically, what $x\mapsto x^n$ does to the complex plane is easy to understand, but many students have little intuition about what this map does, only what polynomials look like on the real line. So, this argument doesn't just say that the statement is true, it is illuminating. The fundamental theorem of algebra is also a result students encountered in algebra, but they usually don't know why it's called a theorem. This is also an opportunity to talk a little about what is studied in more advanced areas of mathematics. It can lead into discussions of topology or the difficulty solving polynomials by radicals.

Answer 14

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.

Answer 15

I've been collecting simple, often one-step, proofs.

http://www.cut-the-knot.org/proofs/index.shtml

Some I judge beautiful - these are listed separately.

Answer 16

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.

Answer 17

Minkowski'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.

Answer 18

Im in high school and i loved the proof of the fermat-toriccelli point of a triangle.

Answer 19

For someone in high school, I think it's good to prove that the sum of the interior angles of a triangle is $\pi$ if they don't know why. Personally, I was never shown why this fact is true, and I feel that it's generally a bad idea to not know why something in math is true, especially when the answer is pretty. My favorite proof is to think about how the normal vector changes as you walk around the triangle -- it's nice because it generalizes to other shapes (which may not even be polygons).

Answer 20

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)).

Answer 21

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.

Answer 22

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 $F$.

Answer 23

Here is one that I like and used it for different purposes, e.g. introduction to proofs, algebraic thinking, beauty, and so one. Shuffle a deck of cards. Divide it into two halves. Magic: The number of the red cards in one of the halves is exactly equal to the number of black cards in the other half. It has a simple algebraic proof. However, The bigger magic is yet to come.

Theorem: Vertically opposite angles are equal.

Bigger Magic: The two theorems/proofs are essentially the same!

Answer 24

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

Answer 25

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.

Answer 26

Using Euler's formula ($F-E+V=2$, as mentioned earlier), it can be proved that the graphs $K_5$ and $K_{3,3}$ are not planar. I think these proof and also the proof of Euler's formula are simple enough to be understood by an interested high school student.

Answer 27

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.

Answer 28

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

Answer 29

Those are pretty nice, and can be done at a fairly low level (say, from 12 years old onwards) :

• The proof of the formula "half base times height" for the area of a triangle, by first considering a right triangle and completing a rectangle, then considering an arbitrary triangle and breaking it in two along an height (two cases : inside(+) or outside(-)) : it is a nice example of how mathematicians treat the general case by reduction to particular cases ;
• Euclid's proof of the pythagorean theorem using the previous formula, as in this animation.

Answer 30

Fermat's Little Theorem: If $p$ is prime and does not divide $a$, then $a^{p-1} \equiv 1 (\mbox{mod } p)$.

Proof: List the multiples of $a$ up to $a(p-1)$:

$$ a, a2, a3, \dots , a(p-1).$$

For any $r$ and $s$ with, $ra \equiv sa (\mbox{mod } p)$, we have $r \equiv s (\mbox{mod } p)$, so that the list above contains $p-1$ many distinct numbers.

Thus, the list above is some ordering of the list $1, 2, 3, \dots p-1$ modulo $p$. This gives us $$ a\cdot a2 \cdot a3 \cdot \cdots a(p-1) \equiv (p-1)! (\mbox{mod } p) $$

Finally we see

$$ a^{p-1} \equiv 1 (\mbox{mod } p). $$

Answer 31

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.

Answer 32

I like the lovely theorem in 19th century Euclidean Geometry as follows.

Let ABC be a triangle. let D,E,F be points on BC,CA,AB respectively. Then the circumcircles of AFE, BDF, CDE meet at a point.

I like this because the proof uses the property of the angles of cyclic quadrilaterals, and its converse. Also if one wants to convince students of the necessity of proof, then one should start with a result which is surprising.

It is a good thing that this situation can we worked on for more implications. Let P,Q,R be the centres of the three circles just given. Then the triangle PQR is similar to the triangle ABC.

For all these reasons I think it is a pity that some of Euclidean Geometry is not in University courses, or often school courses, in order to acquaint students with something important in our mathematical heritage. Should a student get a degree in maths without knowing why the angle in a semicircle is a right angle?

Answer 33

I actually think that Hilbert's Third problem is one of the explainable for school guys. It's even more cool that it exists in such a famous list close to the problems that are so tempting and not yet solved. The question is: can one cut the cube in some polyhedral pieces, reglue them and get a regular tetrahedron? The answer is no and the theorem was proved by Dehn using so-called Dehn invariant. It uses some algebra and number theory but can be understood by high-school level guys. The time you need to explain this is 3-4 hours, so maybe it could be a little and nice course. See, for example, Lectures on Discrete and Polyhedral Geometry by I. Pak

Answer 34

If each brick in a tiling of a rectangle has an integer side, the rectangle does too. This has various generalizations and lots of proofs, some very accessible, like number 7 here:

https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/Wagon601-617.pdf

Answer 35

The most success I have ever had teaching proofs at secondary school level is with the Peaucellier–Lipkin linkage. The proof relies on nothing more than basic geometry, namely similar triangles, but the outcome really is amazing. I found it reading Tom Körner's book called Fourier Analysis.

You can get the proof from Wikipedia, and there are some videos around if you google for them. Körner goes into the history of the problem, he believes Tchebychev was of the opinion that the problem couldn't be solved! And there is a great quote attributed to Kelvin, which I'll leave you with. When Sylvester showed a working model to him...

'[he] nursed it as if it had been his own child, and when a motion was made to relieve him of it, replied "No! I have not had nearly enough of it - it is the most beautiful thing I have ever seen in my life."'

Answer 36

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.

Answer 37

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...)

Answer 38

There are a lot of good suggestions in this feed, but here are a few problems that let you introduce modular arithmetic.

First, one can easily prove that an integer mod 9 is equal to the sum of its digits mod 9.

Second, you can prove Fermat's little theorem k^p mod p = k where p is prime.

I suppose that even (a+b) mod n = (a mod n + b mod n) mod n is kind of neat too.

You can prove that the calendar repeats itself every 28 years.

Answer 39

I love the incredibly clever proof of Sylvester’s theorem (https://en.m.wikipedia.org/wiki/Sylvester–Gallai_theorem) by Kelly. It is described nicely in the wiki page.

Answer 40

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.

Answer 41

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.

Answer 42

Quite elementary and quite beautiful: There exist two (infinitely many) irrational numbers $a,b$ such that $a^b$ is rational; and the usual proof with $\sqrt{2}^\sqrt{2}$. (Maybe starting with the proof of $\sqrt{p}$ being irrational for any prime $p>0$.)

Answer 43

I would put on such a list the proof of the Cauchy-Schwarz inequality directly from the axioms of an inner product using the discriminant criterion for the classification of the roots of a second-order polynomial with real coefficients. The latter is something all high school students either learn or already know, but it is a common target in those jokes about the parts of school math that most people find "useless"... The axioms for an inner product themselves are easy to motivate from $\mathbb{R}^2$ and $\mathbb{R}^3$.

Let me recall the proof here because it is that simple... Let $V$ be a (real) vector space with an inner product $\langle\cdot,\cdot\rangle$, and let $\vec{x},\vec{y}\in V$. We want to prove that $$\langle\vec{x},\vec{y}\rangle^2\leq\langle\vec{x},\vec{x}\rangle\langle\vec{y},\vec{y}\rangle\ ,$$ with equality iff $\vec{x}$ and $\vec{y}$ are collinear (i.e. linearly dependent). Since $\langle\vec{y},\vec{y}\rangle=0$ iff $\vec{y}=\vec{0}$ (nondegeneracy axiom) and in that case $\langle\vec{x},\vec{y}\rangle=0$ for all $\vec{x}\in V$ (due to the bilinearity axiom) the Cauchy -Schwarz inequality holds (with equality) if $\vec{y}=\vec{0}$, so there is no loss of generality if we assume $\vec{y}\neq\vec{0}$. In that case, define for $t\in\mathbb{R}$ $$P(t)=\langle\vec{x}+t\vec{y},\vec{x}+t\vec{y}\rangle=\langle\vec{y},\vec{y}\rangle t^2+2\langle\vec{x},\vec{y}\rangle t+\langle\vec{x},\vec{x}\rangle\ ,$$ where the last equality follows from the bilinearity and symmetry axioms. Now, due to the positive definiteness axiom (of which, let us recall, the nondegeneracy axiom is a consequence), we must have $P(t)\geq 0$ for all $t\in\mathbb{R}$. Recall now that the roots $t_\pm$ of a second-order polynomial $Q(t)=at^2+bt+c$ with real coefficients $a,b,c$ ($a\neq 0$) are given by the discriminant formula $$t_\pm=\frac{-b\pm\sqrt{\Delta}}{2a}\ ,$$ where the discriminant $\Delta$ of $Q(t)$ is given by $\Delta=b^2-4ac$. In the case that $Q(t)=P(t)$, one has that $\Delta=4(\langle\vec{x},\vec{y}\rangle^2-\langle\vec{x},\vec{x}\rangle\langle\vec{y},\vec{y}\rangle)$. Since $P(t)\geq 0$ for all $t$, this implies that $P(t)$ has at most one real root, which is only possible if $\Delta\leq 0$, whence the above inequality follows. The equality case happens if and only if $t=t_+=t_-$ is the only real root of $P(t)$ - for such a $t$, the nondegeneracy axiom implies that $\vec{x}=-t\vec{y}$, that is, $\vec{x}$ and $\vec{y}$ are collinear.

The above proof is completely general (works also for complex vector spaces and in infinite dimensions) and the actual result lies at the basis of Euclidean geometry, for one can get from it all the properties of the Euclidean distance and it also guarantees that the definition of the (Euclidean) angle between vectors makes sense through the cosine rule $$\angle(\vec{x},\vec{y})=\arccos\left(\frac{\langle\vec{x},\vec{y}\rangle}{\sqrt{\langle\vec{x},\vec{x}\rangle\langle\vec{y},\vec{y}\rangle}}\right)\ ,$$ so one can also use the latter to tell the students: "Do you see? The discriminant formula plays a role whenever you are doing any kind of Euclidean geometry..." In other words, it displays a deep connection about two seemingly completely different areas of mathematics (geometry and algebra).

Answer 44

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?

Answer 45

Well, I can't guarantee that I can make you happy, but atleast guess that I can. A simple problem: Determine the set of all points lying in the plane of a triangle ABC (say P), for which (PA)^2+(PB)^2+(PC)^2 is minimum. We all know that this set is the singleton set: {Centroid G of ABC}. Unfortunately, a previous knowledge of the answer makes it easy to prove the assertion, but more unfortunately, even after stating:"We will show that the centroid is the only such point", most books give a long, boring proof, involving non-trivial and non-motivating constructions and lengthy calculations. Right, I'm going to give a what-I-think beautiful proof, which is due to me! I solved it while preparing for the Indian National Mathematical Olympiad last year. Just join P with C1,A1,B1, the midpoints of AB,BC,CA,resp and form the triangle A1B1C1. Note that A1B1C1 is the image of ABC under a homothety of factor -0.5 about their common centroid. Next, applying the Apollonius' theorem on the 3 triangles APB, BPC & CPA and adding the 3 relations, note that PA^2+PB^2+PC^2 is minimum, if and only if PA1^2+PB1^2+PC1^2 is minimum. That the set above is singleton, is immediate from the extremal principle applied to 2 possible points having the same property and showing that their midpoint has the sum of squares less than them. Now, let P' be the image of P under the homothety -0.5. Then, by properties of homothety, P' and P both have minimum sum-of-squares with respect to A1B1C1. But the set is singleton, so P=P'. However, the only point that remains invariant under a homothety is the centre, which in this case, is the centroid!!:) Right, whenever one sees the sum-of-squares form one guesses to apply Apollonius' theorem and this proof doesn't even require a pre-knowledge of the answer! Please tell me if you've enjoyed it or not!

Answer 46

Well I personally liked the Euler's Theorem when I first saw, and I feel one can easily understand it at the High-School level.

There is a Youtube video with William Dunham's lecture on Euler which contains this theorem at 32:30 : https://www.youtube.com/watch?v=fEWj93XjON0

Answer 47

Figuring out the Lagrange interpolation polynomial was a pretty awesome moment for me as a high school nerd.

I was amazed a while later that you can simulate a Turing machine with just two counters, but that takes a bit of technical stuff to explain what a Turing machine is.

$x+1/x\ge 2$ if $x > 0$. Proof: $(\sqrt x-\sqrt{1/x})^2$ must be >=0, so expanding, $(x + {1\over x} - 2) \ge 0$. Not very deep, but kind of an aha moment in seeing reasoning appear from nowhere and immediately look obvious, getting rid of a calculus problem.

Proof of the triangle inequality in R**n, using Schwarz's inequality. Again, maybe the proof isn't beautiful in itself, but it was eye-opening in connecting geometry to analysis.

Answer 48

I would also like to add a different proof of Fermats little theorem ($p|(a^p-a)$ for prime $p$) to the list.

Suppose you have a colors of pearls and you want to produce pearl-chains of length $p$.

First you put $p$ pearls on a string. There are $a^p$ possibilities. Next you discard the mono-colored ones, they are boring. This leads to $a^p-a$ choices.

Next you put a knot into the string to turn it to a circular chain.

Now you have each type of chain multiple times, since cyclic permutations give you the same chain.

Finally you have to convince yourself that each type of chain arises in exactly $p$ ways using that $p$ is prime. Thus there are $(a^p-a)/p$ such different chains and hence that number has to be an integer.

Answer 49

Pick up any of the three volumes in Roger Nelsen's Proofs Without Words series and turn to any page. Many of the summation identities mentioned in this thread are included in these books.

Answer 50

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.

Answer 51

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/