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I've just been working with my 12-year-old daughter on Cantor's diagonal argument, and countable and uncountable sets.

Why? Because the maths department at her school is outrageously good, and set her the task of researching a mathematician, and understanding some of the maths they did - the real thing.

So what else could we have done - thinking that we know our multiplication tables and fractions, but aren't yet completely confident with equations which have letters for unknown numbers?

I did think of proving that there are infinitely many primes - we can follow an argument - other suggestions welcome.

And incidentally, tell your local high school to do this ...

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    $\begingroup$ How about how Archimedes was too smart to invent Calculus. I believe, at 12, you'r daughter know volumes and their formulae right? $\endgroup$
    – Dinesh
    Jun 1, 2013 at 18:33
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    $\begingroup$ I think Euler and the 7 bridges of Königsberg problem could be a good choice. $\endgroup$
    – user1242967
    Jun 1, 2013 at 18:47
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    $\begingroup$ The Euler bridges problem I will be using with 10/11-year-olds next month as a physical exercise, laid out on the playing field - thanks for reminding me. $\endgroup$
    – Mark Bennet
    Jun 1, 2013 at 18:51
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    $\begingroup$ @ShuhaoCao You've just reminded me how good some of the problems are in the Professor Layton series for Nintendo DS. $\endgroup$
    – Mark Bennet
    Jun 1, 2013 at 18:56
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    $\begingroup$ The irrationality of $\sqrt2$ would be a useful introduction to number theory and variables. And all it really takes is that the square of an even number is even and the square of an odd number is odd. $\endgroup$
    – Eric Jablow
    Jun 2, 2013 at 2:52

28 Answers 28

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Six people at a dinner party is sufficient to ensure that there are either three mutual strangers or three mutual acquaintances. In fact, six is the smallest number that ensures this phenomenon. This is the diagonal Ramsey number $R(3,3)$, and the proof can be demonstrated with a couple pictures and just a dash of the pigeonhole principle. There are lots of directions she could go after understanding $R(3,3)$ (though most of it is not due to Ramsey).

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  • $\begingroup$ Out of curiosity, what other non-Ramsey directions were you thinking about? $\endgroup$
    – user10575
    Dec 17, 2014 at 6:36
  • $\begingroup$ @Shahab I meant that the interested individual could move to larger classic Ramsey numbers, multicolor versions, infinite versions, non-clique versions, and so on. These all fall under the umbrella of "Ramsey Theory", but none of it was work of Frank P. Ramsey. $\endgroup$
    – Austin Mohr
    Dec 18, 2014 at 6:34
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Even though it implies a paradox, I like the Hilbert hotel. It can be explained to everyone, no matter the age. It deals with the concept of infinity, cardinality can be explained in an easy way, if all rooms are occupied and all costumers are in a room, then the "cardinality" is equal. And so on.

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  • $\begingroup$ We've discussed the Hilbert Hotel on car journeys - I have an older daughter too, and that hit her inquisitive years. Thanks for the suggestion - I'd forgotten. $\endgroup$
    – Mark Bennet
    Jun 1, 2013 at 19:57
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I would suggest Euler and his characteristic - for example, use it to show that there are only five regular polyhedra. One advantage of this subject is that one has to only work with pictures and integer numbers.

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    $\begingroup$ That's a great one too. There are all sorts of practical applications too - soccer balls make an interesting study. And why the Eden Project (which has domes in it) has pentagons in its domes as well as hexagons ... $\endgroup$
    – Mark Bennet
    Jun 1, 2013 at 18:59
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    $\begingroup$ One can even prove the five-colour theorem if the concentration span of the young person is sufficient ... $\endgroup$
    – Mark Bennet
    Jun 1, 2013 at 19:00
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    $\begingroup$ I just discovered that a very nice application of Euler's characteristic is the "3 houses" problem, where you have to connect 3 houses to 3 factories without crossing lines, which is impossible (in other words, show that the graph $K_{3,3}$ is not planar). $\endgroup$
    – Denis
    Jun 1, 2013 at 21:49
  • $\begingroup$ One can also use the Euler characteristic to study the numbers of hexagons and pentagons on a soccer ball! $\endgroup$
    – Adam Saltz
    Jun 3, 2013 at 2:40
  • $\begingroup$ This example admits interesting generalizations. $\endgroup$
    – J.G.
    Sep 22, 2020 at 19:14
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If we are talking about cardinals, there are two similar, but important things one can try and understand (which I believe are within the grasp of a 12 year old):

  1. There is a bijection between the integers and the rationals.
  2. There is no bijection between the rationals and the real numbers.

Equally great, Cantor's theorem.

Outside set theory, the existence of irrational numbers, including the irrationality of $\sqrt2$.

Then one can give the most exquisite proof that there are two irrational numbers that one is raised to the power of the second give us a rational number:

$$\sqrt2^\sqrt2\quad\text{ or }\quad\left(\sqrt2^\sqrt2\right)^\sqrt2.$$

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    $\begingroup$ Call it 'full pairing' or 'one-to-one correspondence'. $\endgroup$
    – Berci
    Jun 1, 2013 at 19:25
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    $\begingroup$ I don't think "one-to-one" is a good choice of words to describe a bijection (more just an injection), but "full pairing" does. Just my two pennies worth. $\endgroup$
    – Andrew D
    Jun 1, 2013 at 19:34
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    $\begingroup$ @Andrew: I don’t agree. The term one-to-one is symmetric: when we say in everyday speech that we can match these objects with those one to one or (more idiomatically) one for one, we’re talking about a bijection, not an injection. Moreover, it’s a well-established term that one must eventually learn anyway. $\endgroup$
    – Brian M. Scott
    Jun 2, 2013 at 6:20
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    $\begingroup$ @BrianM.Scott That makes a bit more sense now you explain it - I can't say I'd ever probably use it still, especially seeing as a "one-to-one function" is an injection while a "one-to-one correspondence" is a bijection; they seem like statements which take too long to write out for little gain. $\endgroup$
    – Andrew D
    Jun 2, 2013 at 8:34
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    $\begingroup$ @Mark: So two years later, can your daughter prove the independence of CH from ZFC by now? :-) $\endgroup$
    – Asaf Karagila ♦
    Jul 14, 2015 at 8:17
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One topic that might be accessible and interesting to explore are straightedge and compass constructions. Actually 'proving' certain things -- like the impossibility of trisecting an angle -- would probably be a little to advanced there are certainly many interesting things that can be done like bisecting an angle, constructing the square root of an integer, or the existence of irrational numbers. Additionally, this is something that would be quite hands on. After laying out the basic rules and a few basic techniques you could just let your daughter explore and mess around and see what she can come up with.

While this is not necessarily due to one mathematician you could tie this to how mathematicians, especially the ancient Greeks actually viewed numbers.

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  • $\begingroup$ Great suggestion! $\endgroup$
    – Mark Bennet
    Jun 1, 2013 at 19:31
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A group of $500$ students eagerly awaited their professor inside a university lecture hall. As the professor took the stage he posed a question to the students.

How many students here today share a birthday with another student in this lecture hall?

A) Less than 10%

B) Less than 25%

C) More than 25%

The professor counted a show of hands. A clear majority of students chose B. “No!”, proclaimed the professor; “C is the correct answer”. The professor then stated he could prove it without knowing a single student’s birthday. Perplexed, the students looked at each other and wondered if a few screws needed tightening.

Simply apply the Pigeonhole Principle, the professor explained. In the extreme, there will be $366$ students, each with a different birthday. The remaining $134$ students will share a birthday with another student. Therefore at least $135=134+1$ students share a birthday with another student. The students erupted in applause!

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    $\begingroup$ Dear professor: what if 364 of the students each have distinct birthdays and the other 136 all share the 'last' day of the year as their birthday? $\endgroup$
    – Steven Stadnicki
    Jun 3, 2013 at 1:01
  • $\begingroup$ @StevenStadnicki: Then each of those 136 students is counted 135 times - well over 50%. I think maybe the question might be rephrased as "How many birthdays are shared by two students in this lecture hall?" That would cover cases like this. $\endgroup$
    – Darrel Hoffman
    Jun 3, 2013 at 3:40
  • $\begingroup$ @StevenStadnicki. Very good point, I'll revise the answer to more than 25%. All the screws seem tight now :) $\endgroup$
    – cyclochaotic
    Jun 3, 2013 at 4:04
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A number of combinatorial games may be quite suitable.

  1. Move mountains. Start with any permutation of $[n]$. A legal move consists in moving any number that is a local maximum to the front of the permutation. E.g., the legal moves from $214365$ are to $421365$ and to $621435$. The first player to have no legal move loses. (It’s not too hard to see that this happens precisely when the position $n(n-1)\dots21$ is reached.) This is a good game for getting an understanding of $\mathscr{N}$ and $\mathscr{P}$ positions: the key insight is that if you can move $n$ to the front, either you win outright, or your opponent must make a move that lets you move $n$ to the front again on your next move.

  2. One-heap takeaway. Start with a heap of $n$ counters and a finite set $T$ of positive integers. At each move the player whose turn it is must take $t$ counters from the heap for some $t\in T$. The first player who cannot move loses. If $T=\{1,\dots,m\}$ for some $m$, the game is quite straightforward to analyze fully, and for any $T$ it leads naturally to the equivalence classes of modular arithmetic.

  3. Nim. Start with two-heap Nim; she can probably discover the winning strategy on her own with little or no help. She might well be able to follow the argument for the general strategy for any number of heaps, which is a nice application of binary representation.

A couple of questions to think about:

  1. If you have a two-pan balance and are required to put the object to be weighed in one pan and the balancing weights in the other, what is the most efficient set of standard weights that will allow you to weigh any whole number of ounces from $1$ through some $n$? (Most efficient requires some explanation and probably some examples, and it’s best to use $n$ of the form $2^k-1$.) This leads naturally to binary notation.

  2. If you have a two-pan balance and may put the standard weights on either pan, what is the most efficient set of standard weights that will let you weigh any whole number of ounces from $1$ through some $n$? (Here the most informative values of $n$ are of the form $\frac12(3^k-1)$. This leads naturally to balanced ternary notation, which is kind of fun to play with. E.g., subtraction is just inversion followed by addition, so it doesn’t require learning a separate subtraction table.

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The story of the grains of wheat on a chessboard (sometimes told as rice) offers several possible candidates for a mathematician, or at least a mathematically aware poet!

This story leads several interesting places, including the Law Of Big Numbers, which is more about economics than mathematics.

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Shuffle a standard 52-card poker deck and deal the cards into thirteen piles of four cards each. According to Hall's Marriage Theorem, it is always possible to draw one card from each pile in such a way that the thirteen chosen cards represent every denomination.

To see why, set up a bipartite graph on vertex set $U \cup V$ where

  • $U$ is the set of all thirteen denominations,
  • $V$ is the set of all thirteen piles, and
  • $uv$ is an edge if and only if denomination $u$ can be found in pile $v$.

One can show that this graph satisfies the marriage condition, which requires that the number of vertices in any subset $S$ of $U$ is no more than the number of vertices in its open neighborhood. According to Hall's Theorem, the graph contains a perfect matching, which corresponds to the way in which the denominations should be chosen from each pile.

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  • $\begingroup$ I really like this as an idea, and of course it can be built up from simpler examples and the ideas developed to explain what is going on as abstraction is increased. It is the best of the ideas I would never have thought of - thank you for sharing it. $\endgroup$
    – Mark Bennet
    Dec 15, 2014 at 21:33
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I immediately thought of the formula for the sum of arithmetic progression. It is easy to understand, really practical and I think I've read somewhere it was proven by Gauss when he was in the 3rd grade.

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Quick Sort can be a funny exercise if you like Information Technology. Show her how to order ten cards using a recursive quick sort strategy. It is quite simple to show it is the fastest algorithm, just because its operations are a comparasion with some arithmetic inside.

It is a pragmatic stuff because you play cards often, and it is a golder ring connecting deep math to Information Technology in a very elegant way, indeed.

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    $\begingroup$ Unless you're using a very different definition of 'quicksort' than I am, I'm not sure how to easily show that it's the fastest algorithm; even the notion of $O(n\log n)$ is probably beyond the average 12-year-old. Radix sorts/merge sorts are, at least for me, much more practical to do by hand, and the natural sort for cards is arguably a form of radix sort... $\endgroup$
    – Steven Stadnicki
    Jun 3, 2013 at 0:59
  • $\begingroup$ @StevenStadnicki Quicksort isn't actually $O(n~\log n)$, try running quicksort on an already sorted list. $\endgroup$
    – DanielV
    Dec 9, 2014 at 15:51
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    $\begingroup$ @DanielV That depends entirely on your implementation. There are versions of Quicksort that can guarantee an $O(n\log n)$ runtime (at the expense of a higher constant) by running median-selection algorithms (which are $O(n)$ in each subinterval) to find a 'perfect' pivot. $\endgroup$
    – Steven Stadnicki
    Dec 9, 2014 at 16:14
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Have you heard of Erik Demaine? He's a mathematician at MIT who researches mathematical origami. Some of the stuff he does might be more interesting to a 12 year old.

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Another fun one is the tessellation of the sphere by regular polygons. You characterize all tessellations of the plane first by way of interior angles, and then show that the same can be done on the sphere. This is a great way of showing that Euclidean geometry isn't the only way to do geometry!

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The Monty Hall Problem has stumped smart people: it's fun and you can run experiments.

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I'm going to add the Russell Paradox to the list - simply because the consequences of considering it lead to some interesting places.

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Show her Colin Adam's knot book! Seriously interesting and accessible in the first part.

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When trying to teach someone about "equations which have letters for unknown numbers" I like to ditch the symbols and instead focus on the concepts they represent. Like this: http://worrydream.com/LadderOfAbstraction/

The have a look at some of the other the amazing work being done at: http://worrydream.com/KillMath/

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If she's into chess, explore various upper bounds for the number of games that can be played, constrained by a rule that if the board reaches the same configuration three times, that the game ends in a draw. (In real chess, if this happens a player has to choose to end it in a draw, but forget about that.)

An extremely naive upper bound is something like $2^{2\cdot65^{32}}(2\cdot65^{32})!$, taking the power set of possible board configurations and assigning an order. But then start considering things like disallowing two pieces on the same square, and the actual constraints on where pieces can be given the game rules, and in what order moves may be made...

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  • $\begingroup$ See this link for a discussion chess.com/blog/kurtgodden/the-longest-possible-chess-game. This takes the 50-move-rule into account. There is some counting and use of parity involved. It would be an interesting task to set a bright group. $\endgroup$
    – Mark Bennet
    Jun 2, 2013 at 19:33
  • $\begingroup$ Before things get better (via the constraints you mention), they get worse because you need to take into account the possibility of promoted pieces. Also, the configuration of pieces may not be enough to determine whether castling is legal (did the king or rook move previously and then return to its starting position?) and whether an en passant capture is legal (did the pawn to be captured move two squares on the immediately previous move?); so, in addition to the locations of pieces, you sometimes need a little bit of the game's history. $\endgroup$
    – Andreas Blass
    Jun 2, 2013 at 22:02
  • $\begingroup$ @Andreas All these things make it an interesting problem to work on because there is no nice and tidy solution. There is always going to be some complication that you can factor in that will allow you to reduce the bound, if you work hard enough. You have alerted me to a miswording in my answer though. $\endgroup$
    – 2'5 9'2
    Jun 2, 2013 at 23:07
  • $\begingroup$ @Andreas Regarding promoted pawns, they are still the same pieces they always were, just with a new face. So possibly such a pawn can be in some specific spot four times, rather than two. And we can replace the $2\cdot65^{32}$ with $4\cdot65^{32}$, if I am counting right. $\endgroup$
    – 2'5 9'2
    Jun 2, 2013 at 23:13
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Some things that can capture fanciful imagination at younger age, to keep up enthusiasm later on:

1) V + F = E + 2 , leading up to the Euler characteristic, topology..

2) The Moebius Band has only one side.

3) After basically understanding what the imaginary number $ i$ is, how come $ i^i $ could be not just one, but an infinite set of real numbers !

4) How so, unrelated numbers are beautifully integrated by the Euler relation $ e^{2 \pi n i} = 1? $

5) When a thin plastic ball is bent, the product of the radii of curvatures cannot change, just as in a gas, pressure volume product cannot change ( isometry , isothermic expansions ).

6) The ever strange link between physics and maths.

7) A browsing through a well illustrated intermediate level popular math books will offer several pictorial themes.

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I'm adding another - the 17 Wallpaper Groups, which the link here says

"A proof that there were only 17 possible patterns was first carried out by Evgraf Fedorov in 1891 and then derived independently by George Pólya in 1924."

Because there is a really nice exposition in "The Symmetries of Things" by John H. Conway, Heidi Burgiel & Chaim Goodman-Strauss. The book has some great illustrations.

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    $\begingroup$ I think the book is one of the best mathematical books of the last ten years - I've give four copies away. $\endgroup$
    – Mark Bennet
    Dec 14, 2013 at 21:27
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  1. Anything from N.Vilenkin's "In Search of Infinity" will do: http://www.amazon.com/In-Search-Infinity-N-Ya-Vilenkin/dp/0817638199/ref=sr_1_2?ie=UTF8&qid=1393976309&sr=8-2&keywords=Vilenkin (problems related to infinity, like a space hotel with infinite number of rooms).
  2. "Word substitution" problem: there are 3 words A,B,C, we can substitute any subword B in the word A with the word C. When this process (B->C) can be repeated infinitely for given A,B,C? I'm not sure there is a known solution in general, but even finding any A,B,C which provide an infinite example is interesting.
  3. Graph theory problems (Königsberg Bridge Problem,...).
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Why not expand into some other interesting diagonal arguments? I believe a proof of the halting problem could be quite accessible. If you are feeling more daring, you could try Gödel Incompleteness.

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I think that it is best to be a combinatorial problem. In my opinion, the problem should be very easy to understood, not-so-easy to approach, and could have a lot of potential "connected problems" or generalization. But I think that it should not be a problem where there is no resolution to (the simple variant).

This leads to to thinking about paths on a graph (konigsberg bridge). Another thing is Ramsey theory. They are both easy to understand but not so easy to approach and a certain kind of mathematical thinking is needed for both.

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How about a cool sieve? An easy to explain and in my opinion fascinating sieve to generate powers of natural numbers is stated in this question. You can show connections with the Pascal triangle and you can talk about algorithms. And, maybe most important, you can encourage your daughter to detect interesting connections of natural numbers by herself! :-)

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Here is a mathematical "magic trick" I learned from TV when I was about 8:

  • Pick any number
  • Add 5 to the number
  • Multiply by 2
  • Subtract 8
  • Divide by 2
  • Subtract the original number you picked

The result is 1. It's like a magic trick that is an easy introduction to algebra.

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  • $\begingroup$ Program test, for disbelievers: var one = x => ((((x+5)*2)-8)/2)-x (JavaScript). $\endgroup$
    – Conor O'Brien
    Dec 9, 2014 at 1:27
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How about teaching her some introductory group theory? Its relation to symmetry makes it very accesible via geometric examples.

To make it more appealing you could use:

The Rubik's cube (2x2x2, 3x3x3 and so on).

Crystallography. Maybe taking her to some museum or geology department to see some crystals.

Music! Some people like to mix music an mathematics and see what happens, I'm not familiar with this work, but you could do some digging yourself.

Some western symbols like mandalas or islamic art.

I can't think of any other examples ATM, but this pretty cool MO link might be helpful.

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Pythagoras' theorem!

Proof using a rotated square within a square.

This gets straight to the essence of mathematics!

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  1. Show her that 2 + 2 may not equal 4, but for example may equal 12, and explore the implications of that fact with her! Subtracting from each side 2 and then 2 again you'd get 0 = 8, and this would be a great intro to modular arithmetic.

Moreover, this is a great example of a situation in which dismissing what's known and allowing a seemingly wrong answer can open new doors. And it is also a way of showing that, in maths, there are no wrong ways (unless you contradict yourself, of course), but only roads that may lead nowhere (but that are worth exploring though).

  1. An intro to quotients with simple and concrete examples would be mind-blowing as well, I believe. Let her come up with her own answer to what happens if, on the [0, 1] segment, we make 1 and 0 be the same element (we get a circle!). Then you can go on with more difficult examples and build a cylinder, a torus and a mobius string from a unit square! Funny enough, modular arithmetic is based upon quotients!

  2. Show that the commutative property may not hold in infinite cases. Take the $\sum_{i=0}^\infty(-1)^n$ and ask her what she thinks it may sum up to. That sum waves up and down never approaching anything, but rearranging its terms it can be made divergent.

This is a great example of how things may not always be true in more difficult cases, even if they are in simpler ones.

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  • $\begingroup$ @downvoter care to explain why these are not good examples? $\endgroup$
    – Ste_95
    Dec 2, 2016 at 19:59

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