For example, you have 10 black and 10 white socks in a drawer. If you pick out 3 socks at random, you’re definitely going to have a matching pair. Because even if the first 2 socks are of different colours, the third one will match up with one of them, right?

Let’s extend that sock example. You have 10 black, 10 white, 10 blue, 10 green, 10 red and 10 yellow socks. What’s the maximum number you need to pick out at random this time? The answer is 7, because even if you don’t get a matching pair in the first 6 (i.e., all of them are different colours), the seventh one will match up with one of them.

Okay, now let’s cut to the chase. The pigeonhole principle, though so simple, can be used to prove things that aren’t very intuitive at all. A famous example is proving that at least two people in London have exactly the same number of hairs on their head. This is because the number of hairs on a human head is definitely less than a million (10⁶), and the number of people in London is more than a million. Using the pigeonhole principle, we get the result. (how?)

How about another example? You have a set of integers. Prove that a subset of these can be chosen whose sum is divisible by n.

• +Solution!

  Take the n+1 integers given by: , ,  … until .

  Two of these integers have a difference that is divisible by n.
  (why? hint: how many possible remainders can you get when you divide a number by, say, 2? 3? n?)

  Taking the difference of those two integers, we get the answer.

• +Solution!

  Place an equilateral triangle on the plane anywhere, with side length equal to the distance given.

  Now since there are 2 colours and 3 points, at least two of those three points have to be the same colour (remember the first socks example?).
  So, we have found two points of the same colour at that distance apart.

Consider the set of numbers {-4, -3, -2, -1, 0, 1, 2, 3, 4}. Two players alternately choose one number at a time from the set (without replacement).

The first player who obtains any three out of his selected numbers (this may happen after the player has chosen 3, 4 or even 5 numbers) that sum to zero wins the game.

• +Okay, so here's a hint.

  Transform the problem to another very simple two-player game we all know so well. How many numbers do you have to pick from? Arrange them in a square.

• +And here's the solution. Give it a try first!

  Going through all the possible plays by both players might work, but why put in so much effort?

  This ingenious solution combines tic-tac-toe and magic squares.

  You have nine numbers, arrange them in a square such that each row and column sums to zero. It may not seem obvious at first, but a little trial and error will get you there. Zero, as intuition should tell you, is at the center. One such square is the following:

   1   2  -3
  -4   0   4
   3  -2   -1

  An arrangement of numbers in a square where all the rows, all the columns and the two diagonals add up to the same number, is called a magic square. That sum in question is called the magic sum of the square. Here, it’s zero.

  There are eight ways possible for three numbers from the set to sum in the original question to zero, and all of them have been taken care of in the above square.

  Now, whenever a person picks a number from the set, have them mark an X or an O (as in tic-tac-toe) on the magic square-board we’ve made above. As you can see, the game indeed ends when one person gets three-in-a-row, or equivalently, three numbers that sum to zero.

  Coming to forced winning, you might know that if both players play without making “mistakes,” then any tic-tac-toe game will result in a draw. Which essentially means, the answer to the original question is that: NO, neither player has a forced win in this game.
  

The first time I laid my eyes on it, I sat and thought for a moment. All I had to do was provide an example which satisfied the premise, and I was done. Or prove mathematically that it would never occur, of course, which seemed infinitely tougher.

Gut instinct would tell us that the latter would be true. I mean, how can ? Our basic knowledge (which is very basic) of the real number system leads us to guess so. However, as some of you may be knowing, in math, instinct can get the better of you when the answer is otherwise.

So I kept thinking, kept trying out trivial example to see whether I could find the easy way out. Unfortunately, my (relatively insignificant) efforts were in vain, and I couldn’t find an example as such. Then, almost convinced that my instinct was right, I moved on to proving that it would never happen. This was a more herculean task, as you would imagine, than providing a single example, and I had no idea where to start. And I gave up on that, too.

Having given my two cents to an attempt at answering the question, I reluctantly skimmed to the end of the book containing the solutions. I wasn’t worried that I couldn’t solve it myself, because you don’t get everything all the time, but there’s always that feeling of regret when you have to concede to that fact. As I read through the solution, I was impressed.

• +Click for the solution!

  First off, the answer is that it is possible. How? And what are those two irrational numbers? I may be able to answer the first question, but perhaps not the second. What do I mean? Well:

  Consider . We all know that it’s an irrational number. (the interwebs have many, many proofs)

  Raise this to the power of  and let the resulting number be denoted by , i.e., we get

  

  Any real number is either a rational or an irrational.

  1. Let’s assume that is rational. Then we’re done here, because we’ve found a case of a rational equal to an irrational raised to an irrational power.

  2. If, however, is irrational, raise x to the power , and call it . That is,

  

  This is now an irrational raised to an irrational power, since we considered to be irrational in the second case.

  And this turns out to be the

  

  And this is obviously rational.

  What do we conclude from this? In both cases (whether is rational or irrational), we can always find an example of an irrational to an irrational power resulting in a rational. And since one of the cases must be true, the answer is YES, it is possible, but we don’t know which case holds.

  This kind of proof is called a non-constructive proof, as it proves the existence of something, yet doesn’t provide a concrete example. This is why I mentioned before that I probably wouldn’t be able to answer the question “And what are those two irrational numbers?”

  But for the record,  is an irrational number. In fact, it is a transcendental number.

Countless people have fallen into the clever trap set by… whoever first proposed the question, I guess. Few have cleared this hurdle in their life, and some have even descended into a chronically deeply disturbed state, unable to come to terms with the answer that everyone else seems to have come to terms with. (alright, maybe I’m exaggerating a tiny little bit)

But what is the correct answer to this question? Think for a moment before you read on, although I know you’re not going to do it anyway.

• +The answer, to put it simply, is

  Yes. They are perfectly equal. Not happy? Read on.

Now, for the math. Here are three ways:

Note, there are also a few other methods on how to prove this. A simple one is based on geometric progressions, and you can find a more rigorous one using limits. I’m not gonna do that here.

All in all, I hope you’re finally convinced (if you weren’t before) that 0.999… = 1.

In fact, it’s just another way of writing a number with a finite decimal representation. For example, you know that

1/4 = 0.25

But, you can also write this as:

1/4 = 0.24999…

It’s exactly the same thing. Turns out it’s just a second notation, with minor differences. You learn something new everyday.

Here, we’re gonna play one game in particular, that we define ourselves.

Rook on the chessboard

A chessboard contains a lone rook, initially on the left-bottom corner square of the board. As you might know, a rook can only move in the horizontal or the vertical direction in a single move.

However, we’re going to add another restriction to the rook’s movement in this game:

The rook is allowed to move only towards the right, or upwards, in a single move. It is not allowed to move ‘back’ (i.e., left and/or down).

With this in mind, two players take turns in moving the rook. The player who moves the rook onto the square in the opposite corner, the one marked by the star, wins.

Note that a player cannot decide not to make a move; a move must be made every turn. So, it is clear that the game will end in a finite number of moves, and not go on forever.

• +Want a hint?

  Is there a specific position of the rook from where a player will surely win?

• +The solution!

  You might have already got the answer. The hint tells you to find a position from which one player would win. The key here is that once you find such a position, the aim of the game is now to get to that position, since you’re gonna win from there, anyway.

  Which position is this? It’s the square diagonal to the starred square. Why?

  Because: say the rook is at that position. Now, the next player (whichever that is) moves. The only possible moves are up or right. And obviously, the other player can move the rook accordingly onto the corner square and win.

  To get to the starred square, we need to get to the one diagonal to it. How do we do that? Wait… what about analysing it the same way again? The square diagonal to the one we just found?

  Well, it’s slightly different this time, since the next player can move the rook either one or two squares in a direction, but if he moves two squares, the other player wins anyway. So yes, our logic still holds. We need to get to the square diagonal to the square diagonal to the starred square.
  

  You must be getting the hang of it by now. Can you conclude something simple from this?

  To sum it up: if you can move the rook onto the diagonal when it’s your move, you win. Why?

  Because: the other player must move it out of the diagonal, and you’ll always move it back. So eventually, you will be the winner.

  And since the first player to play always has to move the rook out of the diagonal, the second player can always win the game.

So that’s a very simple example of a forced win in a two-player game. Games like chess are, of course, much more complicated. Which is why people still play chess. I mean, do you really think there would be international “Rook Game” competitions?

For the next in this series, go to Y(S)L – 2.

Let’s start off with something simple. Everyone knows how to reduce fractions to their simplest form, right? But what if I was annoyed at the conventional simplification, and started doing this?

16 / 64 = 1 / 4

Not good enough? You want to see more?

26 / 65 = 2 / 5
19 / 95 = 1 / 5
49 / 98 = 4 / 8

Why only two digits, you ask? Okay then. (from Wolfram Mathworld)

364 / 637 = 4 / 7

Still not satisfied? Maybe four digits is what you’re looking for. (note, got this one from here)

4277 / 2730 = 47 / 30

Not yet? Alright, you asked for it.

143185 / 1701856 = 1435 / 17056

But wait. This is only base 10. Why not try and find this in other bases?

15₆ / 53₆ = 1₆ / 3₆
2C₁₅ / C9₁₅ = 2₁₅ / 9₁₅

But how? Well, that’s something to think about. This phenomenon is called anomalous cancellation. Perhaps you can find a way to generate these strange cancelling fractions?