1. Activity 1-12 : Multiple-free sets www.carom-maths.co.uk

2. Let’s call a set of numbers multiple-free if there is no pair of numbers in the set such that one is a multiple of the other. So { 3, 4, 5} is multiple-free, while { 3, 4, 6} is not.

3. Task: you are given the whole numbers from 1 to 20. What is the largest multiple-free set of these you can pick?

4. The primes? {19, 17, 13, 11, 7, 5, 3, 2} 8 numbers. If we take off 2 and 3, and add some others... {19, 18, 17, 13, 12, 11, 8, 7, 5} 9 numbers.

5. Or we could take the top half... {19, 18, 17, 16, 15, 14, 13, 12, 11} 10 numbers.

6. Conjecture: if we pick n + 1 of the numbers from 1 to 2n, this set will never be multiple-free. Call this statement S n. Task: prove S n for all n  1.

7. So S 5 says: if we pick 6 of the numbers from 1 to 10, this set will never be multiple-free. So S 14 says: if we pick 15 of the numbers from 1 to 28, this set will never be multiple-free. if we pick 106 of the numbers from 1 to 210, this set will never be multiple-free. So S 105 says:

8. Task: you are given a 2 by 2 square, and you are asked to pick five points. Prove that two of the points must lie within √2 of each other. And now for something apparently unrelated...

9. This looks tough initially, but The Pigeon-hole Principle makes it easy. What does this say? If you have n + 1 letters and n pigeon-holes to post them into, then some pigeonhole must contain at least two letters. This sounds obvious, but it is a wonderfully useful tool for proving all sorts of things.

10. Task: we know n people enter a room and some handshakes take place. No one shakes their own hand! Show that there is always some pair of people who have shaken hands the same number of times. Could The Pigeon-hole Principle help us here? There are two cases: 1.Someone shakes everybody else’s hand 2.Nobody shakes everybody else’s hand

11. If somebody shakes everybody else’s hand, then the possible handshake-scores around the room are 1, 2,..., n  1. No one can score 0, since everybody’s hand has been shaken. We have n  1 possible scores and n people, so by the Pigeon-hole Principle, two people have the same score. If nobody shakes everybody else’s hand, then the possible handshake-scores around the room are 0, 1, 2,..., n  2. Once again we have n  1 possible scores and n people, so by the Pigeon-hole Principle, some two people have the same score.

12. Let’s return to our five points problem. Divide the square into four equal ones. We have to pick five points, so by the Pigeon-hole Principle, two must be in the same square. The furthest apart that two points in the square can be is √2, so we are done.

13. Proof: Every integer can be written as 2 u v, where v is odd, and u  0. So we can write the 2n numbers in this form, a power of 2 times an odd number. Returning to our original problem...

14. Now pick n + 1 numbers from these 2n numbers. There are only n possibilities for v, so we must (by the Pigeon-Hole Principle) pick two numbers with the same v. So we write the 2n numbers in this form, 2 u v, where v is odd, and u  0. But 2 p v always divides 2 q v, if q > p, so we now have a pair in our set where one number divides the other, and so the set cannot be multiple-free.

15. We will use a method of DESCENT. We have a conjecture about whole numbers. If this is false, there must be a smallest example contradicting this. So we cannot find such an example at all, And we have proved our conjecture. It’s a powerful technique! The Minimal Criminal... Now construct from this a smaller such example. But a smaller example cannot exist! Alternative proof:

16. Suppose our conjecture is false. Let n be the first number for which S k fails, so we can pick n + 1 numbers from the first 2n so that our set is multiple-free, and we cannot do this for any smaller k. We are now going to attempt to show that (if the above is true) we must be able to find n - 1 numbers from 1 to 2n - 2 that are multiple-free. In which case, we have a contradiction! Conjecture: if we pick n + 1 of the numbers from 1 to 2n, this set will never be multiple-free. Call this statement S n. Task: prove this for all n  1.

17. There are 3 cases: Case 3: delete one number from the set of n + 1 you have chosen – you now have a set of n numbers from 1 to 2n - 2 which is multiple-free. Case 2: delete either 2n or 2n - 1 from the set of n + 1 you have chosen – you now have a set of n numbers from 1 to 2n - 2 which is multiple-free.

18. Case 1: neither n nor any factor of n can be in your set, because 2n is. So delete 2n and 2n - 1 from your list and add n. You now have n numbers between 1 and 2n - 2 that are multiple-free. So for Cases 1, 2, and 3, we have found a multiple-free set for the numbers from 1 to 2n - 2. But S k was meant to be impossible for any number less than n... We thus have a contradiction, and so choosing a multiple-free set of size n + 1 from the numbers from 1 to 2n must be impossible. What about Case 1?

19. With thanks to: Nrich, and Mazzafarius. Carom is written by Jonny Griffiths, hello@jonny-griffiths.nethello@jonny-griffiths.net