Riddles in Mathematics: A Book of Paradoxes is a popular mathematics book written in 1944, revised in 1961. I purchased a 1975 reprint at Hay, and finally got round to reading it on holiday.
It is good fun, if a little dry and could be accused of being a little like a text book: there is a discussion, and couple of examples and then some paradoxes for you to resolve (or exercises). To get the most out of the book you do need to engage with the maths: you’ll need pencil and paper and a willingness to get your algebra out.
The dated bits are in geometry: sadly Euclidean proof has dropped off the school curriculum (in favour of more statistics). I think there is nothing in the section that isn’t in GCSE Maths, but probably in a different style. Also the four colour theorem has now been proved, albeit in an aesthetically unsatisfactory way.
The chapter on the infinite is great, and the chapter on probability has provoked me to get Grimmett and Walsh out to revise my probability.
A couple of my favourite paradoxes. Firstly one from p169 that I’ve put into steps and changed slightly.
1. If something has probability 0 it is impossible.
2. It is known there are an infinite number of primes
3. We only know a finite number of them (the highest known prime has just under thirteen million digits). Call it P.
4. The probability of picking a known prime is P dividied by the number of primes.
5. Any finite number divided by infinity is 0.
6. Therefore the probability that an arbitrary prime is known is 0.
7. Therefore it is impossible that an arbitrary prime is known.
8. Therefore there are no known primes
A round of applause to anyone who can pick the (many) holes in that. It is the same basic construction as Douglas Adam’s proof that the population of the universe is 0.
Also worth a look are the geometric probabilities (p172-3).
Some good ol’ algebraic sleight of hand is included, like this from p86:
1. Let b and c be two different positive numbers.
2. Let a = b + c. Note a must be greater than b.
3. Multiply by (a-b) to get a^2 – ab = ab – b^2 + ac – bc
4. Take off ac: a^2 – ab – ac = ab – b^2 – bc
5. Factorise: a(a-b-c) = b (a-b-c)
6. Divide by a-b-c to get a = b, but this contradicts a being bigger than c. So any two different numbers are the same (chose to be c=a-b)
7. In fact as b+c=a we have b+c=b, which means c=0.
8. But c was an arbitrary positive numbers, so all arbitrary positive numbers are 0!
Again a round of applause to those who can hit that one into touch. A bit easier I think.
I think you’d need to be a bit odd to enjoy the book: you’d need solid secondary school mathematics, and a willingness to play with the maths. Like me.