Back when I was in school, my favourite mathematical game was to use a false axiom to rigorously prove a false conclusion. We as a species are not very good at handling either logic or untruths so its fun to try to kick a logical hacky sack around while you try to follow the rules.

If 2012 = 2011, show 11/23 = 23/11.

2011 = 2012 408*2011 = 408*2012 (408*2011)-820367 = (408*2012)-820367 121 = 529 11*11 = 23*23 11/23 = 23/11QED

Most purely arithmetic examples work out similarly, the only effort is stretching the falsehood to the correct size and then doing the legwork. But the same principle applies to much more interesting questions of the form “Assuming [mildly bizarre thing] show [very bizarre thing]”. Since one slightly wrong axiom poisons the whole soup, it’s interesting to take something that barely seems noticeable and stretch it with your wrong-fu:

**Assuming that the square root of two is rational, show that there’s an odd number of bytes in a kilobyte**

Since sqrt(2) is rational, let's write it as a/b where a and b are co-prime. (a/b)^{2}= 2 a^{2}/b^{2}= 2 a^{2}= 2b^{2}2b^{2}is even, so a^{2}must be as well odd*odd is odd, so a must be even let's take half of a and call it j, a=2j 2b^{2}=a^{2}= (2j)^{2}By the same process, b is even, so for some k, b=2k 2(2k)^{2}= (2j)^{2}

Normally we’d stop there because that’s ridiculous since 2m and 2k can’t possibly co-prime and so sqrt(2) isn’t rational. Instead I want to use that falsehood as a building block; we have a building block for building unlimited co-prime multiple powers of 2.

After 10 iterations we end up with: b=2^{10}x is co-prime with a=2j 2^{10}=1024 is co-prime with 2 2 does not divide 1024 1024 is prime

The difficulty is not commutative of course: if we assume 6 is the largest natural number, proving that the square root of two is not rational is easy (I mean it’s not 3/2, 4/3 or 6/5 and there aren’t that many possibilites^{1}) but show that 6 is the largest natural number in a world where the square root of two is rational is left as an exercise for the reader ^{2}.

And generally the more wild the false axiom, the easier the problem, it’s trivial to show that “all rectangles are squares” given that “all angles are 90^{o}“^{3} — it’s much harder working from “all parallel lines cross each other 2 times on a plane”

They’re hard to come up with though, since you don’t know how hard it’s going to be before you solve it:

- Assuming that division is commutative, (that is a/b=b/a for all b & a) prove that all numbers are rational
- Assuming there are a finite number of prime numbers, show there are a finite number of composite numbers
- Assuming all isoceles triangles are equilateral show that parallel lines always touch
- All numbers of the form a*2
^{b}-1 and a*2^{b}+1 are prime: prove the weak Goldbach Conjecture

The problem really is coming up with problems that are hard enough to be interesting — deciding which mathematical to mix up is far from trivial^{4}. I’ve left out a fair number of the ones I’ve solved since they turned out to be trivial or uninteresting.

So rather than presenting this as “Here’s my game you should play” this is more of a request: **can you think of any puzzles I can play logical-fallacy Calvinball with?**

May 1st 2013 update: naturally there’s a relevant XKCD

- These are games, so I’m not being rigorous in my definitions and we can assume rational numbers where the numerator and denominator are both negative are ruled out by magic ↩
- Seriously, I don’t know how I’d do this one ↩
- draw the X in the middle ↩
- I was hoping a list of paradoxes would be more helpful in coming up with questions ↩