Friday, August 8, 2008

Updated

The angle question seems to be solved. Well done to both Rick and Ben.

The square root of 2 thing, is actually quite easy for anybody interested, you begin by saying let root(2)= m/n where m,n are integers, n non-zero. Without loss of generality you can assume they have no common divisors because you could just divide them out and then start from there. You then find that 2 is a common divisor, thus there is a contradiction so the inition assumption that such an m and n exist is false. It's quite neat.

Have a good solid think at the other one. Its really quite amazing that it is possible I reckon.

1 comment:

Rick said...

Heh, yeah the sqrt(2) thing.

I remembered the fuzzy version of that proof, sort of like this:
"Assume sqrt(2) = m/n, m,n E R ... common divisor ... therefore a contraction, so sqrt(2) is irrational."

Tada!

I'm going back to the original post now to remind myself of the number three problem...

Oh, and do you know about the 4-colour theorem?
http://en.wikipedia.org/wiki/Four_color_theorem