I've no idea how to prove pi is irrational but folks might be interested to
see how root 2 can be shown to be irrational, i.e.
there is are no integers p,q such that (p/q)^2 = 2
The following is a proof by contradiction i.e. we assume something is false
and show it leads to a contradiction.
Suppose the hypothesis sqrt(2) is irrational is false. Then we have two
integers p,q satisfying (p/q)^2 = 2
We further assume that the fraction p/q is in its lowest terms i.e p,q have
no common factors
Then we have
p^2 = 2q^2
hence p^2 is even.
I first show that if p is odd then so is p^2.
First observe that any integer of the form 2k+1 (k an integer) is odd
Let p = 2k+1
then p^2 = (2k+1)^2 = 4k^2 + 2k + 1 = 2(2k^2 + k) + 1
which is also odd because (2k^2 + k) is an integer.
Furthermore if p is even then so is p^2, for
let p = 2k (k an integer) then p^2 = 4k^2 = 2(2k^2) which is even.
Therefore p is even because p^2 = 2q^2
So we can write p = 2k for some integer k then
(p/q)^2 = (2k/q)^2 = 4k^2/q^2 = 2
therefore 2k^2/q^2 = 1
hence 2k^2 = q^2
So q is even as well!
This contradicts our original assumption that p,q have no common factors. If
p,q were not in thier lowest terms then we just keep dividing out the common
factors until they become so. But as the above shows that still leads to the
contradiction that both integers are even.
Therefore p,q cannot exist.
Now what's the big deal I hear you ask.
Well if sqrt(2) is irrational then how do we measure distance accross the
diagonal of a unit square?
The theorem of Pythagoras tells us that, for example, a right triangle of
height 1 m and base 1 m will have a third side of sqrt(2) m - and two
asjacent sides of a square and its diagonal form such a triangle.
So, as with the circle, how do we deal with this?
Phil Hall
