Okay so what I did was first, assume that square root of 2 is rational. let √2 = a/b
2 = a^2/b^2
2b^2 = a^2
a^2 is even therefore a is even so it can be written as 2n...
2b^2 = (2n)^2
2b^b = 4n^2
b^2 = 2n^2
2 = b^2 / n^2
√2 = b/n
Since we started with √2 = a/b and then got √2 = b/n...
Homework Statement
I know how to prove that square root of 2 is irrational using the well ordering principle but what I'm wondering is, how can we use the well ordering principle to prove this when the square root of two isn't even a subset of the natural numbers? Doesn't the well ordering...