The Imperative "Root 2 is Irrational" Post

 Today, I couldn't think of anything better about which to write, so I decided to demonstrate that the square root of 2 is irrational. This is just one of those imperative posts that any blogger that pretends to care about the mathematical nature of reality needs to have.

 Firstly, what does it actually mean to be irrational? Well, there are really two definitions. The first is that you pour milk before cereal (Hah... see my attempt at humor?). The second is that the number can be expressed as a ratio between two integers in simplest form. The "in simplest form" part is really the crux of the argument here.

Alright, so let's first assume that root 2 is rational. This means that integers a and b exist such that sqrt(2) = a/b in simplest form.

This implies that a^2 = 2b^2 (I'll leave it to you to fill in those steps...). This means that a^2 is even, which means that a*a is even. So, we have two possibilities. Either a is even or odd. Suppose a is odd. It could then be written as 2*u + 1, where u is some integer. So, a^2 would be 4*u^2 + 4*u + 1 = even + even + 1. The evens could be represented as 2*k and 2*m for integers (I'll leave it to you to find k and m if you're curious, but it isn't relevant for the proof). So, a^2 = 2*k + 2*m + 1 = 2*(k+m) + 1 = even + 1 = odd. This would lead to a contradiction, so a must be even.

Now, if a is even, again a = 2*y for some integer y. Therefore, a^2 = 4*y^2. Thus, a^2 is a multiple of 4.

Now, b^2 = a^2/2 = 4*y^2/2 = 2*y^2. Thus, b^2 is even. It follows that b is even using the same argument above. So, both a and b are both even.

However, that means a/b can't be in simplest form. In simplest form, the factor of 2 for both a and b should cancel out. Thus, root 2 being rational leads to a contradiction. Therefore, root 2 is irrational.

Q.E.D.