# Proof

To prove irrational numbers exist, we will prove √2 is irrational.
We will do this by assuming √2 is rational and proving that this assumption is false.

By definition, a rational number can be represented as a ratio of two integers. Any such ratio can be reduced to the lowest common denominator, i.e.; having no common factors.

If √2 is rational, the following equation can be proven either true or false.

a/b = √2
Where a and b are integers with no common factors.

(a/b)2 = (√)2
a2/b2 = 2
a2 = 2b2
ergo a2 is even, since it is twice b2
ergo a is even, since only even numbers have even squares

Since a is even, we can replace it with 2c where c is half of a

Restate the equation:

2c/b = √2
(2c/b)2 = (√2)2
4c2/b2 = 2
4c2 = 2b2
2c2 = b2
ergo b2 is even since it is twice c2
ergo b is even, since only even numbers have even squares

We had postulated that a and b had no common factors. However, we proved both are even integers and thus had the common factor of 2. This demonstrates the postulated equation is false.

Ergo, postulating √2 is rational is false, and it must therefore be irrational, Q.E.D.