The Square Root of 2 is irrational

Theorem 7: The Square Root of $2$ cannot be represented by the ratio of two integers

Proof:

(1)  Assume that the squre root of $2$ could be represented by a ratio of two integers with $a$ and $b$ the reduced form.

$$\frac{a}{b} = \sqrt{2}$$  

(2)  Squaring both sides:

$$a^2 = 2b^2$$

(3)  Since $a$ must be even, there exists $c$ such that $a=2c$ and:

$$a^2 = (2c)^2 = 4c^2 = 2b^2$$

(4)  It follows that $b$ must be even since:

$$2c^2 = b^2$$

(5)  Then there exists $d$ such that $b=2d$ and:

$$\sqrt{2} = \frac{2c}{2d} = \frac{c}{d}$$

(6)  But then we have a contradiction since $a$ and $b$ can be reduced to $c$ and $d$.