Square Root of 2 is Irrational/Proof 3

Proof
Seeking a contradiction, assume that $\sqrt 2$ is rational.

Then $\sqrt 2 = \dfrac p q$ for some $p, q \in \Z_{>0}$

Consider the quantity $\left({\sqrt 2 - 1}\right)$:

Now, observe that for any $n \in \Z_{>0}$:

By Power of Number less than One:
 * $\displaystyle \lim_{n \mathop \to \infty} \left({\sqrt 2 - 1}\right)^n = 0$

where $\lim$ denotes limit.

Recall the definition of $a_n$ and $b_n$.

By Lower and Upper Bounds for Sequences:
 * $0 = \displaystyle \lim_{n \mathop \to \infty} \frac {a_n q + b_n p} q \ge \frac 1 q$

which is a contradiction.