Square Root of 2 is Irrational/Proof 3

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Theorem

$\sqrt 2$ is irrational.

Proof

Aiming for a contradiction, suppose that $\sqrt 2$ is rational.

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

Consider the quantity $\paren {\sqrt 2 - 1}$:

 $\displaystyle 1 \ \$ $\displaystyle <$ $\sqrt 2$ $\displaystyle <$ $\displaystyle 2$ Ordering of Squares in Reals $\displaystyle \leadsto \ \$ $\displaystyle 0 \ \$ $\displaystyle <$ $\sqrt 2 - 1$ $\displaystyle <$ $\displaystyle 1$

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

 $\displaystyle \paren {\sqrt 2 - 1}^n$ $=$ $\displaystyle \sum_{k \mathop = 0}^n \binom n k \paren {\sqrt 2}^k \paren {-1}^{n-k}$ Binomial Theorem $\displaystyle$ $=$ $\displaystyle \sum_{\substack {0 \mathop \le k \mathop \le n \\ k \, \text{even} } } \binom n k 2^{k/2} \paren {-1}^{n - k} + \sqrt 2 \sum_{\substack {0 \mathop \le k \mathop \le n \\ k \, \text{odd} } } \binom n k 2^{\paren {k - 1}/2} \paren {-1}^{n-k}$ $\displaystyle$ $=$ $\displaystyle a_n + b_n \sqrt 2$ for some integers $a_n, b_n$ $\displaystyle$ $=$ $\displaystyle a_n + b_n \paren {\frac p q}$ recalling the assumption that $\sqrt 2 = \dfrac p q$ $\displaystyle$ $=$ $\displaystyle \frac {a_n q + b_n p} q$ $\displaystyle$ $\ge$ $\displaystyle \frac 1 q$ as the numerator is an integer and $\sqrt 2 - 1 > 0$
$\displaystyle \lim_{n \mathop \to \infty} \paren {\sqrt 2 - 1}^n = 0$

where $\lim$ denotes limit.

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

$0 = \displaystyle \lim_{n \mathop \to \infty} \frac {a_n q + b_n p} q \ge \frac 1 q$

which is a contradiction.

$\blacksquare$