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 $\sqrt 2 - 1 > 0$


By Sequence of Powers of Number less than One:

$\displaystyle \lim_{n \mathop \to \infty} \paren {\sqrt 2 - 1}^n = 0$

where $\lim$ denotes limit.


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

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.

$\blacksquare$