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 $\paren {\sqrt 2 - 1}$:
\(\ds 1 \ \ \) | \(\ds <\) | \(\sqrt 2\) | \(\ds <\) | \(\ds 2\) | Ordering of Squares in Reals | |||||||||
\(\ds \leadsto \ \ \) | \(\ds 0 \ \ \) | \(\ds <\) | \(\sqrt 2 - 1\) | \(\ds <\) | \(\ds 1\) |
Now, observe that for any $n \in \Z_{>0}$:
\(\ds \paren {\sqrt 2 - 1}^n\) | \(=\) | \(\ds \sum_{k \mathop = 0}^n \binom n k \paren {\sqrt 2}^k \paren {-1}^{n-k}\) | Binomial Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{\substack {0 \mathop \le k \mathop \le n \\ k \, \text{even} } } \binom n k 2^{k/2} \paren {-1}^{n - k} + \sqrt 2 \sum_{\substack {0 \mathop \le k \mathop \le n \\ k \, \text{odd} } } \binom n k 2^{\paren {k - 1}/2} \paren {-1}^{n-k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a_n + b_n \sqrt 2\) | for some integers $a_n, b_n$ | |||||||||||
\(\ds \) | \(=\) | \(\ds a_n + b_n \paren {\frac p q}\) | recalling the assumption that $\sqrt 2 = \dfrac p q$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a_n q + b_n p} q\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds \frac 1 q\) | as the numerator is an integer and $\sqrt 2 - 1 > 0$ |
By Sequence of Powers of Number less than One:
- $\ds \lim_{n \mathop \to \infty} \paren {\sqrt 2 - 1}^n = 0$
where $\lim$ denotes limit.
Recall the definition of $a_n$ and $b_n$.
By Lower and Upper Bounds for Sequences:
- $0 = \ds \lim_{n \mathop \to \infty} \frac {a_n q + b_n p} q \ge \frac 1 q$
which is a contradiction.
$\blacksquare$