Square Root of 2 is Irrational

Theorem

 * $\sqrt 2$ is irrational.

Classic Proof
First we note that, from Parity of Integer equals Parity of its Square, if a number is even, its square root, if an integer, is also even.

Thus it follows that:
 * $(A) \qquad 2 \backslash p^2 \implies 2 \backslash p$

where $2 \backslash p$ indicates that $2$ is a divisor of $p$.

Now, assume that $\sqrt 2$ is rational.

So $\displaystyle \sqrt 2 = \frac p q$ for some $p, q \in \Z$ and $\gcd \left({p, q}\right) = 1$.

Squaring both sides yields:
 * $\displaystyle 2 = \frac {p^2} {q^2} \iff p^2 = 2q^2$

Therefore, $2 \backslash p^2 \implies 2 \backslash p$ (see $(A)$ above).

That is, $p$ is an even integer.

So $p = 2k$ for some $k \in \Z$.

Thus:
 * $2 q^2 = p^2 = \left({2 k}\right)^2 = 4 k^2 \implies q^2 = 2k^2$

so by the same reasoning
 * $2 \backslash q^2 \implies 2 \backslash q$

This contradicts our assumption that $\gcd \left({p, q}\right) = 1$, since $2 \backslash p, q$.

Therefore, from Proof by Contradiction, $\sqrt 2$ cannot be rational.

Proof 2
This is a special case of the result that the square root of any prime is irrational.

Proof 3
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 a Number Less Than One:
 * $\displaystyle \lim_{n \to \infty} \left({\sqrt 2 - 1}\right)^n = 0$

where $\lim$ denotes limit.

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

a contradiction.

Decimal Expansion
The decimal expansion of $\sqrt 2$ starts:
 * $\sqrt 2 \approx 1.41421 \ 35623 \ 73095 \ 0488 \ldots$