Square Root of 2 is Irrational

Theorem

 * $\sqrt 2$ is irrational.

Proof 1: Euclid's proof
First we note that, from Parity of Integer equals Parity of its Square, if a number is even, its integer square root 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 $\sqrt 2$ cannot be rational.

This is a special case of the result that the square root of any prime is irrational, which is proved using the same argument. More generally it is a special case of Eisenstein's Criterion which further generalizes this argument.

Proof 2
Assume that $\sqrt{2} = p/q$ where $p$ and $q$ are positive integers. Let $n$ be a positive integer, then by the Binomial Theorem and collecting terms,
 * $\displaystyle (\sqrt{2} - 1)^n = a_n + b_n\sqrt{2} = \frac{a_n q + pb_n}{q},$

for integers $a_n$ and $b_n$. Since the denominator is $q$ and the numerator is a positive integer this fraction is $\ge 1/q$. However, since $0< \sqrt{2} - 1 < 1$, it follows that $(\sqrt{2} - 1)^n \to 0$ as $n\to \infty$, a contradiction.

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