# 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 an integer is even, its square root, if an integer, is also even.

Thus it follows that:

$(1): \quad 2 \divides p^2 \implies 2 \divides p$

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

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

So:

$\sqrt 2 = \dfrac p q$

for some $p, q \in \Z$, and:

$\gcd \set {p, q} = 1$

where $\gcd$ denotes the greatest common divisor.

Squaring both sides yields:

$2 = \dfrac {p^2} {q^2} \iff p^2 = 2 q^2$

Therefore from $(1)$:

$2 \divides p^2 \implies 2 \divides p$

That is, $p$ is an even integer.

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

Thus:

$2 q^2 = p^2 = \paren {2 k}^2 = 4 k^2 \implies q^2 = 2 k^2$

so by the same reasoning:

$2 \divides q^2 \implies 2 \divides q$

This contradicts our assumption that $\gcd \set {p, q} = 1$, since $2 \divides p, q$.

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

$\blacksquare$

## Proof 2

Special case of Square Root of Prime is Irrational.

$\blacksquare$

## Proof 3

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

where $\lim$ denotes limit.

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

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

$\blacksquare$

## Proof 4

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

Let $n$ be the smallest positive integer such that:

$\sqrt 2 = \dfrac m n$

for some $m \in \Z_{>0}$

Then:

$m = n \sqrt2 > n$

so:

$(1): \quad m - n > 0$

We also have:

$m = n \sqrt2 < 2 n$

so:

$m < 2 n$

and therefore:

$(2): \quad m - n < n$

Finally, we have

 $\ds m^2$ $=$ $\ds 2 n^2$ $\ds \leadsto \ \$ $\ds m^2 - n m$ $=$ $\ds 2 n^2 - n m$ $\ds \leadsto \ \$ $\ds m \paren {m - n}$ $=$ $\ds \paren {2 n - m} n$ $\ds \leadsto \ \$ $\ds \dfrac {2 n - m} {m - n}$ $=$ $\ds \dfrac m n$

It follows that:

$\dfrac {2 n - m} {m - n} = \sqrt2$

By $(1)$, the denominator of $\dfrac {2 n - m} {m - n}$ is positive.

By $(2)$, the denominator of $\dfrac {2 n - m} {m - n}$ is less than $n$.

We have thus written $\sqrt2$ as a fraction with a smaller denominator than $n$, which is a contradiction.

$\blacksquare$

## Decimal Expansion

$\sqrt 2 \approx 1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$

## Binary Expansion

The binary expansion of the square root of $2$ starts:

$\sqrt 2 \approx 1 \cdotp 01101 \, 01000 \, 00 \ldots$

## Historical Note

This result Square Root of 2 is Irrational is attributed to Pythagoras of Samos, or to a student of his.

Some legends have it that it is due to Hippasus of Metapontum who was thrown off a boat by his angry fellow Pythagoreans and drowned.

The ancient Greeks prior to Pythagoras believed that irrational numbers did not exist in the real world.

However, from the Pythagorean Theorem, a square with sides of length $1$ has a diagonal of length $\sqrt 2$.