# Square Root of 2 is Irrational

## Contents

## 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:

- $(1): \quad 2 \mathrel \backslash p^2 \implies 2 \mathrel \backslash p$

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

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

So:

- $\sqrt 2 = \dfrac p q$

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

- $\gcd \left({p, q}\right) = 1$

Squaring both sides yields:

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

Therefore from $(1)$:

- $2 \mathrel \backslash p^2 \implies 2 \mathrel \backslash p$

That is, $p$ is an even integer.

So $p = 2 k$ 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:

- $2 \mathrel \backslash q^2 \implies 2 \mathrel \backslash q$

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

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 $\left({\sqrt 2 - 1}\right)$:

\(\displaystyle 1 \ \ \) | \(\displaystyle <\) | \(\sqrt 2\) | \(\displaystyle <\) | \(\displaystyle 2\) | $\quad$ Ordering of Squares in Reals | $\quad$ | |||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 0 \ \ \) | \(\displaystyle <\) | \(\sqrt 2 - 1\) | \(\displaystyle <\) | \(\displaystyle 1\) | $\quad$ | $\quad$ |

Now, observe that for any $n \in \Z_{>0}$:

\(\displaystyle \left({\sqrt 2 - 1}\right)^n\) | \(=\) | \(\displaystyle \sum_{k \mathop = 0}^n \binom n k \left({\sqrt 2}\right)^k \left({-1}\right)^{n-k}\) | $\quad$ Binomial Theorem | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{\substack{0 \mathop \le k \mathop \le n \\ k \, \text{even} } } \binom n k 2^{k/2} \left({-1}\right)^{n-k} + \sqrt 2 \sum_{\substack{0 \mathop \le k \mathop \le n \\ k \, \text{odd} } } \binom n k 2^{\left({k-1}\right)/2} \left({-1}\right)^{n-k}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a_n + b_n \sqrt 2\) | $\quad$ for some integers $a_n, b_n$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a_n + b_n \left({\frac p q}\right)\) | $\quad$ recall the assumption that $\sqrt 2 = \dfrac p q$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {a_n q + b_n p} q\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(\ge\) | \(\displaystyle \frac 1 q\) | $\quad$ since the numerator is an integer and $\sqrt 2 - 1 > 0$ | $\quad$ |

By Sequence of Powers of Number less than One:

- $\displaystyle \lim_{n \mathop \to \infty} \left({\sqrt 2 - 1}\right)^n = 0$

where $\lim$ denotes limit.

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

By Lower and Upper Bounds for Sequences:

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

which is a contradiction.

$\blacksquare$

## 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

\(\displaystyle m^2\) | \(=\) | \(\displaystyle 2 n^2\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle m^2 - n m\) | \(=\) | \(\displaystyle 2 n^2 - n m\) | $\quad$ | $\quad$ | ||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle m \paren {m - n}\) | \(=\) | \(\displaystyle \paren {2 n - m} n\) | $\quad$ | $\quad$ | ||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \dfrac {2 n - m} {m - n}\) | \(=\) | \(\displaystyle \dfrac m n\) | $\quad$ | $\quad$ |

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

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

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

This sequence is A002193 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).

## Binary Expansion

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

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

This sequence is A004539 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).

## 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 $\sqrt 2$.

## Sources

- 1981: Murray R. Spiegel:
*Theory and Problems of Complex Variables*(SI ed.) ... (previous) ... (next): $1$: Complex Numbers: The Real Number System: $4$ - 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.2$: Pythagoras (ca. $580$ – $500$ B.C.) - 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes - 2008: Ian Stewart:
*Taming the Infinite*... (previous) ... (next): Chapter $2$: The Logic Of Shape