Square Root of 2 is Irrational/Classic Proof

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$\sqrt 2$ is irrational.


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.


$\sqrt 2 = \dfrac p q$

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 $k \in \Z$.


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

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.


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 $\sqrt 2$.


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