# Square root of two

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The square root of two, denoted ${\displaystyle {\sqrt {2}}}$, is the positive number whose square equals 2. It is approximately 1.4142135623730950488016887242097. It provides a typical example of an irrational number.

## In Right Triangles

The square root of two plays an important role in right triangles in that a unit right triangle (where both legs are equal to 1), has a hypotenuse of ${\displaystyle {\sqrt {2}}}$. Thus, ${\displaystyle \sin \left({\frac {\pi }{4}}\right)=\cos \left({\frac {\pi }{4}}\right)={\frac {1}{\sqrt {2}}}={\frac {\sqrt {2}}{2}}}$.

## Proof of Irrationality

There exists a simple proof by contradiction showing that ${\displaystyle {\sqrt {2}}}$ is irrational. This proof is often attributed to Pythagoras. It is an example of a reductio ad absurdum type of proof:

Suppose ${\displaystyle {\sqrt {2}}}$ is rational. Then there must exist two numbers, ${\displaystyle x,y\in \mathbb {N} }$, such that ${\displaystyle {\frac {x}{y}}={\sqrt {2}}}$ and ${\displaystyle x}$ and ${\displaystyle y}$ represent the smallest such integers (i.e., they are mutually prime).

Therefore, ${\displaystyle {\frac {x^{2}}{y^{2}}}=2}$ and ${\displaystyle x^{2}=2\times y^{2}}$,

Thus, ${\displaystyle x^{2}}$ represents an even number; therefore ${\displaystyle x}$ must also be even. This means that there is an integer ${\displaystyle k}$ such that ${\displaystyle x=2\times k}$. Inserting it back into our previous equation, we find that ${\displaystyle (2\times k)^{2}=2\times y^{2}}$

Through simplification, we find that ${\displaystyle 4\times k^{2}=2\times y^{2}}$, and then that, ${\displaystyle 2\times k^{2}=y^{2}}$,

Since ${\displaystyle k}$ is an integer, ${\displaystyle y^{2}}$ and therefore also ${\displaystyle y}$ must also be even. However, if ${\displaystyle x}$ and ${\displaystyle y}$ are both even, they share a common factor of 2, making them not mutually prime. And that is a contradiction, so the assumption must be false, and ${\displaystyle {\sqrt {2}}}$ must not be rational.