Square root of two

The square root of two, denoted ${\sqrt {2}}$, 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 ${\sqrt {2}}$. Thus, $\sin \left({\frac {\pi }{4}}\right)=\cos \left({\frac {\pi }{4}}\right)={\frac {1}{\sqrt {2}}}={\frac {\sqrt {2}}{2}}$.

Proof of Irrationality

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

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

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

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

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

Since $k$ is an integer, $y^{2}$ and therefore also $y$ must also be even. However, if $x$ and $y$ 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 ${\sqrt {2}}$ must not be rational.