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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{2}} . Thus, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 ${\sqrt {2}}$ is irrational:

Assume that there 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

If we take the integer, $k$, such that $k=2\times x$, and insert it back into our previous equation, we find that $(2\times k)^{2}=2\times y^{2}$

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

Since $k$ is an integer, $y$ 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.