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.