Square root of two

From Knowino

Jump to: navigation, search

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.

[edit] 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, $\sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$ .

[edit] 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.

Some content on this page may previously have appeared on Citizendium.

Square root of two

[edit] In Right Triangles

[edit] Proof of Irrationality

Personal tools

Namespaces

Variants

Views

Actions

Search

Navigation

Community

Toolbox