Square root of two

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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, x2 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, y2 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.

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