Square root of two: Difference between revisions

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imported>Sébastien Moulin
m (→‎In Right Triangles: correcting TeX syntax)
imported>Graham Nelson
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There exists a simple proof by contradiction showing that <math>\sqrt{2}</math> is irrational:
There exists a simple proof by contradiction showing that <math>\sqrt{2}</math> is irrational:


Assume that there exists two numbers, <math>x, y \in \mathbb{N}</math>, such that <math>\frac{x}{y} = \sqrt{2}</math> and <math>x</math> and <math>y</math> represent the smallest such [[integer|integers]] (i.e., they are [[mutually prime]]).
Assume that there exist two numbers, <math>x, y \in \mathbb{N}</math>, such that <math>\frac{x}{y} = \sqrt{2}</math> and <math>x</math> and <math>y</math> represent the smallest such [[integer|integers]] (i.e., they are [[mutually prime]]).


Therefore, <math>\frac{x^2}{y^2} = 2</math> and <math>x^2 = 2 \times y^2</math>,
Therefore, <math>\frac{x^2}{y^2} = 2</math> and <math>x^2 = 2 \times y^2</math>,

Revision as of 11:35, 15 April 2007

The square root of two, denoted , 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 . Thus, .

Proof of Irrationality

There exists a simple proof by contradiction showing that is irrational:

Assume that there exist two numbers, , such that and and represent the smallest such integers (i.e., they are mutually prime).

Therefore, and ,

Thus, represents an even number

If we take the integer, , such that , and insert it back into our previous equation, we find that

Through simplification, we find that , and then that, ,

Since is an integer, must also be even. However, if and are both even, they share a common factor of 2, making them not mutually prime. And that is a contradiction.