Square root of two: Difference between revisions

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imported>Graham Nelson
imported>Catherine Woodgold
(→‎In Right Triangles: Adding a step (ratio of two sides of the triangle))
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== In Right Triangles ==
== In Right Triangles ==
The square root of two plays an important role in [[right triangle|right triangles]] in that a unit right triangle (where both legs are equal to 1), has a [[hypotenuse]] of <math>\sqrt{2}</math>. Thus, <math>\sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}</math>.
The square root of two plays an important role in [[right triangle|right triangles]] in that a unit right triangle (where both legs are equal to 1), has a [[hypotenuse]] of <math>\sqrt{2}</math>. Thus, <math>\sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}</math>.


== Proof of Irrationality ==
== Proof of Irrationality ==

Revision as of 11:34, 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.