Square root of two: Difference between revisions

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imported>Catherine Woodgold
(→‎Proof of Irrationality: Essential step in proof: if x^2 is even, then x must be even.)
imported>Jitse Niesen
(→‎Proof of Irrationality: x = 2 * k, not k = 2 * x; also reformulate)
Line 11: Line 11:
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>,


Thus, <math>x^2</math> represents an [[even number]];  therefore <math>x</math> must also be even.
Thus, <math>x^2</math> represents an [[even number]];  therefore <math>x</math> must also be even. This means that there is an integer <math>k</math> such that <math>x = 2 \times k</math>. Inserting it back into our previous equation, we find that <math>(2 \times k)^2 = 2 \times y^2</math>
 
If we take the integer <math>k</math> such that <math>k = 2 \times x</math>, and insert it back into our previous equation, we find that <math>(2 \times k)^2 = 2 \times y^2</math>


Through simplification, we find that <math>4 \times k^2 = 2 \times y^2</math>, and then that, <math>2 \times k^2 = y^2</math>,
Through simplification, we find that <math>4 \times k^2 = 2 \times y^2</math>, and then that, <math>2 \times k^2 = y^2</math>,

Revision as of 23:18, 27 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:

Suppose is rational. Then there must 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; therefore must also be even. This means that there is an integer such that . Inserting it back into our previous equation, we find that

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

Since is an integer, and therefore also 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, so the assumption must be false, and must not be rational.