Square root of two: Difference between revisions

From Citizendium
Jump to navigation Jump to search
imported>Catherine Woodgold
(→‎Proof of Irrationality: Inserting exponent for k^2 in two places to correct the equations; adding words to clarify proof by contradiction)
imported>Catherine Woodgold
(→‎Proof of Irrationality: Essential step in proof: if x^2 is even, then x must be even.)
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]]
Thus, <math>x^2</math> represents an [[even number]];  therefore <math>x</math> must also be even.


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>
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>,


Since <math>k</math> is an integer, <math>y</math> must ''also'' be even. However, if <math>x</math> and <math>y</math> 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 <math>\sqrt{2}</math> must not be rational.
Since <math>k</math> is an integer, <math>y^2</math> and therefore also <math>y</math> must ''also'' be even. However, if <math>x</math> and <math>y</math> 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 <math>\sqrt{2}</math> must not be rational.


[[Category:Mathematics Workgroup]]
[[Category:Mathematics Workgroup]]
[[Category:CZ_Live]]
[[Category:CZ_Live]]

Revision as of 19:40, 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.

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

Through simplification, we find that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 4 \times k^2 = 2 \times y^2} , and then that, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2 \times k^2 = y^2} ,

Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} is an integer, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y^2} and therefore also Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} must also be even. However, if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{2}} must not be rational.