Difference between revisions of "The square root of 2 is irrational"
Tags: Replaced Undo |
|||
| (4 intermediate revisions by 3 users not shown) | |||
| Line 1: | Line 1: | ||
'''The square root of 2 is irrational''' is the claim in [[number theory]] that there is no rational number that when multiplied by itself equals the number 2. | '''The square root of 2 is irrational''' is the claim in [[number theory]] that there is no rational number that when multiplied by itself equals the number 2. | ||
One proof by [[reductio ad absurdum]] consists of first assuming that the square root of 2 can be written as a rational number. Thus, there is a pair of coprime integers <math>p</math> and <math>q</math> such that their ratio <math>\frac{p}q = \sqrt 2</math> ([[Every rational number has an irreducible representation]]). And so, <math>\frac{p^2}{q^2} = 2</math>, and <math>p^2=2q^2</math>. As <math>p^2</math> has a factor of <math>2</math>, it is even, and thus <math>p</math> is also even. Thus <math>p</math> can be written as <math>2k</math>, where <math>k</math> is an integer. It then follows that <math>(2k)^2=2q^2</math>, <math>4k^2=2q^2</math>, <math>2k^2=q^2</math>, and thus <math>q^2</math> also has a factor of <math>2</math>, and thus <math>q</math> is also even. If <math>p</math> and <math>q</math> are both even, then they share a common factor of <math>2</math>, and thus are not coprime, leading to a contradiction as we have already established that <math>p</math> and <math>q</math> are coprime. | One proof by [[reductio ad absurdum]] consists of first assuming that the square root of 2 can be written as a rational number. Thus, there is a pair of coprime integers <math>p</math> and <math>q</math> such that their ratio <math>\frac{p}q = \sqrt 2</math> ([[Every rational number has an irreducible representation]]). And so, <math>\frac{p^2}{q^2} = 2</math>, and <math>p^2=2q^2</math>. As <math>p^2</math> has a factor of <math>2</math>, it is even, and thus <math>p</math> is also even. Thus <math>p</math> can be written as <math>2k</math>, where <math>k</math> is an integer. It then follows that <math>(2k)^2=2q^2</math>, <math>4k^2=2q^2</math>, <math>2k^2=q^2</math>, and thus <math>q^2</math> also has a factor of <math>2</math>, and thus <math>q</math> is also even. If <math>p</math> and <math>q</math> are both even, then they share a common factor of <math>2</math>, and thus are not coprime, leading to a contradiction as we have already established that <math>p</math> and <math>q</math> are coprime. And so, no such <math>p,q</math> can exist, and the square root of 2 is irrational. | ||
{{Claim | {{Claim | ||
| Line 8: | Line 8: | ||
|Nature=Theoretical | |Nature=Theoretical | ||
|Counterclaim= | |Counterclaim= | ||
|DependentOn1 | |DependentOn1=Every rational number has an irreducible representation | ||
|DependencyOf1= | |||
}} | }} | ||
[[Category:Mathematics]] | |||
Latest revision as of 04:40, 23 January 2022
The square root of 2 is irrational is the claim in number theory that there is no rational number that when multiplied by itself equals the number 2.
One proof by reductio ad absurdum consists of first assuming that the square root of 2 can be written as a rational number. Thus, there is a pair of coprime integers [math]\displaystyle{ p }[/math] and [math]\displaystyle{ q }[/math] such that their ratio [math]\displaystyle{ \frac{p}q = \sqrt 2 }[/math] (Every rational number has an irreducible representation). And so, [math]\displaystyle{ \frac{p^2}{q^2} = 2 }[/math], and [math]\displaystyle{ p^2=2q^2 }[/math]. As [math]\displaystyle{ p^2 }[/math] has a factor of [math]\displaystyle{ 2 }[/math], it is even, and thus [math]\displaystyle{ p }[/math] is also even. Thus [math]\displaystyle{ p }[/math] can be written as [math]\displaystyle{ 2k }[/math], where [math]\displaystyle{ k }[/math] is an integer. It then follows that [math]\displaystyle{ (2k)^2=2q^2 }[/math], [math]\displaystyle{ 4k^2=2q^2 }[/math], [math]\displaystyle{ 2k^2=q^2 }[/math], and thus [math]\displaystyle{ q^2 }[/math] also has a factor of [math]\displaystyle{ 2 }[/math], and thus [math]\displaystyle{ q }[/math] is also even. If [math]\displaystyle{ p }[/math] and [math]\displaystyle{ q }[/math] are both even, then they share a common factor of [math]\displaystyle{ 2 }[/math], and thus are not coprime, leading to a contradiction as we have already established that [math]\displaystyle{ p }[/math] and [math]\displaystyle{ q }[/math] are coprime. And so, no such [math]\displaystyle{ p,q }[/math] can exist, and the square root of 2 is irrational.
| Statement of the claim | The square root of 2 is irrational |
| Level of certainty | Proven |
| Nature | Theoretical |
| Counterclaim | |
| Dependent on | |
| Dependency of |
|