Editing Square root (section)

===As periodic continued fractions===
A result from the study of [[irrational number]]s as [[simple continued fraction]]s was obtained by [[Joseph Louis Lagrange]] {{circa|1780}}. Lagrange found that the representation of the square root of any non-square positive integer as a continued fraction is [[Periodic continued fraction|periodic]]. That is, a certain pattern of partial denominators repeats indefinitely in the continued fraction. In a sense these square roots are the very simplest irrational numbers, because they can be represented with a simple repeating pattern of integers.

{|
|- 
|align="right"|<math>\sqrt{2}</math>|| = [1; 2, 2, ...]
|-
|align="right"|<math>\sqrt{3}</math>|| = [1; 1, 2, 1, 2, ...]
|-
|align="right"|<math>\sqrt{4}</math>|| = [2]
|-
|align="right"|<math>\sqrt{5}</math>|| = [2; 4, 4, ...]
|-
|align="right"|<math>\sqrt{6}</math>|| = [2; 2, 4, 2, 4, ...]
|-
|align="right"|<math>\sqrt{7}</math>|| = [2; 1, 1, 1, 4, 1, 1, 1, 4, ...]
|-
|align="right"|<math>\sqrt{8}</math>||= [2; 1, 4, 1, 4, ...]
|-
|align="right"|<math>\sqrt{9}</math>|| = [3]
|-
|align="right"|<math>\sqrt{10}</math>|| = [3; 6, 6, ...]
|- 
|align="right"|<math>\sqrt{11}</math>|| = [3; 3, 6, 3, 6, ...]
|-
|align="right"|<math>\sqrt{12}</math>|| = [3; 2, 6, 2, 6, ...]
|-
|align="right"|<math>\sqrt{13}</math>|| = [3; 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, ...]
|-
|align="right"|<math>\sqrt{14}</math>|| = [3; 1, 2, 1, 6, 1, 2, 1, 6, ...]
|-
|align="right"|<math>\sqrt{15}</math>|| = [3; 1, 6, 1, 6, ...]
|-
|align="right"|<math>\sqrt{16}</math>|| = [4]
|-
|align="right"|<math>\sqrt{17}</math>|| = [4; 8, 8, ...]
|-
|align="right"|<math>\sqrt{18}</math>|| = [4; 4, 8, 4, 8, ...]
|-
|align="right"|<math>\sqrt{19}</math>|| = [4; 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, ...]
|-
|align="right"|<math>\sqrt{20}</math>|| = [4; 2, 8, 2, 8, ...]
|}

The [[square bracket]] notation used above is a short form for a continued fraction. Written in the more suggestive algebraic form, the simple continued fraction for the square root of 11, [3; 3, 6, 3, 6, ...], looks like this:<math display="block">
\sqrt{11} = 3 + \cfrac{1}{3 + \cfrac{1}{6 + \cfrac{1}{3 + \cfrac{1}{6 + \cfrac{1}{3 + \ddots}}}}}
</math>

where the two-digit pattern {3, 6} repeats over and over again in the partial denominators. Since {{math|1=11 = 3<sup>2</sup> + 2}}, the above is also identical to the following [[generalized continued fraction#Roots of positive numbers|generalized continued fractions]]:

<math display="block">
\sqrt{11} = 3 + \cfrac{2}{6 + \cfrac{2}{6 + \cfrac{2}{6 + \cfrac{2}{6 + \cfrac{2}{6 + \ddots}}}}} = 3 + \cfrac{6}{20 - 1 - \cfrac{1}{20 - \cfrac{1}{20 - \cfrac{1}{20 - \cfrac{1}{20 - \ddots}}}}}.
</math>