Editing Pi (section)

=== Continued fractions ===
As an irrational number, {{pi}} cannot be represented as a [[common fraction]]. But every number, including {{pi}}, can be represented by an infinite<!--rationals have infinitely many 0's in the CF representation--> series of nested fractions, called a [[simple continued fraction]]:

<math display=block>
\pi = 3+\textstyle \cfrac{1}{7+\textstyle \cfrac{1}{15+\textstyle \cfrac{1}{1+\textstyle \cfrac{1}{292+\textstyle \cfrac{1}{1+\textstyle \cfrac{1}{1+\textstyle \cfrac{1}{1+\ddots}}}}}}}
</math>

Truncating the continued fraction at any point yields a rational approximation for {{pi}}; the first four of these are {{math|3}}, {{math|{{sfrac|22|7}}}}, {{math|{{sfrac|333|106}}}}, and {{math|{{sfrac|355|113}}}}. These numbers are among the best-known and most widely used historical approximations of the constant. Each approximation generated in this way is a best rational approximation; that is, each is closer to {{pi}} than any other fraction with the same or a smaller denominator.{{sfn|Eymard|Lafon|2004|p=78}} Because {{pi}} is transcendental, it is by definition not [[algebraic number|algebraic]] and so cannot be a [[quadratic irrational]]. Therefore, {{pi}} cannot have a [[periodic continued fraction]]. Although the simple continued fraction for {{pi}} (with numerators all 1, shown above) also does not exhibit any other obvious pattern,{{sfn|Arndt|Haenel|2006|p=33}}<ref name=mollin>{{cite journal |last=Mollin |first=R. A. |issue=3 |journal=Nieuw Archief voor Wiskunde |mr=1743850 |pages=383–405 |title=Continued fraction gems |volume=17 |year=1999}}</ref> several non-simple [[continued fraction]]s do, such as:<ref>{{cite journal |title=An Elegant Continued Fraction for π |first=L. J. |last=Lange |journal=[[The American Mathematical Monthly]] |volume=106 |issue=5 |date=May 1999 |pages=456–458 |jstor=2589152 |doi=10.2307/2589152}}</ref>{{efn|The middle of these is due to the mid-17th century mathematician [[William Brouncker, 2nd Viscount Brouncker|William Brouncker]], see [[William Brouncker, 2nd Viscount Brouncker#Brouncker's formula|§&nbsp;Brouncker's formula]].}}

<math display=block>
\begin{align}
\pi &= 3+ \cfrac
  {1^2}{6+ \cfrac
    {3^2}{6+ \cfrac
      {5^2}{6+ \cfrac
        {7^2}{6+ \ddots}}}}
= \cfrac
  {4}{1+ \cfrac
    {1^2}{2+ \cfrac
      {3^2}{2+ \cfrac
        {5^2}{2+ \ddots}}}}
= \cfrac
  {4}{1+ \cfrac
    {1^2}{3+ \cfrac
      {2^2}{5+ \cfrac
        {3^2}{7+ \ddots}}}}
\end{align}
</math>