Continued Fraction Expansion of Irrational Square Root/Example/2

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Examples of Continued Fraction Expansion of Irrational Square Root

The continued fraction expansion of the square root of $2$ is given by:

$\sqrt 2 = \sqbrk {1, \sequence 2}$

This sequence is A040000 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Proof

\(\displaystyle \sqrt 2\) \(=\) \(\displaystyle 1 + \paren {\sqrt 2 − 1}\)
\(\displaystyle \) \(=\) \(\displaystyle 1 + \frac {\paren {\sqrt 2 − 1} \paren {\sqrt 2 + 1} } {\sqrt 2 + 1}\) multiplying top and bottom by $\sqrt 2 + 1$
\(\displaystyle \) \(=\) \(\displaystyle 1 + \frac {\paren {\sqrt 2}^2 − 1^2} {\sqrt 2 + 1}\) Difference of Two Squares
\(\displaystyle \) \(=\) \(\displaystyle 1 + \frac 1 {1 + \sqrt 2}\) as $\paren {\sqrt 2}^2 − 1^2 = 2 - 1 = 1$


Thus it is possible to replace $\sqrt 2$ recursively:

\(\displaystyle \sqrt 2\) \(=\) \(\displaystyle 1 + \frac 1 {1 + \sqrt 2}\)
\(\displaystyle \) \(=\) \(\displaystyle 1 + \frac 1 {1 + \paren {1 + \cfrac 1 {1 + \sqrt 2} } }\)
\(\displaystyle \) \(=\) \(\displaystyle 1 + \frac 1 {2 + \cfrac 1 {1 + \sqrt 2} }\)
\(\displaystyle \) \(=\) \(\displaystyle 1 + \frac 1 {2 + \cfrac 1 {1 + \paren {1 + \cfrac 1 {1 + \sqrt 2} } } }\)
\(\displaystyle \) \(=\) \(\displaystyle 1 + \frac 1 {2 + \cfrac 1 {2 + \cfrac 1 {1 + \sqrt 2} } }\)


The pattern repeats indefinitely, producing the continued fraction expansion:

$\sqrt 2 = \sqbrk {1, 2, 2, 2, \ldots} = \sqbrk {1, \sequence 2}$

$\blacksquare$