Continued Fraction Expansion of Irrational Square Root/Examples/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).
Convergents
The sequence of convergents to the continued fraction expansion of the square root of $2$ begins:
- $\dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \dfrac {1393} {985}, \dfrac {3363} {2378}, \ldots$
Proof
\(\ds \sqrt 2\) | \(=\) | \(\ds 1 + \paren {\sqrt 2 − 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \frac {\paren {\sqrt 2 − 1} \paren {\sqrt 2 + 1} } {\sqrt 2 + 1}\) | multiplying top and bottom by $\sqrt 2 + 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \frac {\paren {\sqrt 2}^2 − 1^2} {\sqrt 2 + 1}\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \frac 1 {1 + \sqrt 2}\) | as $\paren {\sqrt 2}^2 − 1^2 = 2 - 1 = 1$ |
Thus it is possible to replace $\sqrt 2$ recursively:
\(\ds \sqrt 2\) | \(=\) | \(\ds 1 + \frac 1 {1 + \sqrt 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \frac 1 {1 + \paren {1 + \cfrac 1 {1 + \sqrt 2} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \frac 1 {2 + \cfrac 1 {1 + \sqrt 2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \frac 1 {2 + \cfrac 1 {1 + \paren {1 + \cfrac 1 {1 + \sqrt 2} } } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \frac 1 {2 + \cfrac 1 {2 + \cfrac 1 {1 + \sqrt 2} } }\) |
The pattern repeats indefinitely, producing the continued fraction expansion:
- $\sqrt 2 = \sqbrk {1, 2, 2, 2, \ldots} = \sqbrk {1, \sequence 2}$
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$\blacksquare$