Continued Fraction Expansion of Irrational Square Root/Examples/5

From ProofWiki
Jump to navigation Jump to search

Examples of Continued Fraction Expansion of Irrational Square Root

The continued fraction expansion of the square root of $5$ is given by:

$\sqrt 5 = \sqbrk {2, \sequence 4}$

This sequence is A040002 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 $5$ begins:

$\dfrac 2 1, \dfrac 9 4, \dfrac {38} {17}, \dfrac {161} {72}, \dfrac {682} {305}, \dfrac {2889} {1292}, \dfrac {12238} {5473}, \dfrac {51841} {23184}, \ldots$


Proof

Let $\sqrt 5 = \sqbrk {a_0, a_1, a_2, a_3, \ldots}$

From Partial Denominators of Continued Fraction Expansion of Irrational Square Root, the partial denominators of this continued fraction expansion can be calculated as:

$a_r = \floor {\dfrac {\floor {\sqrt 5} + P_r} {Q_r} }$

where:

$P_r = \begin {cases} 0 & : r = 0 \\

a_{r - 1} Q_{r - 1} - P_{r - 1} & : r > 0 \\ \end {cases}$


$Q_r = \begin {cases} 1 & : r = 0 \\

\dfrac {n - {P_r}^2} {Q_{r - 1} } & : r > 0 \\ \end {cases}$


$r$ $P_r = a_{r - 1} Q_{r - 1} - P_{r - 1}$ $Q_r = \dfrac {n - {P_r}^2} {Q_{r - 1} }$ $a_r = \floor {\dfrac {\floor {\sqrt { 5 } } + P_r} {Q_r} }$
$0$ $0$ $1$ $\floor {\dfrac {\floor {\sqrt { 5 } } + 0} 1} = 2$
$1$ $2 \times 1 - 0 = 2$ $\dfrac { 5 - 2^2} { 1 } = 1$ $\floor {\dfrac {\floor {\sqrt { 5 } }\ + 2 } { 1 } } = 4$
$2$ $4 \times 1 - 2 = 2$ $\dfrac { 5 - 2^2} { 1 } = 1$ $\floor {\dfrac {\floor {\sqrt { 5 } }\ + 2 } { 1 } } = 4$
$3$ $4 \times 1 - 2 = 2$ $\dfrac { 5 - 2^2} { 1 } = 1$ $\floor {\dfrac {\floor {\sqrt { 5 } }\ + 2 } { 1 } } = 4$


and the pattern is established:

$\sequence 4$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Continued_Fraction_Expansion_of_Irrational_Square_Root/Examples/5&oldid=670441"