Continued Fraction Expansion of Irrational Square Root/Examples/13

Example of Continued Fraction Expansion of Irrational Square Root

The continued fraction expansion of the square root of $13$ is given by:

$\sqrt {13} = \sqbrk {3, \sequence {1, 1, 1, 1, 6} }$

This sequence is A010122 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 $13$ begins:

$\dfrac 3 1, \dfrac 4 1, \dfrac 7 2, \dfrac {11} 3, \dfrac {18} 5, \dfrac {119} {33}, \dfrac {137} {38}, \dfrac {256} {71}, \dfrac {393} {109}, \dfrac {649} {180}, \ldots$

Proof

Let $\sqrt {13} = \sqbrk {a_0, a_1, a_2, a_3, \ldots}$

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

$a_r = \floor {\dfrac {\floor {\sqrt {13} } + P_r} {Q_r} }$

where:

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

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

$r$	$P_r = a_{r - 1} Q_{r - 1} - P_{r - 1}$	$Q_r = \dfrac {n - {P_r}^2} {Q_{r - 1} }$	$a_r = \floor {\dfrac {\floor {\sqrt { 13 } } + P_r} {Q_r} }$
$0$	$0$	$1$	$\floor {\dfrac {\floor {\sqrt { 13 } } + 0} 1} = 3$
$1$	$3 \times 1 - 0 = 3$	$\dfrac { 13 - 3^2} { 1 } = 4$	$\floor {\dfrac {\floor {\sqrt { 13 } }\ + 3 } { 4 } } = 1$
$2$	$1 \times 4 - 3 = 1$	$\dfrac { 13 - 1^2} { 4 } = 3$	$\floor {\dfrac {\floor {\sqrt { 13 } }\ + 1 } { 3 } } = 1$
$3$	$1 \times 3 - 1 = 2$	$\dfrac { 13 - 2^2} { 3 } = 3$	$\floor {\dfrac {\floor {\sqrt { 13 } }\ + 2 } { 3 } } = 1$
$4$	$1 \times 3 - 2 = 1$	$\dfrac { 13 - 1^2} { 3 } = 4$	$\floor {\dfrac {\floor {\sqrt { 13 } }\ + 1 } { 4 } } = 1$
$5$	$1 \times 4 - 1 = 3$	$\dfrac { 13 - 3^2} { 4 } = 1$	$\floor {\dfrac {\floor {\sqrt { 13 } }\ + 3 } { 1 } } = 6$

and the cycle is complete:

$\sequence {1, 1, 1, 1, 6}$

$\blacksquare$