# Continued Fraction Expansion of Irrational Square Root/Examples/61

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## Examples of Continued Fraction Expansion of Irrational Square Root

The continued fraction expansion of the square root of $61$ is given by:

$\sqrt {61} = \sqbrk {7, \sequence {1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14} }$

### Convergents

The sequence of convergents to the continued fraction expansion of the square root of $61$ begins:

$\dfrac 7 1, \dfrac {8} 1, \dfrac {39} 5, \dfrac {125} {16}, \dfrac {164} {21}, \dfrac {453} {58}, \dfrac {1070} {137}, \dfrac {1523} {195}, \dfrac {5639} {722}, \dfrac {24079} {3083}, \ldots$

## Proof

Let $\sqrt {61} = \sqbrk {a_0, a_1, a_2, a_3, \ldots}$

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

$a_r = \floor {\dfrac {\floor {\sqrt {61} } + P_r} {Q_r} }$

where:

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

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

$r$ $P_r = a_{r - 1} Q_{r - 1} - P_{r - 1}$ $Q_r = \dfrac {n - {P_r}^2} {Q_{r - 1} }$ $a_r = \floor {\dfrac {\floor {\sqrt { 61 } } + P_r} {Q_r} }$
$0$ $0$ $1$ $\floor {\dfrac {\floor {\sqrt { 61 } } + 0} 1} = 7$
$1$ $7 \times 1 - 0 = 7$ $\dfrac { 61 - 7^2} { 1 } = 12$ $\floor {\dfrac {\floor {\sqrt { 61 } }\ + 7 } { 12 } } = 1$
$2$ $1 \times 12 - 7 = 5$ $\dfrac { 61 - 5^2} { 12 } = 3$ $\floor {\dfrac {\floor {\sqrt { 61 } }\ + 5 } { 3 } } = 4$
$3$ $4 \times 3 - 5 = 7$ $\dfrac { 61 - 7^2} { 3 } = 4$ $\floor {\dfrac {\floor {\sqrt { 61 } }\ + 7 } { 4 } } = 3$
$4$ $3 \times 4 - 7 = 5$ $\dfrac { 61 - 5^2} { 4 } = 9$ $\floor {\dfrac {\floor {\sqrt { 61 } }\ + 5 } { 9 } } = 1$
$5$ $1 \times 9 - 5 = 4$ $\dfrac { 61 - 4^2} { 9 } = 5$ $\floor {\dfrac {\floor {\sqrt { 61 } }\ + 4 } { 5 } } = 2$
$6$ $2 \times 5 - 4 = 6$ $\dfrac { 61 - 6^2} { 5 } = 5$ $\floor {\dfrac {\floor {\sqrt { 61 } }\ + 6 } { 5 } } = 2$
$7$ $2 \times 5 - 6 = 4$ $\dfrac { 61 - 4^2} { 5 } = 9$ $\floor {\dfrac {\floor {\sqrt { 61 } }\ + 4 } { 9 } } = 1$
$8$ $1 \times 9 - 4 = 5$ $\dfrac { 61 - 5^2} { 9 } = 4$ $\floor {\dfrac {\floor {\sqrt { 61 } }\ + 5 } { 4 } } = 3$
$9$ $3 \times 4 - 5 = 7$ $\dfrac { 61 - 7^2} { 4 } = 3$ $\floor {\dfrac {\floor {\sqrt { 61 } }\ + 7 } { 3 } } = 4$
$10$ $4 \times 3 - 7 = 5$ $\dfrac { 61 - 5^2} { 3 } = 12$ $\floor {\dfrac {\floor {\sqrt { 61 } }\ + 5 } { 12 } } = 1$
$11$ $1 \times 12 - 5 = 7$ $\dfrac { 61 - 7^2} { 12 } = 1$ $\floor {\dfrac {\floor {\sqrt { 61 } }\ + 7 } { 1 } } = 14$

and the cycle is complete:

$\sequence {1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14}$

$\blacksquare$