# Continued Fraction Expansion of Irrational Square Root/Examples/8/Convergents

## Convergents to Continued Fraction Expansion of $\sqrt 8$

The sequence of convergents to the continued fraction expansion of the square root of $8$ begins:

$\dfrac 2 1, \dfrac 3 1, \dfrac {14} 5, \dfrac {17} 6, \dfrac {82} {29}, \dfrac {99} {35}, \dfrac {478} {169}, \dfrac {577} {204}, \dfrac {2786} {985}, \dfrac {3363} {1189}, \ldots$

## Proof

Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be its continued fraction expansion.

Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be its numerators and denominators.

Then the $n$th convergent is $\dfrac {p_n} {q_n}$.

By definition:

$p_k = \begin {cases} a_0 & : k = 0 \\ a_0 a_1 + 1 & : k = 1 \\ a_k p_{k - 1} + p_{k - 2} & : k > 1 \end {cases}$
$q_k = \begin {cases} 1 & : k = 0 \\ a_1 & : k = 1 \\ a_k q_{k - 1} + q_{k - 2} & : k > 1 \end {cases}$
$\sqrt 8 = \sqbrk {2, \sequence {1, 4} }$

Thus the convergents are assembled:

$k$ $a_k$ $p_k = a_k p_{k - 1} + p_{k - 2}$ $q_k = a_k q_{k - 1} + q_{k - 2}$ $\dfrac {p_k} {q_k}$ Decimal value
$0$ $2$ $2$ $1$ $\dfrac { 2 } 1$ $2$
$1$ $1$ $2 \times 1 + 1 = 3$ $1$ $\dfrac { 3 } { 1 }$ $3$
$2$ $4$ $4 \times 3 + 2 = 14$ $4 \times 1 + 1 = 5$ $\dfrac { 14 } { 5 }$ $2.8$
$3$ $1$ $1 \times 14 + 3 = 17$ $1 \times 5 + 1 = 6$ $\dfrac { 17 } { 6 }$ $2.8333333333$
$4$ $4$ $4 \times 17 + 14 = 82$ $4 \times 6 + 5 = 29$ $\dfrac { 82 } { 29 }$ $2.8275862069$
$5$ $1$ $1 \times 82 + 17 = 99$ $1 \times 29 + 6 = 35$ $\dfrac { 99 } { 35 }$ $2.8285714286$
$6$ $4$ $4 \times 99 + 82 = 478$ $4 \times 35 + 29 = 169$ $\dfrac { 478 } { 169 }$ $2.8284023669$
$7$ $1$ $1 \times 478 + 99 = 577$ $1 \times 169 + 35 = 204$ $\dfrac { 577 } { 204 }$ $2.8284313725$
$8$ $4$ $4 \times 577 + 478 = 2786$ $4 \times 204 + 169 = 985$ $\dfrac { 2786 } { 985 }$ $2.8284263959$
$9$ $1$ $1 \times 2786 + 577 = 3363$ $1 \times 985 + 204 = 1189$ $\dfrac { 3363 } { 1189 }$ $2.8284272498$

$\blacksquare$