# Continued Fraction Expansion of Irrational Square Root/Example/29/Convergents

## Convergents to Continued Fraction Expansion of $\sqrt {29}$

The sequence of convergents to the continued fraction expansion of the square root of $29$ begins:

$\dfrac 5 1, \dfrac {11} 2, \dfrac {16} 3, \dfrac {27} 5, \dfrac {70} {13}, \dfrac {727} {135}, \dfrac {1524} {283}, \dfrac {2251} {418}, \dfrac {3775} {701}, \dfrac {9801} {1820}, \ldots$

## 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 $p_n / q_n$.

By definition:

$p_k = \begin {cases} a_0 & : k = 0 \\ a_0 a_1 + 1 & : k = 1 \\ a_k p_{k - 1} + p_{k - 2} & : k > 1\end {cases}$
$q_k = \begin {cases} 1 & : k = 0 \\ a_1 & : k = 1 \\ a_k q_{k - 1} + q_{k - 2} & : k > 1\end {cases}$
$\sqrt {29} = \sqbrk {5, \sequence {2, 1, 1, 2, 10} }$

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
$0$ $5$ $5$ $1$ $\dfrac { 5 } 1$ $5$
$1$ $2$ $5 \times 2 + 1 = 11$ $2$ $\dfrac { 11 } { 2 }$ $5.5$
$2$ $1$ $1 \times 11 + 5 = 16$ $1 \times 2 + 1 = 3$ $\dfrac { 16 } { 3 }$ $5.3333333333$
$3$ $1$ $1 \times 16 + 11 = 27$ $1 \times 3 + 2 = 5$ $\dfrac { 27 } { 5 }$ $5.4$
$4$ $2$ $2 \times 27 + 16 = 70$ $2 \times 5 + 3 = 13$ $\dfrac { 70 } { 13 }$ $5.3846153846$
$5$ $10$ $10 \times 70 + 27 = 727$ $10 \times 13 + 5 = 135$ $\dfrac { 727 } { 135 }$ $5.3851851852$
$6$ $2$ $2 \times 727 + 70 = 1524$ $2 \times 135 + 13 = 283$ $\dfrac { 1524 } { 283 }$ $5.3851590106$
$7$ $1$ $1 \times 1524 + 727 = 2251$ $1 \times 283 + 135 = 418$ $\dfrac { 2251 } { 418 }$ $5.3851674641$
$8$ $1$ $1 \times 2251 + 1524 = 3775$ $1 \times 418 + 283 = 701$ $\dfrac { 3775 } { 701 }$ $5.3851640514$
$9$ $2$ $2 \times 3775 + 2251 = 9801$ $2 \times 701 + 418 = 1820$ $\dfrac { 9801 } { 1820 }$ $5.3851648352$

$\blacksquare$