Continued Fraction Expansion of Irrational Square Root/Examples/13/Convergents

Convergents to Continued Fraction Expansion of $\sqrt {13}$
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 $\sqbrk {a_0, a_1, a_2, \ldots}$ 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}$

From Continued Fraction Expansion of $\sqrt {13}$:
 * $\sqrt {13} = \sqbrk {3, \sequence {1, 1, 1, 1, 6} }$

Thus the convergents are assembled:


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