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}$

From Continued Fraction Expansion of $\sqrt 8$:
 * $\sqrt 8 = \sqbrk {2, \sequence {1, 4} }$

Thus the convergents are assembled:


 * }

[[Category:8]]