Continued fraction of the square root of eleven

Rationalizing successive complete quotients gives

11 = [3; \overline{3, 6}] .

Table of contents
- Alternating multiplier recurrence for convergents of the square root of eleven Continued fraction of the square root of eleven
- Pell-type norm from a periodic continued fraction Continued fraction of the square root of eleven

Alternating multiplier recurrence for convergents of the square root of eleven

If

p_{n} / q_{n}

are the convergents of

11

, then

p_{n + 1} + q_{n + 1} 11 = ⎩ ⎨ ⎧ (3 + 11) (p_{n} + q_{n} 11), \frac{3 + 11}{2} (p_{n} + q_{n} 11), n odd, n even .

Pell-type norm from a periodic continued fraction

For the even convergents of

11

,

p_{2 m} + q_{2 m} 11 = (3 + 11) (10 + 311)^{m} .

Since the second factor has norm one, every such convergent satisfies

p_{2 m}^{2} - 11 q_{2 m}^{2} = - 2

.

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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 1G / Solution

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