ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

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The continued fraction [3; overline 1,2,1,6] for sqrt(14)

Example

The complete quotients of $\sqrt{14}$ cycle through the states $\frac{\sqrt{14} + 3}{5}, \frac{\sqrt{14} + 2}{2}, \frac{\sqrt{14} + 2}{5}, \sqrt{14} + 3,$ so $\sqrt{14} = [3; \overline{1, 2, 1, 6}] .$

Facts & Assumptions

Given: The real number $\sqrt{14}$ .

[F1]

The continued-fraction algorithm is deterministic: each complete quotient $α_{n}$ determines the digit $a_{n}$ and, when $α_{n} \neq a_{n}$ , the next complete quotient $α_{n + 1}$ (Complete quotients in the continued-fraction algorithm).

Verification

technique · direct

1.1

Since $3 < \sqrt{14} < 4$ , the first digit is $a_{0} = 3$ . Then [given, algebra] $α_{1} = \frac{1}{\sqrt{14} - 3} = \frac{\sqrt{14} + 3}{5}, α_{2} = \frac{1}{α_{1} - 1} = \frac{\sqrt{14} + 2}{2},$ $α_{3} = \frac{1}{α_{2} - 2} = \frac{\sqrt{14} + 2}{5}, α_{4} = \frac{1}{α_{3} - 1} = \sqrt{14} + 3,$ and finally $α_{5} = \frac{1}{α_{4} - 6} = \frac{1}{\sqrt{14} - 3} = α_{1} .$ So the digits from $α_{1}$ onward are $1, 2, 1, 6$ and then repeat.

givenalgebra

2.1

Step 1.1 shows that $α_{5} = α_{1}$ . By the determinism in [F1], the same four digits $1, 2, 1, 6$ therefore repeat from $α_{1}$ onward. Hence $\sqrt{14} = [3; \overline{1, 2, 1, 6}] .$

F1step 1.1∎

Depends on

Complete quotients in the continued-fraction algorithm

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Peter Hackman, Elementary Number Theory (standard reference, not scraped)
Bruce Ikenaga, Periodic Continued Fractions (standard reference, not scraped)