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The continued fraction of 37/11 matches Euclid and Bezout

Example

For $37 / 11$ , the Euclidean divisions $37 = 3 \cdot 11 + 4, 11 = 2 \cdot 4 + 3, 4 = 1 \cdot 3 + 1, 3 = 3 \cdot 1 + 0$ give the continued fraction $\frac{37}{11} = [3; 2, 1, 3],$ and the penultimate convergents encode the Bezout relation $37 \cdot 3 - 11 \cdot 10 = 1.$

Facts & Assumptions

Given: The integers $37$ and $11$ .

[F1]

For a rational number, the continued-fraction algorithm terminates and its digits are exactly the Euclidean quotient digits (The continued-fraction algorithm terminates exactly on rational numbers).

[F2]

Bezout's identity characterizes $\gcd (a, b)$ as an integer linear combination of $a$ and $b$ (Bézout's identity: for integers $a, b$ not both zero, $\gcd (a, b)$ is the least positive element of ${a x + b y : x, y \in Z}$ ; in particular $a x + b y = \gcd (a, b)$ has an integer solution).

Verification

technique · direct

1.1

The Euclidean quotient digits are $3, 2, 1, 3$ , so [F1] gives. [F1, given, algebra] $\frac{37}{11} = [3; 2, 1, 3] .$ Its convergents are $3, 3 + \frac{1}{2} = \frac{7}{2}, 3 + \frac{1}{2 + \frac{1}{1}} = \frac{10}{3}, 3 + \frac{1}{2 + \frac{1}{1 + \frac{1}{3}}} = \frac{37}{11} .$

F1givenalgebra

2.1

The penultimate convergent is $10 / 3$ , and. [F2, step 1.1, algebra] $37 \cdot 3 - 11 \cdot 10 = 111 - 110 = 1.$ So $1$ is an explicit integer linear combination of $37$ and $11$ , which is exactly the Bezout identity for $\gcd (37, 11) = 1$ in the sense of [F2].

F2step 1.1algebra∎

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Used by

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Dependency tree · two levels

21 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)
William Stein, Elementary Number Theory: Primes, Congruences, and Secrets (standard reference, not scraped)