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Determinant identity for consecutive convergents

Statement

Let pn/qn be the convergents of a regular continued fraction. Then for every n0, pnqn1pn1qn=(1)n1. Consequently, for every n1, pnqnpn1qn1=(1)n1qnqn1.

Facts & Assumptions

Given: A regular continued fraction with convergent sequences pn,qn.

[F1]

The convergents satisfy p2=0, p1=1, q2=1, q1=0, and pn=anpn1+pn2, qn=anqn1+qn2 for n0. (Convergents of a regular continued fraction).

[F2]

If a subset of N contains 0 and is closed under successor, then it is all of N (The principle of mathematical induction).

Proof

technique · direct
1.1

At n=0 one has. [given, F1, base, algebra] p0q1p1q0=a0011=1=(1)1.

givenF1basealgebra
1.2

If Dn:=pnqn1pn1qn, then the recurrences of [F1] give. [F1, induction, algebra] Dn+1=(an+1pn+pn1)qnpn(an+1qn+qn1)=Dn. So the sign flips at each successor step.

F1inductionalgebra
2.1

Steps 1.1 and 1.2 imply by induction that. [F2, step 1.1, step 1.2, discharge-induction] Dn=(1)n1 for every n0.

F2step 1.1step 1.2discharge-induction
3.1

For n1. [step 2.1, algebra] pnqnpn1qn1=pnqn1pn1qnqnqn1=(1)n1qnqn1 by step 2.1.

step 2.1algebra

Depends on

Used by

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