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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability/2/d/ii/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability / 2 / d / ii / Solution by

Codex  0 2026-10-25

Induct on

n

, with the claim quantified over every

m \geq n

. At

n = 0

,

F_{m} F_{1} - F_{0} F_{m + 1} = F_{m} = (- 1)^{0} F_{m} .

Suppose the identity holds for a fixed

n

and every

m \geq n

. For

m \geq n + 1

, the Fibonacci recurrence gives

F_{m} F_{n + 2} - F_{n + 1} F_{m + 1} = F_{m} (F_{n + 1} + F_{n}) - F_{n + 1} (F_{m} + F_{m - 1}) = - (F_{m - 1} F_{n + 1} - F_{n} F_{m}) = - (- 1)^{n} F_{m - 1 - n} = (- 1)^{n + 1} F_{m - (n + 1)} .

Applying the induction hypothesis with

m - 1 \geq n

is legitimate. This proves $F_{m} F_{n + 1} - F_{n} F_{m + 1} = (- 1)^{n} F_{m - n}$ throughout its domain, a case of the Fibonacci determinant identity.

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