structural:expansion, quality:corruption, stylistic:style_transfer
Example: Fibonacci numbers

Complete induction is most useful when several instances of the inductive hypothesis are required for each induction step. For example, complete induction can be used to show that F n = φ n − ψ n φ − ψ {\displaystyle F_{n}={\frac {\varphi ^{n}-\psi ^{n}}{\varphi -\psi }}} where F n {\displaystyle F_{n}} is the n -th Fibonacci number, and φ = 1 2 (1 + 5) {\textstyle \varphi ={\frac {1}{2}}(1+{\sqrt {5}})} (the golden ratio) and ψ = 1 2 (1 − 5) {\textstyle \psi ={\frac {1}{2}}(1-{\sqrt {5}})} are the roots of the polynomial x 2 − x − 1 {\displaystyle x^{2}-x-1}. By using the fact that F n + 2 = F n + 1 + F n {\displaystyle F_{n+2}=F_{n+1}+F_{n}} for each n ∈ N {\displaystyle n\in \mathbb {N} }, the identity above can be verified by direct calculation for F n + 2 {\textstyle F_{n+2}} if one assumes that it already holds for both F n + 1 {\textstyle F_{n+1}} and F n {\textstyle F_{n}}. To complete the proof, the identity must be verified in the two base cases: n = 0 {\displaystyle n=0} and n = 1 {\textstyle n=1}.