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91Ó°ÊÓ

The Fibonacci numbers F0,F1,...are given by the recurrenceFn+1=Fn+Fn-1 ,F0=0,F1=1. Show that for anyn≥1,gcd(Fn+1,Fn)=1.

Short Answer

Expert verified

For any n≥1,gcd(Fn+1,Fn)=1

Step by step solution

01

Explain Fibonacci Series

Fibonacci Series is defined as every element of the series is the sum of two previous number of the series. Series starts from 0 and 1.

Fn+1=Fn+Fn-1 forn=1Here,Fnisthe nthFibonacci numberF0=0F1=F2=1

02

Calculate the gcd by two methods

GCDfortheFibonacciseries,gcd(F1,F0)=gcd(1,0)=1Letthegcdbecorrectforn=k,thatisgcd(Fk,Fk-1)=1Provethatthegcdisalsocorrectforn=k+1,i.e.gcd(Fk+1,Fk)=1gcd(Fk+1,Fk)=gcd(Fk+Fk-1,Fk)=1So,gcd(Fk,Fk-1)=1

Letgcd(p,q)=1 and let gcd(p+q,q)=x.Provethatx=1gcd(p+q,q)=x=x|p+±çÙ³æ|q=x|(p+q)-q=x|pandsincex|q=x|1=x=1gcd(p+q,q)=1

Thus, gcd(Fn+1,Fn)=1. Hence proved.

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