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Chapter 1: Q19E (page 49)

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+qÙx|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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Most popular questions from this chapter

Wilson's theorem says that a numberis prime if and only if
(N-1)!=-1(modN).

(a) If is prime, then we know every number1≤x<p is invertible modulo . Which of thesenumbers is their own inverse?
(b) By pairing up multiplicative inverses, show thatrole="math" localid="1658725109805" (p-1)!=-1(modp) for prime p.
(c) Show that if N is not prime, then(N-1)!≢(modN) .(Hint: Considerd=gcd(N,(N-1)!.)
(d) Unlike Fermat's Little Theorem, Wilson's theorem is an if-and-only-if condition for primality. Why can't we immediately base a primality test on this rule?

What is the least significant decimal digit of (1717)17? (Hint: For distinct primesp,q, and any a is not equal to role="math" localid="1658726105638" a≢0(modpq), we proved the formula role="math" localid="1658726171933" a(p-1)≡1(modpq)in Section 1.4.2.)

The grade-school algorithm for multiplying two n-bit binary numbers x and y consist of addingtogethern copies of r, each appropriately left-shifted. Each copy, when shifted, is at most 2n bits long.
In this problem, we will examine a scheme for adding n binary numbers, each m bits long, using a circuit or a parallel architecture. The main parameter of interest in this question is therefore the depth of the circuit or the longest path from the input to the output of the circuit. This determines the total time taken for computing the function.
To add two m-bit binary numbers naively, we must wait for the carry bit from position i-1before we can figure out the ith bit of the answer. This leads to a circuit of depthΟ(m). However, carry-lookahead circuits (seewikipedia.comif you want to know more about this) can add inΟ(logn)depth.

  1. Assuming you have carry-lookahead circuits for addition, show how to add n numbers eachm bits long using a circuit of depth Ο(lognlogm).
  2. When adding three m-bit binary numbers x+y+z, there is a trick we can use to parallelize the process. Instead of carrying out the addition completely, we can re-express the result as the sum of just two binary numbersr+s, such that the ith bits of r and s can be computedindependently of the other bits. Show how this can be done. (Hint: One of the numbers represents carry bits.)
  3. Show how to use the trick from the previous part to design a circuit of depthΟ(logn)for multiplying two n-bit numbers.

Prove that the grade-school multiplication algorithm (page 24), when applied to binary numbers, always gives the right answer.

Calculate 2125mod127using any method you choose. (Hint: 127 is prime.)
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