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Chapter 10: Problem 19

$$ \text { Prove that any two consecutive Fibonacci numbers are relatively prime. } $$

Short Answer

Expert verified
Any two consecutive Fibonacci numbers are relatively prime (i.e. their GCD is 1). The proof for this property has been established using the principle of mathematical induction. The claim is true for initial pairs of Fibonacci numbers (base case), and then it was shown that if the claim is true for some Fibonacci number \( k \), it's also true for the next Fibonacci number \( k+1 \).

Step by step solution

01

Base Case

The first step in mathematical induction is to prove the base case. For Fibonacci numbers, the first few are 0, 1, 1, 2, 3, 5, 8... In these initial numbers, we can observe that the pair (0,1), (1,1), (1,2) and (2,3) are all pairs of consecutive Fibonacci numbers and their greatest common divisor is 1. So the statement holds for these base cases.
02

Induction Hypothesis

Now, assume our claim is true for some arbitrary number \( k \). This means that the greatest common divisor, also known as GCD, of the \( k \)th and \( (k+1) \)th Fibonacci numbers is 1. This is our induction hypothesis.
03

Inductive Step

In this step we want to prove that if the claim holds for an arbitrary number \( k \), then it also holds for \( k+1 \). It is known that the \( (k+2) \)th Fibonacci number is the sum of the \( k \)th and the \( (k+1) \)th Fibonacci numbers. If we denote the GCD of the \( k \)th Fibonacci number \( F_k \) and \( k+1 \)th Fibonacci number \( F_{k+1} \) as \( d \), we assumed that \( d=F_k \) and \( d=F_{k+1} \). Because every number dividing both \( F_k \) and \( F_{k+1} \) will also divide \( F_{k+2} \), have \( d=F_{k+2} \). But this also means that \( d \) simultaneously divides \( F_{k+1} \) and \( F_{k+2}-F_{k+1}=F_k \), hence \( d \) divides \( F_k \) and \( F_{k+1} \). But as per the induction hypothesis, the GCD of \( F_k \) and \( F_{k+1} \) is 1, which means their common divisor, \( d \), is 1. Hence by the principle of mathematical induction, any two consecutive Fibonacci numbers are coprime.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mathematical Induction
Mathematical induction is a proof technique that establishes the truth of an infinite number of statements by proving first a base case and then using a step called induction hypothesis to prove a general statement for all integers. It's like a domino effect; if you can prove that knocking over the first domino causes the second to fall, and if knocking over one domino makes the next fall, then all dominoes will eventually fall.

In the context of Fibonacci numbers, the base case would be checking if the initial pairs of numbers are relatively prime. Once we know this holds for the first few numbers, we can proceed to prove that it will hold for all pairs of consecutive Fibonacci numbers.
Greatest Common Divisor (GCD)
The Greatest Common Divisor (GCD) of two numbers is the largest number that divides both of them without leaving a remainder. If the GCD of two numbers is 1, those numbers are called relatively prime, or coprime, which means they have no prime factors in common.

When looking at the Fibonacci sequence, the technique of mathematical induction requires us to assume that the GCD of any two consecutive numbers is 1 and prove that this assumption implies the same for the next pair of consecutive numbers.
Fibonacci Sequence
The Fibonacci Sequence is a series of numbers in which each number is the sum of the two preceding ones, usually starting with 0 and 1. This sequence relates to various phenomena in art, nature, and mathematics. The sequence begins as 0, 1, 1, 2, 3, 5, 8, 13, 21, and so on. In our exercise, we're particularly interested in the property that any two consecutive Fibonacci numbers are relatively prime.
Proof by Induction
Proof by induction is a two-step process used in mathematical induction. After proving the base case, the induction hypothesis is used to assume the statement is true for a number 'k'. Then, the inductive step aims to prove that the statement is also true for 'k+1'.

In our Fibonacci example, the inductive step involves showing that if the GCD of the kth and (k+1)th numbers is 1 (induction hypothesis), then the GCD of the (k+1)th and (k+2)th numbers is also 1. This process is powerful and can conclude the truth for all natural numbers.
Coprime
Two numbers are coprime, or relatively prime, if their only common positive divisor is 1. This concept is important because it tells us about the relationship between two numbers without needing to factor them fully, which can be quite difficult for large numbers.

When examining the Fibonacci sequence, being able to assert that consecutive numbers are coprime is quite profound. It speaks to the nature of the sequence that despite the simple addition rule it follows, such a deep and consistent pattern of coprimality emerges.

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Most popular questions from this chapter

Find and solve a recurrence relation for the number of ways to park motorcycles and compact cars in a row of \(n\) spaces if each cycle requires one space and each compact needs two. (All cycles are identical in appearance, as are the cars, and we want to use up all the \(n\) spaces.)

Solve the following recurrence relations by the method of generating functions. a) \(a_{n+1}-a_{n}=3^{n}, \quad n \geq 0, a_{0}=1\) b) \(a_{n+1}-a_{n}=n^{2}, \quad n \geq 0, a_{0}=1\) c) \(a_{n}-3 a_{n-1}=5^{n-1}, \quad n \geq 1, \quad a_{0}=1\) d) \(a_{n+2}-3 a_{n+1}+2 a_{n}=0, \quad n \geq 0, \quad a_{0}=1, \quad a_{1}=6\) e) \(a_{n+2}-2 a_{n+1}+a_{n}=2^{n}, \quad n \geq 0, a_{0}=1, a_{1}=2\)

Consider a tennis tournament for \(n\) players, where \(n=2^{k}, k \in \mathbf{Z}^{+}\). In the first round \(n / 2\) matches are played, and the \(n / 2\) winners advance to round 2 , where \(n / 4\) matches are played. This halving process continues until a winner is determined. a) For \(n=2^{k}, k \in \mathbf{Z}^{+}\), let \(f(n)\) count the total number of matches played in the tournament. Find and solve a recurrence relation for \(f(n)\) of the form $$ \begin{aligned} &f(1)=d \\ &f(n)=a f(n / 2)+c, \quad n=2,4,8, \ldots, \end{aligned} $$ where \(a, c\), and \(d\) are constants. b) Show that your answer in part (a) also solves the recurrence relation $$ \begin{aligned} &f(1)=d \\ &f(n)=f(n / 2)+(n / 2), \quad n=2,4,8, \ldots \end{aligned} $$

For \(n \geq 1\), let \(D_{n}\) be the following \(n \times n\) determinant. $$ \left|\begin{array}{cccccccccc} 2 & 1 & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & 0 \\ 1 & 2 & 1 & 0 & 0 & \cdots & 0 & 0 & 0 & 0 \\ 0 & 1 & 2 & 1 & 0 & \cdots & 0 & 0 & 0 & 0 \\ \hdashline & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & 0 & 0 & 0 & \cdots & 1 & 2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & \cdots & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 & \cdots & 0 & 0 & 1 & 2 \end{array}\right| $$ Find and solve a recurrence relation for the value of \(D_{n}\).

A particle moves horizontally to the right. For \(n \in \mathbf{Z}^{+}\), the distance the particle travels in the \((n+1)\) st second is equal to twice the distance it travels during the \(n\)th second. If \(x_{n}, n \geq 0\), denotes the position of the particle at the start of the \((n+1)\) st second, find and solve a recurrence relation for \(x_{n}\), where \(x_{0}=1\) and \(x_{1}=5\).
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