/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 256 Let \(f_{1}=2, f_{k+1}=f_{k}\lef... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 256

Let \(f_{1}=2, f_{k+1}=f_{k}\left(f_{k}+1\right)\). Prove by induction that \(f_{k}\) has at least \(k\) distinct prime factors.

Short Answer

Expert verified
The induction shows that \( f_k \) has at least \( k \) distinct prime factors for all \( k \geq 1 \).

Step by step solution

01

Base Case

For the base case, we need to show that the statement holds for the initial value of the sequence, which is when \( k = 1 \). We have \( f_1 = 2 \). Clearly, \( f_1 \) has 1 distinct prime factor, which is 2 itself. Thus, the statement holds for \( k = 1 \).
02

Inductive Hypothesis

Assume that for some integer \( k = n \), \( f_n \) has at least \( n \) distinct prime factors. This is the inductive hypothesis that we will use to prove the statement for \( k = n+1 \).
03

Inductive Step

Now, according to the sequence definition, \( f_{n+1} = f_n(f_n + 1) \). We need to show that \( f_{n+1} \) has at least \( n+1 \) distinct prime factors. By the inductive hypothesis, \( f_n \) has at least \( n \) distinct prime factors. Notice that \( f_n + 1 \) is coprime to \( f_n \) since any prime factor of \( f_n \) can鈥檛 divide \( f_n + 1 \), otherwise it would divide the difference, which is 1. Therefore, \( f_n + 1 \) introduces at least one new prime factor. Thus, \( f_{n+1} \) has at least \( n + 1 \) distinct prime factors.
04

Conclusion

By proving the base case and the inductive step, we conclude that \( f_k \) indeed has at least \( k \) distinct prime factors for all integers \( k \geq 1 \). Thus, the proof by induction is complete.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Prime Factors
Prime factors are the building blocks of numbers. A prime factor is a prime number that divides another number exactly, without leaving a remainder.
  • A prime number itself cannot be divided exactly by any number other than 1 and the number itself. Examples include numbers like 2, 3, 5, 7, etc.
  • When we say a number has distinct prime factors, we mean that it can be broken down into prime numbers which are unique and different from one another.
In the exercise, we use prime factors to show the complexity of the sequence defined under the problem. At each step of the sequence, the number of distinct prime factors increases, signifying that the sequence values become more complex.
Sequence Definition
A sequence is a list of numbers put together according to some rule or formula. In this exercise, the sequence is defined as follows:- The first term of the sequence is denoted by \( f_1 \) and has a value of 2.- For every subsequent term, \( f_{k+1} \), it is calculated as \( f_k(f_k + 1) \).This recursive definition means that each term depends on the previous term. This particular sequence grows rapidly because we multiply the previous term by a slightly larger number each time.
Inductive Hypothesis
An inductive hypothesis is an assumption we make during proofs by induction. It serves as the basis to prove that a statement holds for all values of a sequence.In proof by induction, which is a common mathematical tool:
  • We begin by proving a base case鈥攐ften the simplest possible case.
  • We make an inductive hypothesis by assuming that a statement is true for a given case, say for \( k = n \).
  • We then demonstrate that, based on this hypothesis, the statement holds for the next value, \( k = n+1 \).
In the exercise, we hypothesize that \( f_n \) has at least \( n \) distinct prime factors and then use this assumption to prove it for \( f_{n+1} \).
Coprime Integers
Coprime integers, also known as relatively prime, are two numbers that have no common prime factors apart from the number 1. This means there isn鈥檛 a single prime number that can divide both numbers evenly.In our exercise:
  • When the inductive step is being applied, we consider \( f_n \) and \( f_n + 1 \).
  • These two numbers are coprime because any divisor of \( f_n \) cannot also divide \( f_n + 1 \); otherwise, it would need to divide their difference, which is 1.
This property ensures that \( f_n + 1 \) brings a new prime factor to the equation, as it introduces a factor that isn鈥檛 shared with \( f_n \). This is crucial for proving the induction step in our sequence definition.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Prove by induction the statement: $$ " 1+3+5+\cdots+(2 n-1)=n^{2} \text { holds, for all } n \geqslant 1 " $$ The summation in Problem 231 was known to the ancient Greeks. The mystical Pythagorean tradition (which flourished in the centuries after Pythagoras) explored the character of integers through the 'spatial figures' which they formed. For example, if we arrange each successive integer as a new line of dots in the plane, then the sum " \(1+2+3+\cdots+n\) " can be seen to represent a triangular number. Similarly, if we arrange each odd number \(2 k-1\) in the sum " \(1+3+5+\cdots+(2 n-1)\) " as a " \(k\) -by- \(k\) reverse L-shape", or gnomon (a word which we still use to refer to the L-shaped piece that casts the shadow on a sundial), then the accumulated L-shapes build up an \(n\) by \(n\) square of dots \(-\) the "1" being the dot in the top left hand corner, the "3" being the reverse L-shape of 3 dots which make this initial "1" into a 2 by 2 square, the "5" being the reverse L-shape of 5 dots which makes this 2 by 2 square into a 3 by 3 square, and so on. Hence the sum " \(1+3+5+\cdots+(2 n-1)\) " can be seen to represent a square number. There is much to be said for such geometrical illustrations; but there is no escape from the fact that they hide behind an ellipsis (the three dots which we inserted in the sum between "5" and " \(2 n-1 "\), which were then summarised when arranging the reverse L-shapes by ending with the words "and so on"). Proof by mathematical induction, and its application in Problem 231, constitute a formal way of avoiding both the appeal to pictures, and the hidden ellipsis.

Problem 236 Prove by induction the statement: $$ "(1+2+3+\cdots+n)^{2}=1^{3}+2^{3}+3^{3}+\cdots+n^{3}, \text { for all } n \geqslant 1 " $$ We now know that, for all \(n \geqslant 1\) : $$ 1+1+1+\cdots+1 \quad(n \text { terms })=n $$ And if we sum these "outputs" (that is, the first \(n\) natural numbers), we get the \(n^{\text {th }}\) triangular number: $$ 1+2+3+\cdots+n=\frac{n(n+1)}{2}=T_{n} $$ The next problem invites you to find the sum of these "outputs": that is, to find the sum of the first \(n\) triangular numbers.

(a)(i) Explain why $$ \frac{1}{3^{2}}<\frac{1}{2^{2}} $$ So $$ \frac{1}{2^{2}}+\frac{1}{3^{2}}<\frac{2}{2^{2}}=\frac{1}{2} $$ (ii) Explain why \(\frac{1}{5^{2}}, \frac{1}{6^{2}}, \frac{1}{7^{2}}\) are all \(<\frac{1}{4^{2}},\) so $$ \frac{1}{4^{2}}+\frac{1}{5^{2}}+\frac{1}{6^{2}}+\frac{1}{7^{2}}<\frac{4}{4^{2}}=\frac{1}{4} $$ (b) Use part (a) to prove that $$ \frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{n^{2}}<2, \text { for all } n \geqslant 1 $$ (c) Conclude that the endless sum $$ \frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{n^{2}}+\cdots \text { (for ever) } $$ has a definite value, and that this value lies somewhere between \(\frac{17}{12}\) and 2 . $$ \Delta $$

The sequence of Fibonacci numbers $$ 0,1,1,2,3,5,8,13, \ldots $$ is defined by its first two terms \(F_{0}=0, F_{1}=1,\) and by the recurrence relation: $$ F_{n+2}=F_{n+1}+F_{n} \text { when } n \geqslant 0 $$ Prove by induction that, for all \(n \geqslant 0\) $$ F_{n}=\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}}, \text { where } \alpha=\frac{1+\sqrt{5}}{2} \text { and } \beta=\frac{1-\sqrt{5}}{2} $$

(Calkin-Wilf tree) The binary tree in the plane has a distinguished vertex as 'root', and is constructed inductively. The root is joined to two new vertices; and each new vertex is then joined to two further new vertices \(-\) with the construction process continuing for ever (Figure 11 ). Label the vertices of the binary tree with positive fractions as follows: \- the root is given the label \(\frac{1}{1}\) \- whenever we know the label \(\frac{i}{j}\) of a 'parent' vertex, we label its 'left descendant' as \(\frac{i}{i+j},\) and its 'right descendant' \(\frac{i+j}{j}\). (a) Prove that every positive rational \(\frac{r}{s}\) occurs once and only once as a label, and that it occurs in its lowest terms. (b) Prove that the labels are left-right symmetric in the sense that labels in corresponding left and right positions are reciprocals of each other. \(\triangle\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.