/*! 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 28 Let \(b\) be a fixed integer and... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 28

Let \(b\) be a fixed integer and \(j\) a fixed positive integer. Show that if \(P(b), P(b+1), \ldots, P(b+j)\) are true and \([P(b) \wedge P(b+1) \wedge \cdots \wedge P(k)] \rightarrow P(k+1)\) is true for every integer \(k \geq b+j,\) then \(P(n)\) is true for all integers \(n\) with \(n \geq b\) .

Short Answer

Expert verified
Using mathematical induction, \(P(n)\) is proven true for all \(n \geq b\).

Step by step solution

01

- Understand the given conditions

We are given that propositions \(P(b), P(b+1), \ldots, P(b+j)\) are true, and it is also given that the implication \([P(b) \wedge P(b+1) \wedge \cdots\wedge P(k)] \rightarrow P(k+1)\) is true for every integer \(k \geq b + j\).
02

- Initial assumptions

Assume that the statements \(P(b), P(b+1), \ldots, P(b+j)\) are true. This gives us a base case to work with.
03

- Use of mathematical induction

To prove \(P(n)\) is true for all \(n \geq b\), we use strong induction. We need to show that if \(P(m)\) is true for all \(m\) such that \(b \leq m \leq k\), then \(P(k+1)\) is also true for \(k \geq b + j\).
04

- Base case confirmation

Since \(P(b), P(b+1),\) ... ,\(P(b+j)\) are true, the base case for our induction is confirmed.
05

- Inductive step application

Assume that for some \(k \geq b+j\), all \(P(m)\) are true for \(m\) with \(b \leq m \leq k\). Then by the given condition, since \([P(b) \wedge P(b+1) \wedge \cdots \wedge P(k)] \rightarrow P(k+1)\) is true, it follows that \(P(k+1)\) is true as well.
06

- Conclusion

By the principle of mathematical induction, \(P(n)\) is true for all integers \(n\) such that \(n \geq b\).

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.

Propositional Logic
Propositional logic forms the foundation of reasoning in mathematics and computer science. It deals with propositions, which are statements that can be true or false. In this exercise, we work with a series of propositions, such as \(P(b), P(b+1), \ldots, P(b+j)\). Each proposition here is a statement that can either hold true or not. Understanding how these individual truth values can be combined and manipulated is key. The use of logical operators like 'and' (\( \wedge \)) and 'implies' (\( \rightarrow \)) helps in formulating complex statements and proving or disproving them systematically.
Principle of Mathematical Induction
The Principle of Mathematical Induction is a powerful method of proof used in mathematics to establish that a statement holds for all natural numbers. The idea is to prove that if a statement is true for some initial number, and if the truth of that statement for one number implies its truth for the next number, then the statement is true for all numbers in some given range. In this exercise, induction is applied:
  • Base Case: Verify that the statement is true for the initial number(s), \(P(b), P(b+1), \ldots,P(b+j)\).
  • Inductive Step: Assume the statement is true for an integer \(k\) (where \(k \geq b+j\)) and then prove it for \(k+1\) using the assumption.
This method ensures the statement is true for all numbers starting from the base case onwards.
Base Case and Inductive Step
In mathematical induction, the proof revolves around two main parts: the base case and the inductive step.

Base Case:
This is where you establish the truth of your statement for the initial value(s). For our problem, we confirm that \(P(b), P(b+1), \cdots, P(b+j)\) are all true. These form the anchor points for our induction.

Inductive Step:
Here, we assume that our statement is true for some integer \(k\) (i.e., \(P(m)\) is true for all \(m\) such that \(b \leq m \leq k\)). We then use this assumption to prove that \(P(k+1)\) is also true. The given condition \([P(b) \wedge P(b+1) \wedge \cdots \wedge P(k)] \rightarrow P(k+1)\) helps us make this leap. By doing so, we extend the truth to the next integer, \(k+1\), and thus by induction, to all integers starting from \(b\).
Discrete Mathematics
Discrete mathematics is the study of mathematical structures that are fundamentally discrete rather than continuous. The concepts covered, such as logic, set theory, combinatorics, graph theory, and more, are crucial for computer science and related fields. This exercise falls within the realm of discrete mathematics because it involves logical reasoning (propositional logic) and mathematical induction, both of which are key topics in this field. Understanding discrete mathematics enables students to work effectively with algorithms, data structures, and perform rigorous proofs, all of which are essential skills in both academia and industry.

Learning how to handle problems where truths need to be propagated through logical steps and inductive reasoning forms a core part of the discrete mathematics curriculum.

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

Devise a rule of inference for verification of partial correctness of statements of the form $$ \begin{array}{c}{\text { if condition } 1 \text { then }} \\ {S_{1}} \\\ {\text { else if condition } 2 \text { then }} \\ {\qquad \begin{array}{c}{S_{2}} \\ {\vdots} \\ {\text { else }} \\\ {S_{n}}\end{array}}\end{array} $$ where \(S_{1}, S_{2}, \ldots, S_{n}\) are blocks.

A partition of a positive integer \(n\) is a way to write \(n\) as a sum of positive integers where the order of terms in the sum does not matter. For instance, \(7=3+2+1+1\) is a partition of \(7 .\) Let \(P_{m}\) equal the number of different partitions of \(m,\) and let \(P_{m, n}\) be the number of different ways to express \(m\) as the sum of positive integers not exceeding \(n\). a) Show that \(P_{m, m}=P_{m}\) . b) Show that the following recursive definition for \(P_{m, n}\) is correct: $$P_{m, n}=\left\\{\begin{array}{ll}{1} & {\text { if } m=1} \\ {1} & {\text { if } n=1} \\ {P_{m, m}} & {\text { if } m< n} \\ {1+P_{m, m-1}} & {\text { if } m=n>1} \\ {P_{m, n-1}+P_{m-n, n}} & {\text { if } m>n>1}\end{array}\right.$$ c) Find the number of partitions of 5 and of 6 using this recursive definition.

Deal with values of iterated functions. Suppose that \(f(n)\) is a function from the set of real numbers, or positive real numbers, or some other set of real numbers, to the set of real numbers such that \(f(n)\) is monotonically increasing [that is, \(f(n)< f(m)\) when \(n< m )\) and \(f(n)< n\) for all \(n\) in the domain of \(f . ]\) The function \(f^{(k)}(n)\) is defined recursively by $$f^{(k)}(n)=\left\\{\begin{array}{ll}{n} & {\text { if } k=0} \\\ {f\left(f^{(k-1)}(n)\right)} & {\text { if } k>0}\end{array}\right.$$ Furthermore, let \(c\) be a positive real number. The iterated function \(f_{c}^{*}\) is the number of iterations of \(f\) required to reduce its argument to \(c\) or less, so \(f_{c}^{*}(n)\) is the smallest nonnegative integer \(k\) such that \(f^{k}(n) \leq c\). Let \(f(n)=\sqrt{n} .\) Find a formula for \(f^{(k)}(n) .\) What is the value of \(f_{2}^{*}(n)\) when \(n\) is a positive integer?

Exercises \(49-51\) present incorrect proofs using mathematical induction. You will need to identify an error in reasoning in each exercise. What is wrong with this "proof"? "Theorem" For every positive integer \(n,\) if \(x\) and \(y\) are positive integers with max \((x, y)=n\) , then \(x=y\) . Basis Step: Suppose that \(n=1 .\) If \(\max (x, y)=1\) and \(x\) and \(y\) are positive integers, we have \(x=1\) and \(y=1\) . Inductive Step: Let \(k\) be a positive integer. Assume that whenever max \((x, y)=k\) and \(x\) and \(y\) are positive integers, then \(x=y .\) Now let max \((x, y)=k+1,\) where \(x\) and \(y\) are positive integers. Then \(\max (x-1, y-1)=k,\) so by the inductive hypothesis, \(x-1=y-1 .\) It follows that \(x=y\) completing the inductive step.

Use the well-ordering property to show that if \(x\) and \(y\) are real numbers with \(x1 /(y-x)\) . Then show that there is a rational number \(r\) with denominator \(A\) between \(x\) and \(y\) by looking at the numbers \(\lfloor x\rfloor+ j / A,\) where \(j\) is a positive integer.]
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

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.