/*! 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 2 Use a proof by cases to show tha... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 2

Use a proof by cases to show that 10 is not the square of a positive integer. [Hint: Consider two cases: \((i) 1 \leq x \leq 3\) , (ii) \(x \geq 4.1\)

Short Answer

Expert verified
10 is not the square of any positive integer.

Step by step solution

01

Understand the problem

The goal is to prove that 10 is not the square of a positive integer. To accomplish this, we will use proof by cases based on the hint provided: we will consider two cases, (i) when the integer is between 1 and 3, and (ii) when the integer is greater than or equal to 4.1.
02

Case 1 - Check integers from 1 to 3

In this step, consider possible values of \(x\) where \(1 \leq x \leq 3\).Since \(x\) must be a positive integer, consider \(x = 1, 2, 3\):When \(x = 1\), \(x^2 = 1\).When \(x = 2\), \(x^2 = 4\).When \(x = 3\), \(x^2 = 9\).None of these equal 10.
03

Case 2 - Check integers greater than or equal to 4.1

In this step, consider values of \(x\) where \(x \geq 4.1\).The next integer is \(x = 4\):When \(x = 4\), \(x^2 = 16\).Since squaring any number greater than 4 will only result in numbers larger than 16, squaring any number \(x \geq 4.1\) further confirms that no integer \(x\) meets \(x^2 = 10\).
04

Conclusion

Since no integers \(x = 1, 2, 3\) result in \(x^2 = 10\) and any \(x\) greater than or equal to 4 produces a square greater than 10, we conclude that 10 is not the square of any positive integer.

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.

discrete mathematics
Discrete mathematics is a branch of mathematics dealing with discrete elements that use distinct values. It is crucial in computer science, logic, and combinatorics. Unlike continuous mathematics, which deals with continuous variables, discrete mathematics focuses on countable objects. This means the elements can be counted individually. Examples include integers, graphs, and logical statements.
In our exercise, we use discrete mathematics to explore the properties of integer squares. We work with a finite set of integers to determine if 10 can be one of them by using proof by cases.
integer squares
Integer squares involve squaring whole numbers. For any integer value of 饾懃, we find the square by computing 饾懃 脳 饾懃. Square numbers grow quickly, as each increase in 饾懃 results in a much larger change in 饾懃虏.
Example: If ewline
  • 饾懃 = 1, then 饾懃虏 = 1
  • x = 2, then 饾懃虏 = 4
  • x = 3, then 饾懃虏 = 9
  • and so on.

In the given problem, we investigate integer squares to determine if 10 can be expressed as 饾懃虏 for any positive integer 饾懃. By checking smaller values of 饾懃 and understanding the rapid growth of squares, we conclude that 10 does not belong to the set of integer squares.
mathematical proof
Mathematical proof is a logical argument demonstrating the truth of a mathematical statement. Proofs are essential in ensuring that mathematical results are reliable and verified.
There are various methods of mathematical proof, including:
  • direct proof
  • indirect proof
  • proof by contradiction
  • proof by induction
  • and proof by cases.

In our exercise, we use proof by cases to conclude that 10 is not the square of any positive integer. This method involves analyzing distinct scenarios separately and showing that the statement holds true in each case. By splitting our problem into cases where 1 鈮� 饾懃 鈮� 3 and 饾懃 鈮� 4.1, we systematically demonstrate that no positive integer 饾懃 exists such that 饾懃虏 = 10.
positive integers
Positive integers are all whole numbers greater than zero. These include numbers like 1, 2, 3, and so on. Positive integers are the foundation of many mathematical concepts due to their simplicity and basic counting properties.
When dealing with positive integers, particularly in the context of squares and proofs, we work within a set that is easy to count and analyze.
In the provided exercise, we focus on positive integers to determine if any squared value could equal 10. By evaluating each possible positive integer within specified cases, we carefully prove that no positive integer square equals 10.

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

Translate each of these statements into logical expressions using predicates, quantifiers, and logical connectives. a) Something is not in the correct place. b) All tools are in the correct place and are in excellent condition. c) Everything is in the correct place and in excellent condition. d) Nothing is in the correct place and is in excellent condition. e) One of your tools is not in the correct place, but it is in excellent condition.

Identify the error or errors in this argument that supposedly shows that if \(\forall x(P(x) \vee Q(x))\) is true then \(\forall x P(x) \vee \forall x Q(x)\) is true. $$ \begin{array}{l}{\text { 1. } \forall x(P(x) \vee Q(x))} \\ {\text { 2. } P(c) \vee Q(c)} \\ {\text { 3. } P(c)} \\ {\text { 4. } \forall x P(x)} \\\ {\text { 5. } Q(c)} \\ {\text { 6. } \forall x Q(x)} \\ {\text { 7. } \forall x(P(x) \vee \forall x Q(x))}\end{array} $$

Let \(Q(x)\) be the statement " \(x+1>2 x\) . If the domain consists of all integers, what are these truth values? $$ \begin{array}{llll}{\text { a) }} & {Q(0)} & {\text { b) } Q(-1)} & {\text { c) }} \quad {Q(1)} \\ {\text { d) }} & {\exists x Q(x)} & {\text { e) } \quad \forall x Q(x)} & {\text { f) } \quad \exists x \neg Q(x)}\end{array} $$ g) \(\quad \forall x \neg Q(x)\)

Suppose that five ones and four zeros are arranged around a circle. Between any two equal bits you insert a 0 and between any two unequal bits you insert a 1 to produce nine new bits. Then you erase the nine original bits. Show that when you iterate this procedure, you can never get nine zeros. [Hint: Work backward, assuming that you did end up with nine zeros.]

Let \(S=x_{1} y_{1}+x_{2} y_{2}+\cdots+x_{n} y_{n},\) where \(x_{1}, x_{2}, \ldots, x_{n}\) and \(y_{1}, y_{2}, \ldots, y_{n}\) are orderings of two different sequences of positive real numbers, each containing \(n\) elements. a) Show that \(S\) takes its maximum value over all orderings of the two sequences when both sequences are sorted (so that the elements in each sequence are in nondecreasing order). b) Show that \(S\) takes its minimum value over all orderings of the two sequences when one sequence is sorted into nondecreasing order and the other is sorted into nonincreasing order.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

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.