/*! 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 9 Prove that for each natural numb... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 9

Prove that for each natural number \(n, \sqrt{3 n+2}\) is not a natural number.

Short Answer

Expert verified
Assuming that there exists a natural number \(n\) such that \(\sqrt{3n+2}\) is a natural number, we derived the equation \(n = \frac{k^2 - 2}{3}\), with \(k\) being a natural number. However, by looking at the possible values of \(k\) modulo 3, we found that in all cases, \(k^2 - 2\) is not divisible by 3, contradicting our assumption that \(n\) is a natural number. Therefore, for each natural number \(n\), \(\sqrt{3n+2}\) is not a natural number.

Step by step solution

01

Assume the opposite of the given statement.

To begin, let's assume the opposite of the given statement: there exists a natural number n for which \(\sqrt{3n+2}\) is a natural number. Let \(k\) be such number, so we have \(\sqrt{3n+2} = k\), where \(k\) is a natural number.
02

Square both sides of the equation.

Since we have \(\sqrt{3n+2} = k\), we can square both sides of the equation to get rid of the square root: \((\sqrt{3n+2})^2 = k^2\) This simplifies to: \(3n + 2 = k^2\)
03

Rearrange the equation.

Rearrange the equation to isolate n: \(3n = k^2 - 2\) \(n = \frac{k^2 - 2}{3}\)
04

Show there is a contradiction.

From the equation derived in step 3, we can observe that \(k^2 - 2\) must be divisible by 3 for n to be a natural number. Now, let's consider the possible values of \(k\) modulo 3. There are three possibilities: 1. \(k \equiv 0 \pmod{3}\), which means that \(k\) is divisible by 3. In this case, \(k^2\) is also divisible by 3, but \(k^2 - 2\) is not divisible by 3. 2. \(k \equiv 1 \pmod{3}\), which means that \(k\) has a remainder of 1 when divided by 3. In this case, \(k^2\) has a remainder of 1 when divided by 3, so \(k^2 - 2\) is not divisible by 3. 3. \(k \equiv 2 \pmod{3}\), which means that \(k\) has a remainder of 2 when divided by 3. In this case, \(k^2\) has a remainder of 1 when divided by 3, so \(k^2 - 2\) is not divisible by 3. In each of these cases, \(k^2 - 2\) is not divisible by 3, contradicting our assumption that n is a natural number because in all cases, n would not be an integer. This means that our original assumption that there exists a natural number \(n\) such that \(\sqrt{3n+2}\) is a natural number is incorrect.
05

Conclude the proof.

Since we have reached a contradiction by assuming that there exists a natural number \(n\) such that \(\sqrt{3n+2}\) is a natural number, we can conclude that there is no such natural number \(n\). Therefore, for each natural number \(n\), \(\sqrt{3n+2}\) is not a natural number, proving the given statement.

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.

Natural Numbers
Natural numbers, often denoted as \( \mathbb{N} \), are the numbers we use for counting and ordering. They are the positive integers starting from 1 and going upwards: 1, 2, 3, 4, and so on indefinitely. These are the numbers we first learn as children and use in everyday life.

In mathematics, natural numbers have significant properties and are used in various proofs and theorems. For example, when we say that \( \sqrt{3n+2} \) is not a natural number for any natural number \(n\), we claim that this square root never results in a countable, whole number, which impacts our understanding of number theory and algebra.
Square Roots
A square root of a number is a value that, when multiplied by itself, gives the original number. For instance, the square root of 4 is 2 because \(2 \times 2 = 4\). Symbolically, it is represented as \(\sqrt{x}\), where \(x\) is the number whose square root we are seeking.

Square roots are intriguing because they do not always result in natural numbers. For instance, while some numbers like 25 have a square root of 5, a natural number, others like 2 do not have a natural number as their square root. This is closely related to the concept of irrational numbers, which cannot be expressed as a simple fraction.
Divisibility
Divisibility is a fundamental concept in arithmetic and number theory. A number \(a\) is divisible by another number \(b\), if there exists a natural number \(k\) such that \(a = k \times b\). When a number is divisible by another, it produces no remainder upon division.

In our proof, we are particularly interested in the divisibility by 3. This is because we end up with an expression \(3n = k^2 - 2\) and we need \(n\) to be a natural number, which it can only be if \(k^2 - 2\) is divisible by 3. Understanding the divisibility rules helps in clarifying why the expression \(k^2 - 2\) cannot be divisible by 3 and hence, why \(n\) cannot be a natural number in our scenario.
Modular Arithmetic
Modular arithmetic is a system of arithmetic for integers, where we focus on the remainder when an integer is divided by another integer, known as the modulus. It is often visualized on a clock, which is why it is sometimes referred to as 'clock arithmetic'.

In the context of our proof, we explore the value of \(k^2 - 2\) in terms of its remainder when divided by 3. This approach, modular arithmetic, allows us to simplify the problem by considering all possible remainders \(k\) could have when divided by 3 (0, 1, or 2) and show that for each case, \(k^2 - 2\) does not result in a multiple of 3, therefore it is not divisible by 3, reinforcing our proof by contradiction.

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

Recall that a Pythagorean triple consists of three natural numbers \(a, b,\) and \(c\) such that \(a

Let \(h\) and \(k\) be real numbers and let \(r\) be a positive number. The equation for a circle whose center is at the point \((h, k)\) and whose radius is \(r\) is $$ (x-h)^{2}+(y-k)^{2}=r^{2} $$ We also know that if \(a\) and \(b\) are real numbers, then \- The point \((a, b)\) is inside the circle if \((a-h)^{2}+(b-k)^{2}r^{2}\). Prove that all points on or inside the circle whose equation is \((x-1)^{2}+\) \((y-2)^{2}=4\) are inside the circle whose equation is \(x^{2}+y^{2}=26\)

See the instructions for Exercise (19) on page 100 from Section 3.1 . (a) Proposition. For all integers \(a\) and \(b\), if \((a+2 b) \equiv 0(\) mod 3 ), then \((2 a+b) \equiv 0(\bmod 3)\) Proof. We assume \(a, b \in \mathbb{Z}\) and \((a+2 b) \equiv 0(\) mod 3\() .\) This means that 3 divides \(a+2 b\) and, hence, there exists an integer \(m\) such that \(a+2 b=3 m .\) Hence, \(a=3 m-2 b .\) For \((2 a+b)=0(\bmod 3),\) there exists an integer \(x\) such that \(2 a+b=3 x .\) Hence, $$ \begin{aligned} 2(3 m-2 b)+b &=3 x \\ 6 m-3 b &=3 x \\ 3(2 m-b) &=3 x \\ 2 m-b &=x . \end{aligned} $$ Since \((2 m-b)\) is an integer, this proves that 3 divides \((2 a+b)\) and hence, \((2 a+b) \equiv 0(\bmod 3)\). (b) Proposition. For each integer \(m, 5\) divides \(\left(m^{5}-m\right)\). Proof. Let \(m \in \mathbb{Z}\). We will prove that 5 divides \(\left(m^{5}-m\right)\) by proving that \(\left(m^{5}-m\right) \equiv 0(\bmod 5)\). We will use cases. For the first case, if \(m \equiv 0(\bmod 5),\) then \(m^{5} \equiv 0(\bmod 5)\) and, hence, \(\left(m^{5}-m\right) \equiv 0(\bmod 5)\) For the second case, if \(m \equiv 1(\bmod 5),\) then \(m^{5} \equiv 1(\bmod 5)\) and, hence, \(\left(m^{5}-m\right) \equiv(1-1)(\) mod 5\()\), which means that \(\left(m^{5}-m\right) \equiv\) \(0(\bmod 5)\) For the third case, if \(m \equiv 2(\bmod 5),\) then \(m^{5} \equiv 32(\bmod 5)\) and, hence, \(\left(m^{5}-m\right)=(32-2)(\bmod 5),\) which means that \(\left(m^{5}-m\right)=\) \(0(\bmod 5)\).

Are the following propositions true or false? Justify your conclusion. (a) There exist integers \(x\) and \(y\) such that \(4 x+6 y=2\). (b) There exist integers \(x\) and \(y\) such that \(6 x+15 y=2\). (c) There exist integers \(x\) and \(y\) such that \(6 x+15 y=9\).

Let \(a\) and \(b\) be natural numbers such that \(a^{2}=b^{3}\). Prove each of the propositions in Parts (a) through (d). (The results of Exercise (1) and Theorem 3.10 may be helpful.) (a) If \(a\) is even, then 4 divides \(a\). (b) If 4 divides \(a,\) then 4 divides \(b\). (c) If 4 divides \(b\), then 8 divides \(a\). (d) If \(a\) is even, then 8 divides \(a\). (e) Give an example of natural numbers \(a\) and \(b\) such that \(a\) is even and \(a^{2}=b^{3},\) but \(b\) is not divisible by 8
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Applied Mathematics

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.