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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.

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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.

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Most popular questions from this chapter

Determine if each of the following statements is true or false. If a statement is true, then write a formal proof of that statement, and if it is false, then provide a counterexample that shows it is false. (a) For all integers \(a, b,\) and \(c\) with \(a \neq 0,\) if \(a \mid b,\) then \(a \mid(b c)\). (b) For all integers \(a\) and \(b\) with \(a \neq 0,\) if \(6 \mid(a b),\) then \(6 \mid a\) or \(6 \mid b\). (c) For all integers \(a, b,\) and \(c\) with \(a \neq 0,\) if \(a\) divides \((b-1)\) and \(a\) divides \((c-1),\) then \(a\) divides \((b c-1)\) (d) For each integer \(n,\) if 7 divides \(\left(n^{2}-4\right),\) then 7 divides \((n-2)\). (e) For every integer \(n, 4 n^{2}+7 n+6\) is an odd integer. ? (f) For every odd integer \(n, 4 n^{2}+7 n+6\) is an odd integer. (g) For all integers \(a, b,\) and \(d\) with \(d \neq 0,\) if \(d\) divides both \(a-b\) and \(a+b,\) then \(d\) divides \(a\) (h) For all integers \(a, b,\) and \(c\) with \(a \neq 0,\) if \(a \mid(b c),\) then \(a \mid b\) or \(a \mid c .\)

In Preview Activity \(2,\) we proved that if \(n\) is an integer, then \(n^{2}+n\) is an even integer. We define two integers to be consecutive integers if one of the integers is one more than the other integer. This means that we can represent consecutive integers as \(m\) and \(m+1,\) where \(m\) is some integer.

Let \(a\) and \(b\) be integers. Prove that if \(a \equiv 2(\bmod 3)\) and \(b \equiv 2(\bmod 3)\), then (a) \(a+b \equiv 1(\bmod 3) ;\) (b) \(a \cdot b \equiv 1(\bmod 3)\).

One of the most famous unsolved problems in mathematics is a conjecture made by Christian Goldbach in a letter to Leonhard Euler in 1742. The conjecture made in this letter is now known as Goldbach's Conjecture. The conjecture is as follows: Every even integer greater than 2 can be expressed as the sum of two (not necessarily distinct) prime numbers. Currently, it is not known if this conjecture is true or false. (a) Write \(50,142,\) and 150 as a sum of two prime numbers. (b) Prove the following: If Goldbach's Conjecture is true, then every integer greater than 5 can be written as a sum of three prime numbers. (c) Prove the following: If Goldbach's Conjecture is true, then every odd integer greater than 7 can be written as a sum of three odd prime numbers.

In Exercise (15) in Section 3.2, we proved that there exists a real number solution to the equation \(x^{3}-4 x^{2}=7\). Prove that there is no integer \(x\) such that \(x^{3}-4 x^{2}=7\)
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