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Chapter 3: Problem 42

Every prime number except 2 and 3 has the form \(6 q+1\) or \(6 q+5\) for some integer \(q\).

Short Answer

Expert verified
Using proof by contradiction, we assumed that there is a prime number not in the form of \(6q + 1\) or \(6q + 5\). However, considering other possible forms, we found that numbers of these forms are divisible by 2, 3, or 6, having more than two distinct positive divisors, meaning they cannot be prime numbers. Thus, we conclude that every prime number except 2 and 3 must be in the form of \(6q + 1\) or \(6q + 5\) for some integer \(q\).

Step by step solution

01

Understand the Concept of Prime Numbers and their Properties

A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself. The smallest prime numbers are 2, 3, 5, and 7. A key property of prime numbers is that they cannot be divided by any other numbers without leaving a remainder.
02

Proof by Contradiction

To prove that every prime number except 2 and 3 has the form \(6q + 1\) or \(6q + 5\) for some integer \(q\), we will use proof by contradiction. We will assume that there is a prime number that is not in the form of \(6q + 1\) or \(6q + 5\) and derive a contradiction.
03

Consider Cases for Prime Numbers

If a number is not in the form of \(6q + 1\) or \(6q + 5\), it can be in one of the following forms: 1. \(6q\) (divisible by 6) 2. \(6q + 2\) (divisible by 2) 3. \(6q + 3\) (divisible by 3) 4. \(6q + 4\) (divisible by 2) In all these cases, since the number is divisible by 2, 3, or 6, it must have more than two distinct positive divisors and cannot be a prime number.
04

Conclude the Proof

Since we reached a contradiction to our assumption, it implies that every prime number except 2 and 3 must be in the form of \(6q + 1\) or \(6q + 5\) for some integer \(q\).

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Key Concepts

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

Proof by Contradiction
The method of proof by contradiction is particularly powerful in mathematics. It starts by assuming the opposite of what you want to prove. If this assumption leads to a logical inconsistency, or a contradiction, then the original statement must be true. For prime numbers, this approach is used to establish definitive forms that prime numbers can take.

For instance, if we want to prove that all prime numbers except 2 and 3 are of the form \(6q + 1\) or \(6q + 5\), we assume that there exists a prime number that does not fit this pattern. By exploring all other forms that such a number could take, we find that each alternative results in a number that is not a prime, since it would be divisible by 2, 3, or 6. This contradiction validates our original claim, reinforcing the strength of proof by contradiction in establishing such mathematical truths.
Divisibility
The concept of divisibility is fundamental to understanding prime numbers. A number is said to be divisible by another if you can divide the first number by the second without any remainder. The key aspect of a prime number is that it is only divisible by 1 and itself. To expand on this, we use the divisibility properties of numbers to show why prime numbers have specific forms.

This is illustrated by analyzing numbers that are not of the form \(6q + 1\) or \(6q + 5\). Numbers like \(6q\), \(6q + 2\), \(6q + 3\), and \(6q + 4\) reveal their composite nature through their inherent divisibility by 2 or 3. Employing the concept of divisibility helps clarify why any other form for a prime number besides the two given would imply the number isn't prime at all.
Natural Numbers
Prime numbers are a special category within the set of natural numbers, which consist of all positive integers starting from 1. In the realm of natural numbers, prime numbers stand out because of their unique properties and the role they play in various mathematical theories and applications. Understanding prime numbers requires a grasp of natural numbers since all primes are, by definition, natural numbers.

When discussing the form of prime numbers in the context of the exercise, we are implicitly referring to natural numbers since negative integers and fractions, which are not natural numbers, cannot be primes. This exclusive representation of primes within the natural numbers illustrates both the simplicity and complexity of these fundamental mathematical entities.
Integers
The term integers refers to the set of whole numbers and their negatives. It's a broader category than natural numbers, which include only positive integers and zero. In the context of our exercise, the integer \(q\) within the expressions \(6q + 1\) and \(6q + 5\) represents any integer value that, when substituted, yields a prime number. This is crucial as it implies that the equations can map out various prime numbers based on different integer values of \(q\).

An important point to note is that while the integer \(q\) is essential in defining the formula for prime numbers, the primes themselves remain positive due to their inherent nature as natural numbers. This intersection between the sets of integers and natural numbers is at the heart of understanding how we categorize different types of numbers in mathematics.

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Most popular questions from this chapter

Use the properties of even and odd integers that are listed in Example \(3.2 .3\) to do Indicate which properties you use to justify your reasoning.True or false? If \(a\) is any odd integer, then \(a^{2}+a\) is even. Explain.

For all integers \(m, m^{2}=5 k\), or \(m^{2}=5 k+1\), or \(m^{2}=\) \(5 k+4\) for some integer \(k\).

Definition: The least common multiple of two nonzero integers \(a\) and \(b\), denoted \(\operatorname{lcm}(a, b)\), is the positive integer \(c\) such that a. \(a \mid c\) and \(b \mid c\) b. for all integers \(m\), if \(a \mid m\) and \(b \mid m\), then \(c \mid m\). Prove that for all positive integers \(a\) and \(b\). \(\operatorname{gcd}(a, b) \cdot \operatorname{lcm}(a, b)=a b\).

Definition: The least common multiple of two nonzero integers \(a\) and \(b\), denoted \(\operatorname{lcm}(a, b)\), is the positive integer \(c\) such that a. \(a \mid c\) and \(b \mid c\) b. for all integers \(m\), if \(a \mid m\) and \(b \mid m\), then \(c \mid m\). Prove that for all positive integers \(a\) and \(b, \operatorname{gcd}(a, b)=\) \(\operatorname{lcm}(a, b)\) if, and only if \(a=b\).

If \(r\) is any rational number and \(s\) is any irrational number, then \(r / s\) is irrational.
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