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. Prove that every prime number greater than 3 is either one more or one less than a multiple of 6 .

Short Answer

Expert verified
Every integer can be written in forms 6k to 6k+5. Only 6k+1 and 6k-1 (or equivalently 6k+5) are not divisible by 2 or 3, hence they can be prime if k is an integer. Therefore, every prime number greater than 3 is either one more or one less than a multiple of 6.

Step by step solution

01

Understanding the Problem

We need to show that any prime number greater than 3 can be expressed in the form of either 6k+1 or 6k-1, where k is an integer. Prime numbers are divisible only by 1 and themselves, so we have to prove that such prime numbers have a relationship with multiples of 6.
02

Consider All Possible Forms of an Integer

Any integer can be written in one of these forms: 6k, 6k+1, 6k+2, 6k+3, 6k+4, or 6k+5, where k is an integer. We'll examine which of these can be prime.
03

Eliminate the Non-Prime Forms

6k is not prime (except for k=1) as it is a multiple of 6. 6k+2 and 6k+4 are multiples of 2, hence not prime if k is an integer. 6k+3 is a multiple of 3 and hence not prime. This leaves us with only two forms that could be prime: 6k+1 and 6k+5.
04

Simplify the Remaining Forms

6k+5 can be written as 6(k+1)-1, which is of the form 6m-1, where m is an integer (m=k+1). Therefore, any prime number greater than 3 can be expressed as either 6k+1 or 6k-1, which are one more or one less than a multiple of 6.
05

Conclusion

We have shown that all integers can be written in one of six forms. Upon eliminating non-prime forms, we are left with only 6k+1 and 6k-1 (or 6k+5). Since these forms are always either one more or one less than a multiple of 6, we have proven the statement that every prime number greater than 3 satisfies this condition.

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Key Concepts

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

Proof Techniques in Mathematics
In mathematics, proof techniques are essential tools that enable us to demonstrate the truth or falsity of statements. The problem of proving that every prime number greater than 3 lies immediately adjacent to a multiple of 6 requires a direct proof, a common method employed in mathematics. A direct proof involves logical reasoning and the use of axioms, definitions, and previously established theorems to arrive at a conclusion that is logically consistent with the premises presented.

Our approach to this problem starts with understanding what is given and what must be shown. Then, we consider all possible forms an integer can take relative to a multiple of 6. By process of elimination, non-prime forms are excluded, leaving us with viable candidates for primes. By logical deduction, remaining forms are shown to be equivalently one more or one less than a multiple of 6. This step-by-step methodology is a straightforward illustration of how direct proofs function in mathematics.
Integer Representation
Representing integers in mathematics is a fundamental concept that allows for the discovery of patterns and properties. In the case of our exercise, integers are represented in relation to multiples of 6. The representation takes the form of 6k, where k is an integer, and possible remainders when divided by 6 are accounted for by adding the values from 0 to 5.

This form of integer representation highlights the idea of modular arithmetic, where we are primarily interested in remainders upon division by a specific number—in this case, the number 6. We can then scrutinize each form to determine whether it can be prime or not. This understanding is pivotal because it allows us to forge a direct connection between the general properties of integers and the specific case of prime numbers relative to multiples of 6.
Properties of Prime Numbers
Prime numbers are integers greater than 1 that are divisible only by 1 and themselves. This property makes them the building blocks of the number system, as every integer can be factored into a product of prime numbers, a concept known as the fundamental theorem of arithmetic.

In the context of our exercise, we are focused on prime numbers greater than 3 and their relationship to multiples of 6. Two key properties come into play: a prime number cannot be a multiple of any other integer (other than 1 and itself), and it cannot be even (other than the prime number 2) or a multiple of 3 (other than the prime number 3 itself). These properties are crucial in the elimination process of our proof, guiding us to the conclusion that the primes in question must fall in the form of 6k+1 or 6k-1 (also represented as 6k+5).
Mathematical Reasoning
Mathematical reasoning is the glue that binds individual concepts and steps of a proof together, ensuring a coherent and logical flow from hypothesis to conclusion. It requires not just an understanding of mathematical properties and representation but also the ability to apply logic to move from one step of the proof to the next. During this process, we must make certain that each step logically follows from the last and leads to a valid conclusion.

In proving that a prime number greater than 3 must be one unit away from a multiple of 6, mathematical reasoning is showcased from the initial setup to the elimination of non-prime integer forms. The careful consideration of possible remainders when an integer is divided by 6 illustrates how deduction is used to pinpoint the forms that can represent primes. Finally, reasoning ensures that these remaining forms are simplified and correctly identified as adjacent to multiples of 6, bringing us to a satisfying logical conclusion.

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Most popular questions from this chapter

Can you chain implications together? That is, if \(P \rightarrow Q\) and \(Q \rightarrow R\), does that means the \(P \rightarrow R ?\) Can you chain more implications together? Let's find out: (a) Prove that the following is a valid deduction rule: $$\begin{aligned}& P \rightarrow Q \\\& Q \rightarrow R \\\\\hline \therefore & P \rightarrow R\end{aligned}$$ (b) Prove that the following is a valid deduction rule for any \(n \geq 2\) : $$\begin{array}{c}P_{1} \rightarrow P_{2} \\\P_{2} \rightarrow P_{3} \\\\\vdots \\\P_{n-1} \rightarrow P_{n} \\\\\hline \therefore \quad P_{1} \rightarrow P_{n}\end{array}$$ I suggest you don'\operatorname{tg} o ~ t h r o u g h ~ t h e ~ t r o u b l e ~ o f ~ w r i t i n g ~ o u t ~ a ~ \(2^{n}\) row (a) and mathematical truth table. Instead, you should use part induction.

Write the negation, converse and contrapositive for each of the statements below. (a) If the power goes off, then the food will spoil. (b) If the door is closed, then the light is off. (c) \(\forall x\left(x<1 \rightarrow x^{2}<1\right)\) (d) For all natural numbers \(n,\) if \(n\) is prime, then \(n\) is solitary. (e) For all functions \(f,\) if \(f\) is differentiable, then \(f\) is continuous. (f) For all integers \(a\) and \(b\), if \(a \cdot b\) is even, then \(a\) and \(b\) are even. (g) For every integer \(x\) and every integer \(y\) there is an integer \(n\) such that if \(x>0\) then \(n x>y\) (h) For all real numbers \(x\) and \(y\), if \(x y=0\) then \(x=0\) or \(y=0\). (i) For every student in Math 228 , if they do not understand implications, then they will fail the exam.

Consider the statement: for all integers \(a\) and \(b\), if \(a\) is even and \(b\) is a multiple of 3 , then \(a b\) is a multiple of 6 . (a) Prove the statement. What sort of proof are you using? (b) State the converse. Is it true? Prove or disprove.

Prove the statement: For all integers \(n\), if \(5 n\) is odd, then \(n\) is odd. Clearly state the style of proof you are using.

Simplifying negations will be especially useful in the next section when we try to prove a statement by considering what would happen if it were false. For each statement below, write the negation of the statement as simply as possible. Don't just say, "it is false that ... ". (a) Every number is either even or odd. (b) There is a sequence that is both arithmetic and geometric. (c) For all numbers \(n,\) if \(n\) is prime, then \(n+3\) is not prime.

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