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
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! 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}
Learning Materials
Features
Discover
Chapter 3: Problem 42
Every prime number except 2 and 3 has the form \(6 q+1\) or \(6 q+5\) for some integer \(q\).
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
The following "proof" that every integer is rational is incorrect. Find the mistake. "Proof (by contradiction): Suppose not. Suppose every integer is irrational, Then the integer 1 is irrational. But \(1=1 / 1\), which is rational. This is a contradiction. [Hence the supposition is false and the theorem is true.]"
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 \(b\). \(\operatorname{gcd}(a, b) \cdot \operatorname{lcm}(a, b)=a b\).
For all real numbers \(x,|-x|=|x|\).
Evaluate the expressionsa. 28 div 5 b. 28 mod 5
Prove that \(\log _{5}(2)\) is irrational.
What do you think about this solution?
We value your feedback to improve our textbook solutions.