Chapter 3: Problem 49
For all real numbers \(x,|-x|=|x|\).
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 49
For all real numbers \(x,|-x|=|x|\).
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
If \(n\) is an integer and \(n>1\), then \(n !\) is the product of \(n\) and every other positive integer that is less than \(n\). For example, \(5 !=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\). a. Write \(6 !\) in standard factored form. b. Write \(20 !\) in standard factored form. c. Without computing the value of \((20 !)^{2}\) determine how many zeros are at the end of this number when it is written in decimal form. Justify your answer.
For all integers \(m, m^{2}=5 k\), or \(m^{2}=5 k+1\), or \(m^{2}=\) \(5 k+4\) for some integer \(k\).
"Proof: Suppose \(r\) and \(s\) are rational numbers. By definition of rational, \(r=a / b\) for some integers \(a\) and \(b\) with \(b \neq 0\), and \(s=a / b\) for some integers \(a\) and \(b\) with \(b \neq 0\). Then \(r+s=a / b+a / b=2 a / b\). Let \(p=2 a\). Then \(p\) is an integer since it is a product of integers. Hence \(r+s=p / b\), where \(p\) and \(b\) are integers and \(b \neq 0\). Thus \(r+s\) is a rational number by definition of rational. This is what was to be shown."
When an integer \(a\) is divided by 7 , the remainder is 4 . What is the remainder when \(5 a\) is divided by \(7 ?\)
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 \(b, a \mid b\) if, and only if, \(\operatorname{lcm}(a, b)=b\).
What do you think about this solution?
We value your feedback to improve our textbook solutions.