Chapter 3: Problem 21
For all real numbers \(x\), if \(x>1\) then \(x^{2}>x\).
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 21
For all real numbers \(x\), if \(x>1\) then \(x^{2}>x\).
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
Observe that $$ \begin{aligned} 7524 &=7 \cdot 1000+5 \cdot 100+2 \cdot 10+4 \\ &=7(999+1)+5(99+1)+2(9+1)+4 \\ &=(7 \cdot 999+7)+(5 \cdot 99+5)+(2 \cdot 9+2)+4 \\ &=(7 \cdot 999+5 \cdot 99+2 \cdot 9)+(7+5+2+4) \\ &=(7 \cdot 111 \cdot 9+5 \cdot 11 \cdot 9+2 \cdot 9)+(7+5+2+4) \\ &=(7 \cdot 111+5 \cdot 11+2) \cdot 9+(7+5+2+4) \\ &=(\text { an integer divisible by } 9) \end{aligned} $$ \(\begin{aligned} 7524 &=7 \cdot 1000+5 \cdot 100+2 \cdot 10+4 \\\ &=7(999+1)+5(99+1)+2(9+1)+4 \\ &=(7.999+7)+(5 \cdot 99+5)+(2 \cdot 9+2)+4 \\\ &=(7 \cdot 999+5 \cdot 99+2 \cdot 9)+(7+5+2+4) \\ &=(7 \cdot 111 \cdot 9+5 \cdot 11 \cdot 9+2 \cdot 9)+(7+5+2+4) \\ &=(7 \cdot 111+5 \cdot 11+2) \cdot 9+(7+5+2+4) \\ &=(\text { an integer divisible by } 9) \\ &+(\text { the sum of the digits of } 7524) \end{aligned}\) Since the sum of the digits of 7524 is divisible by 9,7524 can be written as a sum of two integers each of which is divisible by 9 . It follows from exercise 15 that 7524 is divisible by \(9 .\) Generalize the argument given in this example to any nonnegative integer \(n\). In other words, prove that for any nonnegative integer \(n\), if the sum of the digits of \(n\) is divisible by 9 , then \(n\) is divisible by 9 ,
Find the greatest common divisor of each of the pairs of integers in \(9-12\). (Use any method you wish.) 7 and 21
Assume that \(r\) and \(s\) are particular integers. a. Is \(4 r s\) even? b. Is \(6 r+4 s^{2}+3\) odd? c. If \(r\) and \(s\) are both positive, is \(r^{2}+2 r s+s^{2}\) composite?
Write an algorithm that accepts the numerator and denominator of a fraction as input and produces as output the numerator and denominator of that fraction written in lowest terms. (The algorithm may call upon the Euclidean algorithm as needed.)
State a necessary and sufficient condition for the floor of a real number to equal that number.
What do you think about this solution?
We value your feedback to improve our textbook solutions.