## Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning

# Chapter 1 - Speaking Mathematically - Exercise Set 1.2: 8

#### Answer

a. No, because $j$ is an element of $B$ that is not in $A$. b. Yes, because every element of $C$ is an element of $C$. c. Yes, because every element of $C$ also appears in $A$. d. Yes, because every element of $C$ also appears in $C$, but there are elements of $A$ that do not appear in $C$.

#### Work Step by Step

Note that every set is an improper subset of itself. A subset of $X$ is a set $Y$ such that every element in $Y$ appears in $X$. $Y$ is an improper subset of $X$ if $X$ is simultaneously a subset of $Y$. $Y$ is a proper subset of $X$ if $X$ is not a subset of $Y$.

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