Chapter 2, Problem 1RE

If $A=\left\{x\in N|x<7\right\},B=\left\{x\in Z|\left|x-5\right|<3\right\}\text{ and }C=\left\{2,3\right\}$, find $\left(A\oplus B\right)\backslash C$.

To determine

$\left(A\oplus B\right)\backslash C$

Answer to Problem 1RE

Solution: $\left(A\oplus B\right)\backslash C=\left\{1,7\right\}$

Explanation of Solution

Given Information: $A=\left\{x\in N|x<7\right\},B=\left\{x\in Z|\left|x-5\right|<3\right\}\text{ and }C=\left\{2,3\right\}$

Explanation:

The values of x satisfies the condition $x\in N$ such that $x<7$ are $\left\{1,2,3,4,5,6\right\}$.

Therefore, $A=\left\{1,2,3,4,5,6\right\}$.

The values of x satisfies the condition $x\in Z$ such that $\left|x-5\right|<3$ are $\left\{3,4,5,6,7\right\}$.

Therefore, $B=\left\{3,4,5,6,7\right\}$.

The symmetric difference of two sets A and B in the set $A\oplus B$ and is given as $A\oplus B=\left(A-B\right)\cup \left(B-A\right)$.

$\begin{array}{l}A\oplus B=\left(A-B\right)\cup \left(B-A\right)\\ =\left(\left\{1,2,3,4,5,6\right\}-\left\{3,4,5,6,7\right\}\right)\cup \left(\left\{3,4,5,6,7\right\}-\left\{1,2,3,4,5,6\right\}\right)\\ =\left\{1,2\right\}\cup \left\{7\right\}\\ =\left\{1,2,7\right\}\end{array}$

$\left(A\oplus B\right)\backslash C$ is the set difference of the sets $A\oplus B$ and C, and is the set of those elements which are in $A\oplus B$ but not in C.

$\begin{array}{l}\left(A\oplus B\right)\backslash C=\left\{1,2,7\right\}-\left\{2,3\right\}\\ =\left\{1,7\right\}\end{array}$