Question

Let E denote the set of even integers and O denote the set of odd integers. As usual, let Z denote the set of all integers. Determine each of these sets.The set of English words that do not contain the letter x.

$$\begin{gathered} \left(a\right)\mathbf{}\mathit{E}\mathbf{\cup }\mathit{O} \\ \left(b\right)\mathbf{}\mathit{E}\mathbf{\cap }\mathit{O} \\ \left(c\right)\mathbf{}\mathit{Z}\mathbf{-}\mathit{E} \\ \left(d\right)\mathbf{}\mathit{Z}\mathbf{-}\mathit{O} \end{gathered}$$

Short answer

$$\begin{gathered} \left(a\right)Z \\ \left(b\right)\cancel{O} \\ \left(c\right)O \\ \left(d\right)E \end{gathered}$$

Step-by-step solution

Step 1:

Union $\mathit{A}\mathbf{\cup }\mathit{B}$: all elements that are either in A or in B.

Intersection $\mathit{A}\mathbf{\cap }\mathit{B}$: all elements that are both A and in B.

Difference A-B: all elements in A that are NOT in B(complement of B with respect to A).

Step 2:

E= set of even integers.

O= set of odd integers.

Z= set of all integers.

(a) The union of E and O is the set of all integers that are even OR odd. Since every integer is either even or odd, the union is thus the set of all integers.

$E\cup O=\left\{setofallintezers\right\}=Z$

(b) The union of E and O is the set of all integers that are even AND odd. Since an integer cannot be both even and odd, the intersection cannot contain any integers and thus the intersection has to be the empty set.

$E\cap O=\cancel{O}$

(c) The difference of Z and E is the set of all integers that NOT even. An integer that is not even is odd and thus the difference is the set of all odd integers.

$Z-E=\left\{setofallintezers\right\}=O$

(d) The difference of Z and O is the set of all integers that NOT odd. An integer that is even and thus the difference is the set of all even integers.

$Z-O=\left\{setofallintezers\right\}=E$

Hence, the solution is

$$\begin{gathered} \left(a\right)Z \\ \left(b\right)\cancel{O} \\ \left(c\right)O \\ \left(d\right)E \end{gathered}$$