Answer:
[tex]\mathbf{X_E (2) = 1}[/tex]
[tex]\mathbf{X_E (-2) = 0 }[/tex]
[tex]\mathbf{\{ x \in Z: X_E(x) = 1\} = E}[/tex]
Step-by-step explanation:
Let E be the set of all even positive integers in the universe Z of integers,
i.e
E = {2,4,6,8,10 ....∞}
[tex]X_E : Z \to R[/tex] be the characteristic function of E.
∴
[tex]X_E(x) = \left \{ {{1 \ if \ x \ \ is \ an \ element \ of \ E} \atop {0 \ if \ x \ \ is \ not \ an \ element \ of \ E}} \right.[/tex]
For XE(2)
[tex]\mathbf{X_E (2) = 1}[/tex] since x is an element of E (i.e the set of all even numbers)
For XE(-2)
[tex]\mathbf{X_E (-2) = 0 }[/tex] since - 2 is less than 0 , and -2 is not an element of E
For { x ∈ Z: XE(x) = 1}
This can be read as:
x which is and element of Z such that X is also an element of x which is equal to 1.
∴
[tex]\{ x \in Z: X_E(x) = 1\} = \{ x \in Z | x \in E\} \\ \\ \mathbf{\{ x \in Z: X_E(x) = 1\} = E}[/tex]
E = {2,4,6,8,10 ....∞}
