Question

Let Ebe the set of all even positive integers in the universe Zof integers, and XE : Z R be the characteristic function of E.

Then
1. XE(2) =
2. XP(-2) =
3. {z 62: XE() = 1} =

Answer 1

`\mathbf{X_E (2) = 1}`

`\mathbf{X_E (-2) = 0 }`

`\mathbf{\{ x \in Z: X_E(x) = 1\} = E}`

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 ....∞}

`X_E : Z \to R` be the characteristic function of E.

∴

`X_E(x) = \left \{ {{1 \ if \ x \ \ is \ an \ element \ of \ E} \atop {0 \ if \ x \ \ is \ not \ an \ element \ of \ E}} \right.`

For XE(2)

`\mathbf{X_E (2) = 1}` since x is an element of E (i.e the set of all even numbers)

For XE(-2)

`\mathbf{X_E (-2) = 0 }` 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.

∴

`\{ x \in Z: X_E(x) = 1\} = \ x \in E\ \\ \\ \mathbf{\{ x \in Z: X_E(x) = 1\} = E}`

E = {2,4,6,8,10 ....∞}