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.

Answer 1

Answer:

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

$\mathbf{X_E (-2) = 0 }$  

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

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 ....∞}

$X_E : Z \to R$ 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.

∴

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

E = {2,4,6,8,10 ....∞}