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Bumek [7]
3 years ago
7

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

Mathematics
1 answer:
AnnZ [28]3 years ago
5 0

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 ....∞}

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