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Choose the correct option:

Let N be the set of non-negative integers, I the set of integers $N_{p}$ the set of non-positive integers, E the set of even integers and P the set of prime numbers. Then which of the following is correct.

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I - N = 4 N_{p}

N \cap N_{p} = ϕ

E \cap P = ϕ

NΔ N_{p} = 1 - {0}

answer is D.

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Detailed Solution

Given,
Non negative integers = N
Assume that

n = {0, 1, 2, 3, \dots \dots}

[all positive integer is 0]
Sets of integer =

I

I = {\dots \dots, - 3, - 2, - 1, 0, 1, 2, 3, \dots \dots}

Set of non positive integer =

N_{p}

N_{p} = {\dots \dots, - 3, - 2, - 1, 0}

A set of even integer = E

E = \{0, 2, 4, 6 \dots . .\}

A set of prime number = P

P = {2, 3, 5, 7 \dots \dots}

By option verification,

I - N = N_{p}

I - N = \{\dots \dots ., - 3, - 2, - 1, 0, 1, 2, 3, \dots \dots\}

- \{0, 1, 2, 3, \dots \dots\}

= \{\dots \dots \dots, - 3, - 2, - 1\}

Therefore

I - N \neq N

Again, we have

N \cap N_{p} = \emptyset

N \cap N_{p}

[set of common element

N and N_{p}

]

N - Np = \{0, 1, 2, 3, \dots \dots\} - \{\dots \dots \dots \cdot - 3, - 2, - 1, 0\}

= {1, 2, 3, \dots \dots}

Therefore,

N \cap N_{p} \neq ⟨0⟩ N \cap N_{p}

Again we have

E \cap P = \emptyset

E \cap P

[set of common element of E and P]

N Δ N_{p} = I - ⟨0⟩

E \cap P = \{0, 2, 4, 6 \dots \dots\} \cap \{2, 3, 5, 7 \dots \dots\} = \{2\}

E \cap P

is not empty
Again we have

N \cap N_{p} = 1 - \{0\}

N Δ N_{p}

[symmetric difference of N and

Np

]

N Δ N_{p} = (N - N_{p}) \cup (N_{p} - N)

When

N_{p} is removed

N - N_{p} = \{0, 1, 2, 3, \dots \dots\} - \{\dots \dots - 3, - 2, - 1, 0\}

= \{1, 2, 3, \dots \dots\}

When N is removed

N_{p} - N = \{\dots \dots, - 3, - 2, - 1, 0⟩ - ⟨0, 1, 2, 3 \dots . .\}

= \{\dots - 3, - 2, - 1\}

Therefore

N Δ N_{p} = \{1, 2, 3, \dots\} \cup \{\dots \dots - 3, - 2, - 1\}

= \{\dots \dots - 3, - 2, - 1, 1, 2, 3, \dots\}

Also,

I - \{0\} = \{\dots \dots - 3 . - 2 . - 1, \dots \dots\} - \{0\}

= \{\dots ., - 3, - 2, - 1, 1, 2, 3, \dots \dots\}

Hence

N Δ N_{p} = I - \{0\}

Correct option is 1.

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