Question

Let A be the set of non-negative integers I is the set of integers B is the set of non-positive integers E is the set of even integers and P is the set of prime numbers then.​

Answer 1

Step-by-step explanation:

I= {A- (P+E)/2}+P+E/2+B

Here,A represents all whole numbers,

(P+E)/2 represents numbers(3,4,5-----)

B represents all negative integers along with 0

Answer 2

Answer:

If A be the set of non-negative integers I is the set of integers B is the set of non-positive integers E is the set of even integers and P is the set of prime numbers then N△Np=I−{0}.

Step-by-step explanation:

In the given question,

N is a non-negative integer, that is all such integers which are non-negative.

Now, we know that, 0 is an integer which is neither positive nor negative. And integers are positive, negative or zero.

So, N will be zero and all positive integers. Therefore,

n={0,1,2,3,......}

Next given that, I is a set of integers. So, it will contain all, positive, negative and zero. Therefore,

I={.......,−3,−2,−1,0,1,2,3,.....}

Next given that, Np is a set of non-positive integers, that is, all such integers which are not positive. So, Np will contain all negative integers and zero. Therefore

Np={......,−3,−2,−1,0}

Next, given that, E is a set of even integers, that is multiple of 2. Therefore

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

Next, given that, P is the set of prime numbers. Therefore,

P={2,3,5,7......}

Let us now verify the given option.

We have

I−N=Np

In left hand side of this equation, we have I−N, that is set of I from which all element of N are removed, so we, get,

I−N={.......,−3,−2,−1,0,1,2,3,......}−{0,1,2,3,.....}={........,−3,−2,−1}

This is not equal to N as it does not contain element 0

Therefore, I−N≠N

Next, we have,

N∩Np=∅

In the left hand side of this equation, we have N∩Np, that is set of common elements of N and Np, so we get,

N−Np={0,1,2,3,......}−{............−3,−2,−1,0}={1,2,3,.....}

So, N∩Np is not empty

ThereforeN∩Np≠⟨0⟩

Next, we have E∩P=∅

In the left hand side of this equation we have E∩P that is common element of E and P

And N△Np=I−⟨0⟩ is those elements of Np from which elements of N are removed.

So, we get

E∩P={0,2,4,6......}∩{2,3,5,7......}={2}

So, E∩P is not empty

Therefore,

Next, we have is N∩Np=1−{0}........(i)

In the left hand side we have NΔNp, that is symmetric difference of N and Np which is given as,

NΔNp=(N−Np)∪(Np−N)

Here, N−Np is those element of N from which elements Np are removed

So, N−Np={0,1,2,3,......}−{.......−3,−2,−1,0}

={1,2,3,.....}

And Np is those elements of Np from which element of N are removed.

So, Np−N={.......,−3,−2,−1,0⟩−⟨0,1,2,3.....}

={....−3,−2,−1}

Therefore

N△Np={1,2,3,....}∪{.....−3,−2,−1}={......−3,−2,−1,1,2,3,....}

Also, right hand side of equation (i), we have

I−{0}={.........−3.−2.−1,.....}−{0}

={.....,−3,−2,−1,1,2,3,.....}

Hence, we get that

N△Np=I−{0}