Question

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 members. Then which of the following is true?
(a) $I-N={{N}_{p}}$
(b) $N\cap P=\varnothing $
(c) $E\cap P=\varnothing $
(d) $N\Delta {{N}_{p}}=1-\left\{ 0 \right\}$

Answer 1

Hint: In this question, we will write sets in tabular form and then solve the question by verification of options. We shall check each option and find which one is the correct one.

Complete step-by-step answer:
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,
$n=\left\{ 0,1,2,3,...... \right\}$
Next given that, $I$ is a set of integers. So, it will contain all, positive, negative and zero. Therefore,
$I=\left\{ .......,-3,-2,-1,0,1,2,3,..... \right\}$
Next given that, ${{N}_{p}}$ is a set of non-positive integers, that is, all such integers which are not positive. So, ${{N}_{p}}$ will contain all negative integers and zero. Therefore
\[{{N}_{p}}=\left\{ ......,-3,-2,-1,0 \right\}\]
 Next, given that, $E$ is a set of even integers, that is multiple of 2. Therefore
$E=\left\{ 0,2,4,6..... \right\}$
Next, given that, $P$ is the set of prime numbers. Therefore,
$P=\left\{ 2,3,5,7...... \right\}$
Let us now verify the given option.
We have
$I-N={{N}_{p}}$
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,
\[\begin{align}
  & I-N=\left\{ .......,-3,-2,-1,0,1,2,3,...... \right\} \\
 & -\left\{ 0,1,2,3,..... \right\} \\
 & =\left\{ ........,-3,-2,-1 \right\} \\
\end{align}\]
This is not equal to $N$ as it does not contain element $0$
Therefore, $I-N\ne N$
Next, we have,
$N\cap {{N}_{p}}=\varnothing $
In the left hand side of this equation, we have $N\cap {{N}_{p}}$, that is set of common elements of $N$ and ${{N}_{p}}$, so we get,
$\begin{align}
  & N-Np=\left\{ 0,1,2,3,...... \right\}-\left\{ ............-3,-2,-1,0 \right\} \\
 & =\left\{ 1,2,3,..... \right\} \\
\end{align}$
So, \[N\cap {{N}_{p}}\] is not empty
Therefore\[N\cap {{N}_{p}}\ne \langle 0\rangle \]$N\cap {{N}_{p}}$
Next, we have $E\cap P=\varnothing $
In the left hand side of this equation we have $E\cap P$ that is common element of $E$ and $P$
And \[N\vartriangle {{N}_{p}}=I-\langle 0\rangle \] is those elements of ${{N}_{p}}$ from which elements of $N$ are removed.
So, we get
$E\cap P=\left\{ 0,2,4,6...... \right\}\cap \left\{ 2,3,5,7...... \right\}=\left\{ 2 \right\}$
So, $E\cap P$ is not empty
Therefore,
Next, we have is
$N\cap {{N}_{p}}=1-\left\{ 0 \right\}........(i)$$N\cap {{N}_{p}}=1-\left\{ 0 \right\}........(i)$
In the left hand side we have $N\Delta {{N}_{p}}$, that is symmetric difference of $N$ and ${{N}_{p}}$ which is given as,
$N\Delta {{N}_{p}}=\left( N-{{N}_{p}} \right)\cup \left( {{N}_{p}}-N \right)$
Here, \[N-{{N}_{p}}\] is those element of $N$ from which elements \[{{N}_{p}}\] are removed

So, \[N-{{N}_{p}}=\left\{ 0,1,2,3,...... \right\}-\left\{ .......-3,-2,-1,0 \right\}\]
$=\left\{ 1,2,3,..... \right\}$
And ${{N}_{p}}$ is those elements of ${{N}_{p}}$ from which element of $N$ are removed.
So, \[{{N}_{p}}-N=\left\{ .......,-3,-2,-1,0\rangle -\langle 0,1,2,3..... \right\}\]
\[=\left\{ ....-3,-2,-1 \right\}\]
Therefore
\[\begin{align}
  & N\vartriangle {{N}_{p}}=\left\{ 1,2,3,.... \right\}\cup \left\{ .....-3,-2,-1 \right\} \\
 & =\left\{ ......-3,-2,-1,1,2,3,.... \right\} \\
\end{align}\]
Also, right hand side of equation (i), we have
\[I-\left\{ 0 \right\}=\left\{ .........-3.-2.-1,..... \right\}-\left\{ 0 \right\}\]
\[=\left\{ .....,-3,-2,-1,1,2,3,..... \right\}\]
Hence, we get that
\[N\vartriangle {{N}_{p}}=I-\left\{ 0 \right\}\]
Therefore option (d) is the correct option.

Note: In this type of equation, always write tabular form first with all common elements of the set and then apply the operations.The options that match the final answer will be the required.