Question

If $U$ is the universal set of all natural numbers, then which of the following options are true?

• A. $\text{If}A=\{x:x\text{is an odd natural number}\},$
  $\text{then}{A}^{\prime }=\{x:x\text{is an even natural number}\}$
• B. $\text{If}B=\{y:y\text{is a prime number}\},$ $\text{then}{B}^{\prime }=\{y:y\text{is a composite number}\}$
• C. $\text{If}C=\{z:z\in N,z+8=9\}$
  $\text{then}{C}^{\prime }=\left\{1\right\}$
• D. $\text{If}D=\{p:p\in N,3p-1>11\},$
  $\text{then}{D}^{\prime }=\{p:p\in N,p\in 4\}$

Answer 1

Answer: (A), (B), (D)

$\text{If}D=\{p:p\in N,3p-1>11\},$

$\text{then}{D}^{\prime }=\{p:p\in N,p\in 4\}$
$U=N$
$N=\text{collection of both even and odd natural numbers}$
So if $A$ is the collection of odd natural nos., then $A^{\prime }$ is the collection of even natural nos.

If $B$ is the set of all the prime numbers, then $B^{\prime }$ is the set of all composite numbers ${}^{\prime }+1^{\prime }$ as ${}^{\prime }{1}^{\prime }$ is neither a prime nor a composite number.

If $C=\{z:z\in N,z+8=9\},\text{then}z=1,$ $\Rightarrow C^{\prime }$ is the set of all natural numbers other than $1$.

If $D=\{p:p\in N,3p-1>11\},p>4.$This means that the compliment of will all those natural numbers that are less than or equal to $4.$