Answer: The elements of the given sets are listed below.
Step-by-step explanation: We are given to list the elements of the following sets :
[tex](a)~\left\{\dfrac{1}{n}:n\in \{3,4,5,6\}\right\},\\\\\\(b)~\{x\in \mathbb{Z}|x=x+1\},\\\\\\(c)~\{n\in P|\textup{n is a factor of 24}\}.[/tex]
The listing of the elements of the given sets is done below :
[tex](a)~\left\{\dfrac{1}{n}:n\in \{3,4,5,6\}\right\}.[/tex]
Here, the elements of the set will be the reciprocal of 3, 4, 5 and 6.
So, the list of elements of the set is
[tex]\left\{\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5},\dfrac{1}{6}\right\}.[/tex]
[tex](b)~\{x\in \mathbb{Z}|x=x+1\}.[/tex]
For x to be an element of this set, we have
[tex]x=x+1\\\\\Rightarrow 0=1,[/tex] which is NEVER possible. So, the given set is empty.
That is, [tex]\{x\in \mathbb{Z}|x=x+1\}=\phi.[/tex]
[tex](c)~\{n\in P|\textup{n is a factor of 24}\}.[/tex]
The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
Since we need prime factors, so the elements of the given set are
[tex]\{2,3\}.[/tex]
Thus, all the elements of the given sets are listed.