Question

For the universal set\\{ $4,5,6,7,8,9,10,11,12,13$ \\} , find its subsets $A,B,C$and $D$
$A$) \\{ even numbers\\}
$B$) \\{ odd numbers greater than $8$ \\}
$C$) Prime numbers
$D$) even numbers less than $10$

Answer 1

First we have to define what the terms we need to solve the problem are.
A number which is divisible by $2$and generates a remainder of $0$is called an even number.
Odd numbers are whole numbers that cannot be divided exactly into pairs. Odd numbers, when divided by$2$, leave a remainder of $1,3,5,7,9,11,13,15{\text{ }} \ldots$are sequential odd numbers. Odd numbers have the digits $1,3,5,7or9$ in their one’s place.
Prime numbers are whole numbers greater than$1$ , that have only two factors $1$and the number itself. Prime numbers are divisible only by the number $1$or itself.

Complete step by step answer:
Since we know the definition of even number, odd number and prime number we further approach to find option $A$ which is the set of all even numbers in the given set.
Thus $\\{ 4,6,8,10,12\\}$are the numbers which are divisible by $2$and generates a remainder zero
Hence $A$=$\\{ 4,6,8,10,12\\}$
Now for option $B$which is the set of all odd numbers also greater than $8$
Thus, odd numbers also greater that $8$ as seen $\\{ 9,11,13\\}$are the numbers which are cannot divided exactly two pairs and leaves a remainder $1,3,5, \ldots$
Hence $B$= $\\{ 9,11,13\\}$
Now for option $C$ which is a set of all prime number to find from universal set
Since Prime numbers are greater than $1$, that have only two factors $1$and the number itself which are $C$= $\\{ 5,7,11,13\\}$
And finally, $D$ is the even numbers also less that $10$
which are $D$= $\\{ 4,6,8\\}$even numbers also less that $10$
Hence $A$=$\\{ 4,6,8,10,12\\}$, $B$=$\\{ 9,11,13\\}$, $C$= $\\{ 5,7,11,13\\}$and $D$= $\\{ 4,6,8\\}$

Note: We find even numbers, odd numbers, prime numbers all in the universal set only
And $A,B,C$and $D$ are the subsets of the given universal set. Also, the universal set does not contain any repeated elements.