Question

If \[A=\left\{ x:x\in N,x\text{ }is\text{ }a\text{ }factor\text{ }of\text{ }6 \right\}\] and \[B=\left\{ x\in N:x\text{ }is\text{ }a\text{ }factor\text{ }of\text{ 8} \right\}\] then find:
(i) A-B
(ii) B-A

Answer 1

Hint: We have been given the set in set builder form. First of all we will change it into Roster form or Tabular form. In set builder form, a set is represented by a statement or some rule whereas in roster form, all the elements of a set are listed inside a pair of braces. Also, (A-B), where both A and B are set means a set which contains elements of A which are not the elements of B.

Complete step-by-step answer:
We have been given the set A and set B in set builder form as follows:
\[A=\left\{ x:x\in N,x\text{ }is\text{ }a\text{ }factor\text{ }of\text{ }6 \right\}\]
\[B=\left\{ x\in N:x\text{ }is\text{ }a\text{ }factor\text{ }of\text{ 8} \right\}\]
Now, factors of \[6=1,6,2,3\] and factors of \[8=1,8,2,4\].
So the set in roster form is as follows:
\[A=\left\{ 1,2,3,6 \right\}\]
\[B=\left\{ 1,2,4,8 \right\}\]
Now as we know that (A-B) means a set that contains elements of A which are not the elements of B.
\[\Rightarrow A-B=\left\{ 1,2,3,6 \right\}-\left\{ 1,2,4,8 \right\}=\left\{ 3,6 \right\}\]
Hence \[\left( A-B \right)=\left\{ 3,6 \right\}\].
Similarly, B-A means that we have to form a set such that it contains elements of B which are not the elements of A.
\[\Rightarrow B-A=\left\{ 1,2,4,8 \right\}-\left\{ 1,2,3,6 \right\}=\left\{ 4,8 \right\}\]

Note: While finding the factors of numbers 6 and 8, we must be careful not to miss out any factor and also not to include any number which is not a factor. We must remember that a set is a well-defined collection of distinct objects. Also, be careful while finding the difference between the sets as (A-B) is a set of those elements of set ‘A’ which are not the elements of set ‘B’. So here we will get it as \[\left\{ 3,6 \right\}\] and not \[\left\{ 4,8 \right\}\]. This is the most common mistake that we can make.