Question

If \[A=\left\{ \dfrac{1}{x}:x\in N,x<8 \right\}\] and \[B=\left\{ \dfrac{1}{2x}:x\in N,x\le 4 \right\}\], 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:
\[A=\left\{ \dfrac{1}{x}:x\in N,x<8 \right\}\]
\[B=\left\{ \dfrac{1}{2x}:x\in N,x\le 4 \right\}\]
For \[x<8\] the values of \[\dfrac{1}{x}\]are \[\dfrac{1}{1},\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5},\dfrac{1}{6},\dfrac{1}{7}\].
So \[A=\left\{ \dfrac{1}{1},\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5},\dfrac{1}{6},\dfrac{1}{7} \right\}\] in roster form.
For \[x\le 4\] the values of \[\dfrac{1}{2x}\] are \[\dfrac{1}{2\times 1},\dfrac{1}{2\times 2},\dfrac{1}{2\times 3},\dfrac{1}{2\times 4}\] which can be simplified as,
\[B=\left\{ \dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{6},\dfrac{1}{8} \right\}\] in roster form.
Now as we know that (A-B) means a set which contains all elements of A which are not present in set B.
\[\Rightarrow A-B=\left\{ \dfrac{1}{1},\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5},\dfrac{1}{6},\dfrac{1}{7} \right\}-\left\{ \dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{6},\dfrac{1}{8} \right\}\]
Since the elements \[\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{6}\] of A are also present in B. We can subtract those elements from set A and we will get,
\[\Rightarrow A-B=\left\{ \dfrac{1}{1},\dfrac{1}{3},\dfrac{1}{5},\dfrac{1}{7} \right\}\]
Similarly, \[B-A=\left\{ \dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{6},\dfrac{1}{8} \right\}-\left\{ \dfrac{1}{1},\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5},\dfrac{1}{6},\dfrac{1}{7} \right\}\]
Since, \[\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{6}\] elements of B are present in set A.
\[\Rightarrow B-A=\left\{ \dfrac{1}{8} \right\}\]
Therefore, the values of:
(i) A-B \[=\left\{ \dfrac{1}{1},\dfrac{1}{3},\dfrac{1}{5},\dfrac{1}{7} \right\}\]
(ii) B-A \[=\left\{ \dfrac{1}{8} \right\}\]

Note: Be careful while converting the set A from set builder form to roster form as you might take the equality for x < 8. Hence x = 1, 2, 3, 4, 5, 6 and 7 only. If you consider values of x till 8, then you will get a different set. Then you must remember that natural numbers don’t include 0. So while writing the values of x, you must be aware of the same. Also, be careful while finding the difference between the set. We know that a set is a well-defined collection of distinct objects so check it once that all elements must be distinct after the difference.