Question

If A = {1,2,3,4,5}, B = {2,4,6,8} and C= {3,4,5,6}, then verify: $A - \left( {B \cup C} \right) = \left( {A - B} \right) \cap \left( {A - C} \right)$.
A. True
B. False

Answer 1

Hint: Given that, A = {1,2,3,4,5}, B = {2,4,6,8} and C= {3,4,5,6}. We have to prove that $A - \left( {B \cup C} \right) = \left( {A - B} \right) \cap \left( {A - C} \right)$. Note that, the union of two sets will contain all the elements of both sets and the intersection of two sets will contain the elements common in both sets. Therefore simplify both LHS and RHS separately to check if they are the same.

Complete step-by-step solution:
A, B and C are three sets. A = {1,2,3,4,5}, B = {2,4,6,8} and C= {3,4,5,6}, given.
We have to prove that $A - \left( {B \cup C} \right) = \left( {A - B} \right) \cap \left( {A - C} \right)$
The left hand side is given by
$A - \left( {B \cup C} \right)$
$$ = \left\{ {1,2,3,4,5} \right\} - \left\{ {\left\{ {2,4,6,8} \right\} \cup \left\{ {3,4,5,6} \right\}} \right\}$$
$$ = \left\{ {1,2,3,4,5} \right\} - \left\{ {2,3,4,5,6,8} \right\}$$ …… since the union of two sets will contain the elements of both sets
$$ = \left\{ 1 \right\}$$ ….since it is the only element present in A but absent in $\left( {B \cup C} \right)$
Now, the RHS is:
$\left( {A - B} \right) \cap \left( {A - C} \right)$
$ = \left\{ {\left\{ {1,2,3,4,5} \right\} - \left\{ {2,4,6,8} \right\}} \right\} \cap \left\{ {\left\{ {1,2,3,4,5} \right\} - \left\{ {3,4,5,6} \right\}} \right\}$
$ = \left\{ {1,3,5} \right\} \cap \left\{ {1,2} \right\}$
$ = \left\{ 1 \right\}$ …..since, the intersection of two sets will contain the elements common in both sets.
Therefore, LHS = RHS
Hence, the expression $A - \left( {B \cup C} \right) = \left( {A - B} \right) \cap \left( {A - C} \right)$ is true.

So the correct answer is option (A).

Note: The union of two sets will contain all the elements of both sets.
The intersection of two sets will contain only the elements common in both sets.
Subtraction of two sets denotes the elements that are present in the former set but not in the latter.