Question

If A = { a, b, c, d, e }, B = { a, c, e, g } and C = { b, e, f, g }, verify that:
(a) $A\cap \left( B-C \right)=\left( A\cap B \right)-\left( A\cap C \right)$
(b) $A-\left( B\cap C \right)=\left( A-B \right)\cup \left( A-C \right)$

Answer 1

Hint: First we will write the required definition of the terms like union and intersection of sets, and then we will substitute the value of A, B and C in the following options and check whether the equation is true or not.

Complete step-by-step answer:
Union: The union (denoted by $\cup $ ) of a collection of sets is the set of all elements in the collection. It is one of the fundamental operations through which sets can be combined and related to each other.
Intersection: The intersection of two sets has only the elements common to both sets. If an element is in just one set it is not part of the intersection. The symbol is an upside down $\cap $ .
When we subtract two sets the common elements of the set which is being subtract is removed and the remaining elements is the final answer.
For (a):
Let’s first solve LHS,
Let’s first find the value of B – C,
B = { a, c, e, g } and C = { b, e, f, g }, now we have to take the elements of B that are not present in C.
Therefore, we get
B – C = { a, c }
Let’s find the value of $A\cap \left( B-C \right)$,
A = { a, b, c, d, e } and B – C = { a, c }, now the common elements between these two is a and c.
Therefore, we get
$A\cap \left( B-C \right)$ = { a, c }
Now for RHS,
Let’s find $A\cap B$,
A = { a, b, c, d, e }, B = { a, c, e, g }, the common elements in these two are a, c, e.
Therefore, we get
$A\cap B$ = { a, c, e }
Let’s find $A\cap C$,
A = { a, b, c, d, e }, C = { b, e, f, g }, the common elements in these two are b and e.
Therefore, we get
$A\cap C$ = { b, e }
Now we will find $\left( A\cap B \right)-\left( A\cap C \right)$,
We have $A\cap B$ = { a, c, e } and $A\cap C$ = { b, e }, now we have to take the elements of $A\cap B$ that are not present in $A\cap C$.
Therefore, we get
$\left( A\cap B \right)-\left( A\cap C \right)$ = { a, c }
Hence for (a) we have shown that LHS = RHS.
Therefore, $A\cap \left( B-C \right)=\left( A\cap B \right)-\left( A\cap C \right)$.
For (b):
Let’s first solve LHS,
Let’s find $B\cap C$,
B = { a, c, e, g } and C = { b, e, f, g }, now the common elements between B and C is e and g.
Therefore, we get
$B\cap C$ = { e, g }
Now we will find $A-\left( B\cap C \right)$,
Now we have $B\cap C$ = { e, g } and A = { a, b, c, d, e } , now we have to take the elements of A that are not present in $B\cap C$.
Therefore, we get
$A-\left( B\cap C \right)$ = { a, b, c, d }
Now for RHS,
Let’s find A – B,
A = { a, b, c, d, e }, B = { a, c, e, g } , now we have to take the elements of A that are not present in B.
Therefore, we get
A – B = { b, d }
Let’s find A – C,
A = { a, b, c, d, e }, C = { b, e, f, g } , now we have to take the elements of A that are not present in C.
Therefore, we get
A – C = { a, c, d }
Now we will find $\left( A-B \right)\cup \left( A-C \right)$,
We have A – B = { b, d } and A – C = { a, c, d }, now we have to take all the elements that are present in both of these sets.
Therefore, we get
$\left( A-B \right)\cup \left( A-C \right)$ = { a, b, c, d }
Hence for (b) we have shown that LHS = RHS.
Therefore, $A-\left( B\cap C \right)=\left( A-B \right)\cup \left( A-C \right)$

Note: The definitions of the terms that we have used should be kept in mind while solving the question and one more important point is that the subtraction of two sets means removing the common elements.