
If \[A=\left\{ a,b,c,d,e \right\}\] and \[B=\left\{ a,c,e,g \right\}\] and \[C=\left\{ b,e,f,g \right\}\]then find:
(i) \[A\cap \left( B-C \right)\]
(ii) \[A-\left( B\cup C \right)\]
(iii) \[A-\left( B\cap C \right)\]
Answer
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Hint: We know that the difference between two sets A and B i.e. (A-B) is a set of those elements of A which are not the elements of set B. Also, the union of any two sets A and i.e. \[\left( A\cup B \right)\] is a set of all elements of A and B and their intersection i.e. \[\left( A\cap B \right)\] means a set of all elements which are common to both A as well as B.
Complete step-by-step answer:
We have been given the sets as follows:
\[A=\left\{ a,b,c,d,e \right\}\], \[B=\left\{ a,c,e,g \right\}\] and \[C=\left\{ b,e,f,g \right\}\]
Now we have been asked to find the value of (i) \[A\cap \left( B-C \right)\].
We know that (B-C) is a set which contains all elements of B which are not present in C.
\[\Rightarrow B-C=\left\{ a,c,e,g \right\}-\left\{ b,e,f,g \right\}=\left\{ a,c \right\}\]
Since the elements of e and g are also present in set C.
Now the intersection of any two sets means a set which contains the common element of the two sets.
\[\Rightarrow A\cap \left( B-C \right)=\left\{ a,b,c,d,e \right\}\cap \left\{ a,c \right\}=\left\{ a,c \right\}\]
Hence \[\Rightarrow A\cap \left( B-C \right)=\left\{ a,c \right\}\]
Now we have been asked to find (ii) \[A-\left( B\cup C \right)\]
We know that the union of any two sets is a set which contains all the elements of the two sets.
\[\Rightarrow \left( B\cup C \right)=\left\{ a,c,e,g \right\}\cup \left\{ b,e,f,g \right\}=\left\{ a,c,e,g,b,f \right\}\]
Also, the difference between two sets A and B i.e. (A-B) is a set that contains all elements of A which are not present in set B.
\[\Rightarrow A-\left( B\cup C \right)=\left\{ a,b,c,d,e \right\}-\left\{ a,c,e,g,b,f \right\}\]
We can see that elements a, b, c and e are present in \[\left( B\cup C \right)\], so cancelling them, we are left with,
\[\Rightarrow A-\left( B\cup C \right)=\left\{ d \right\}\]
Hence, \[A-\left( B\cup C \right)=\left\{ d \right\}\]
Now we have been asked to find (iii) \[A-\left( B\cap C \right)\]
We have, \[\left( B\cap C \right)=\left\{ a,c,e,g \right\}\cap \left\{ b,e,f,g \right\}=\left\{ e,g \right\}\]
Since we know the intersection of two sets is a set which contains common elements of the given two sets.
\[\Rightarrow A-\left( B\cap C \right)=\left\{ a,b,c,d,e \right\}-\left\{ e,g \right\}=\left\{ a,b,c,d \right\}\]
Since the element ‘e’ is present in \[\left( B\cap C \right)\]. Thus we exclude it from A.
Hence \[\Rightarrow A-\left( B\cap C \right)=\left\{ a,b,c,d \right\}\]
Therefore,
(i) \[A\cap \left( B-C \right)=\left\{ a,c \right\}\]
(ii) \[A-\left( B\cup C \right)=\left\{ d \right\}\]
(iii) \[A-\left( B\cap C \right)=\left\{ a,b,c,d \right\}\]
Note: Be careful while finding the value of difference, union and intersection of the two sets and take care that we won’t miss any elements of the set during calculating these things. Also remember that a set is a well-defined collection of distinct objects so take care that same element won’t repeat in a set.
Complete step-by-step answer:
We have been given the sets as follows:
\[A=\left\{ a,b,c,d,e \right\}\], \[B=\left\{ a,c,e,g \right\}\] and \[C=\left\{ b,e,f,g \right\}\]
Now we have been asked to find the value of (i) \[A\cap \left( B-C \right)\].
We know that (B-C) is a set which contains all elements of B which are not present in C.
\[\Rightarrow B-C=\left\{ a,c,e,g \right\}-\left\{ b,e,f,g \right\}=\left\{ a,c \right\}\]
Since the elements of e and g are also present in set C.
Now the intersection of any two sets means a set which contains the common element of the two sets.
\[\Rightarrow A\cap \left( B-C \right)=\left\{ a,b,c,d,e \right\}\cap \left\{ a,c \right\}=\left\{ a,c \right\}\]
Hence \[\Rightarrow A\cap \left( B-C \right)=\left\{ a,c \right\}\]
Now we have been asked to find (ii) \[A-\left( B\cup C \right)\]
We know that the union of any two sets is a set which contains all the elements of the two sets.
\[\Rightarrow \left( B\cup C \right)=\left\{ a,c,e,g \right\}\cup \left\{ b,e,f,g \right\}=\left\{ a,c,e,g,b,f \right\}\]
Also, the difference between two sets A and B i.e. (A-B) is a set that contains all elements of A which are not present in set B.
\[\Rightarrow A-\left( B\cup C \right)=\left\{ a,b,c,d,e \right\}-\left\{ a,c,e,g,b,f \right\}\]
We can see that elements a, b, c and e are present in \[\left( B\cup C \right)\], so cancelling them, we are left with,
\[\Rightarrow A-\left( B\cup C \right)=\left\{ d \right\}\]
Hence, \[A-\left( B\cup C \right)=\left\{ d \right\}\]
Now we have been asked to find (iii) \[A-\left( B\cap C \right)\]
We have, \[\left( B\cap C \right)=\left\{ a,c,e,g \right\}\cap \left\{ b,e,f,g \right\}=\left\{ e,g \right\}\]
Since we know the intersection of two sets is a set which contains common elements of the given two sets.
\[\Rightarrow A-\left( B\cap C \right)=\left\{ a,b,c,d,e \right\}-\left\{ e,g \right\}=\left\{ a,b,c,d \right\}\]
Since the element ‘e’ is present in \[\left( B\cap C \right)\]. Thus we exclude it from A.
Hence \[\Rightarrow A-\left( B\cap C \right)=\left\{ a,b,c,d \right\}\]
Therefore,
(i) \[A\cap \left( B-C \right)=\left\{ a,c \right\}\]
(ii) \[A-\left( B\cup C \right)=\left\{ d \right\}\]
(iii) \[A-\left( B\cap C \right)=\left\{ a,b,c,d \right\}\]
Note: Be careful while finding the value of difference, union and intersection of the two sets and take care that we won’t miss any elements of the set during calculating these things. Also remember that a set is a well-defined collection of distinct objects so take care that same element won’t repeat in a set.
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