Answer
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Hint: First we will write the definitions of required terms like intersection and union of sets and then we will use them to verify $A-\left( B\cup C \right)=\left( A-B \right)\cap \left( A-C \right)$ by substituting the values of A, B and C.
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 $ .
Let’s first solve the LHS,
First we will find the term $B\cup C$,
We have the sets B = { 2, 3, 5, 6 }, C = { 4, 5, 6, 7 }
Now for $B\cup C$ we have considered all the elements that are present in both B and C.
Therefore, we get
$B\cup C$ = { 2, 3, 4, 5, 6, 7 }
Now we will find the value of $A-\left( B\cup C \right)$ ,
It indicates that we have to get the elements of set A that are not present in $B\cup C$.
Now A = { 1, 2, 4, 5 } and $B\cup C$ = { 2, 3, 4, 5, 6, 7 }, we have to remove 2, 4, 5 from the set A as it is present in both of them.
Therefore, we get
$A-\left( B\cup C \right)$ = { 1 }
Now we will solve for the RHS,
Let’s compute A – B and A – C first,
A = { 1, 2, 4, 5 }, B = { 2, 3, 5, 6 } the common elements are 2 and 5, so we have to remove them from A.
After removing we get,
A – B = { 1, 4 }
Now A = { 1, 2, 4, 5 }, C = { 4, 5, 6, 7 } the common elements are 4 and 5, so we have to remove them from A.
After removing we get,
A – C = { 1, 2 }
Now we will find the value of $\left( A-B \right)\cap \left( A-C \right)$ ,
Now we have A – B = { 1, 4 } and A – C = { 1, 2 }, so the common element in both of them is 1.
Therefore,
$\left( A-B \right)\cap \left( A-C \right)$ = { 1 }
Hence, we can see that LHS = RHS = { 1 }
Hence, the relation $A-\left( B\cup C \right)=\left( A-B \right)\cap \left( A-C \right)$ has been verified.
Note: The definitions of union and intersection of sets must be kept in mind while solving the question. One point that should be kept in mind is 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.
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 $ .
Let’s first solve the LHS,
First we will find the term $B\cup C$,
We have the sets B = { 2, 3, 5, 6 }, C = { 4, 5, 6, 7 }
Now for $B\cup C$ we have considered all the elements that are present in both B and C.
Therefore, we get
$B\cup C$ = { 2, 3, 4, 5, 6, 7 }
Now we will find the value of $A-\left( B\cup C \right)$ ,
It indicates that we have to get the elements of set A that are not present in $B\cup C$.
Now A = { 1, 2, 4, 5 } and $B\cup C$ = { 2, 3, 4, 5, 6, 7 }, we have to remove 2, 4, 5 from the set A as it is present in both of them.
Therefore, we get
$A-\left( B\cup C \right)$ = { 1 }
Now we will solve for the RHS,
Let’s compute A – B and A – C first,
A = { 1, 2, 4, 5 }, B = { 2, 3, 5, 6 } the common elements are 2 and 5, so we have to remove them from A.
After removing we get,
A – B = { 1, 4 }
Now A = { 1, 2, 4, 5 }, C = { 4, 5, 6, 7 } the common elements are 4 and 5, so we have to remove them from A.
After removing we get,
A – C = { 1, 2 }
Now we will find the value of $\left( A-B \right)\cap \left( A-C \right)$ ,
Now we have A – B = { 1, 4 } and A – C = { 1, 2 }, so the common element in both of them is 1.
Therefore,
$\left( A-B \right)\cap \left( A-C \right)$ = { 1 }
Hence, we can see that LHS = RHS = { 1 }
Hence, the relation $A-\left( B\cup C \right)=\left( A-B \right)\cap \left( A-C \right)$ has been verified.
Note: The definitions of union and intersection of sets must be kept in mind while solving the question. One point that should be kept in mind is 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.