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Hint: First we will draw the Venn diagram and then we will use that diagram to prove the above relation. We will also give the definition of complement, intersection and union of sets.
Complete step by step solution:
Let’s first look at the Venn diagram of RHS.
Complete step by step solution:
Let’s first look at the Venn diagram of RHS.
This is the Venn diagram and the shaded portion is $A\cap B$ .
Complement A’ will be all the regions in the Venn diagram except A.
Complement A’ will be all the regions in the Venn diagram except A.
The shaded part is A’.
Now for LHS let’s first look at $\left( A'\cup B \right)$
For $A'\cup B$ we have to take all the regions that are present in A’ or B, we get all the region except A, but in that we have to add the shaded region $A\cap B$ also.
Hence, its venn diagram is:
Now for LHS let’s first look at $\left( A'\cup B \right)$
For $A'\cup B$ we have to take all the regions that are present in A’ or B, we get all the region except A, but in that we have to add the shaded region $A\cap B$ also.
Hence, its venn diagram is:
Now $A\cap \left( A'\cup B \right)$ will be the common part between A and $A'\cup B$, which is $A\cap B$.
Now the diagram for $A\cap \left( A'\cup B \right)$ will be,
Now the diagram for $A\cap \left( A'\cup B \right)$ will be,
Hence, we have proved that $A\cap \left( A'\cup B \right)=A\cap B$.
Complement of a set: Complement of a set A, denoted by A c, is the set of all elements that belongs to the universal set but does not belong to set A.
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 $ .
By the above definition of complement A’ will be all the regions in the Venn diagram except A.
And for $A'\cup B$ we have to take all the regions that are present in A’ or B, we get all the regions except A, but in that we have to add the shaded region $A\cap B$ also.
Now $A\cap \left( A'\cup B \right)$ will be the common part between A and $A'\cup B$, which is $A\cap B$.
Hence, we have proved that $A\cap \left( A'\cup B \right)=A\cap B$.
Let’s take some values of A, U and B to check $A\cap \left( A'\cup B \right)=A\cap B$.
A = { 1, 2, 3 }
B = { 3, 4 }
U = { 1, 2, 3, 4, 5 }
Now the common element between A and B is 3.
Therefore,
$A\cap B$ = { 3 }
Now A = { 1, 2, 3 } and U = { 1, 2, 3, 4, 5 }, A’ will be elements in U that are not in A.
Therefore,
A’ = { 4, 5 }
Now A’ = { 4, 5 } and B = { 3, 4 }, taking all the elements in A’ and B we get,
$A'\cup B$ = { 3, 4, 5 }
Now A = { 1, 2, 3 } and $A'\cup B$ = { 3, 4, 5 } the common element is 3.
Therefore,
$A\cap \left( A'\cup B \right)$ = { 3 }
Hence, from this we can see that $A\cap \left( A'\cup B \right)=A\cap B$ is true.
Note: Here we have used the Venn diagram and the definitions of the given term to prove. One can also check by putting some different values of A, B and U so that they can understand it without having some issues.
Complement of a set: Complement of a set A, denoted by A c, is the set of all elements that belongs to the universal set but does not belong to set A.
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 $ .
By the above definition of complement A’ will be all the regions in the Venn diagram except A.
And for $A'\cup B$ we have to take all the regions that are present in A’ or B, we get all the regions except A, but in that we have to add the shaded region $A\cap B$ also.
Now $A\cap \left( A'\cup B \right)$ will be the common part between A and $A'\cup B$, which is $A\cap B$.
Hence, we have proved that $A\cap \left( A'\cup B \right)=A\cap B$.
Let’s take some values of A, U and B to check $A\cap \left( A'\cup B \right)=A\cap B$.
A = { 1, 2, 3 }
B = { 3, 4 }
U = { 1, 2, 3, 4, 5 }
Now the common element between A and B is 3.
Therefore,
$A\cap B$ = { 3 }
Now A = { 1, 2, 3 } and U = { 1, 2, 3, 4, 5 }, A’ will be elements in U that are not in A.
Therefore,
A’ = { 4, 5 }
Now A’ = { 4, 5 } and B = { 3, 4 }, taking all the elements in A’ and B we get,
$A'\cup B$ = { 3, 4, 5 }
Now A = { 1, 2, 3 } and $A'\cup B$ = { 3, 4, 5 } the common element is 3.
Therefore,
$A\cap \left( A'\cup B \right)$ = { 3 }
Hence, from this we can see that $A\cap \left( A'\cup B \right)=A\cap B$ is true.
Note: Here we have used the Venn diagram and the definitions of the given term to prove. One can also check by putting some different values of A, B and U so that they can understand it without having some issues.
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