Question
Question: From Venn diagram show that \(n\left( A\cup B \right)+n\left( A\cap B \right)=n\left( A \right)+n\le...
From Venn diagram show that n(A∪B)+n(A∩B)=n(A)+n(B).
Solution
In this question we have been given with a Venn diagram and using the Venn diagram we have to show that n(A∪B)+n(A∩B)=n(A)+n(B). We will first write all the sets which include the intersection and union set of the given sets A and B. We will then consider the left-hand side of the expression and find its value and then consider the right-hand side and find its value and check whether the condition holds true.
Complete step by step answer:
We can see from the Venn diagram that:
\Rightarrow A=\left\\{ a,b,c,d \right\\}
\Rightarrow B=\left\\{ a,b,e,f,g \right\\}
Therefore, we get:
⇒n(A)=4
⇒n(B)=5
The union set which is represented as A∪B represents the set of all the distinct elements that are present in the sets A or B. We have the intersection as:
\Rightarrow A\cup B=\left\\{ a,b,c,d,e,f,g \right\\}
Therefore, we have:
⇒n(A∪B)=7
The intersection set which is represented as A∩B represents the set of all the distinct elements that are present in the both sets A and B. We have the intersection as:
\Rightarrow A\cap B=\left\\{ a,b \right\\}
Therefore, we have:
⇒n(A∩B)=2
Now we have to prove that n(A∪B)+n(A∩B)=n(A)+n(B).
Consider the left-hand side of the expression, we have:
⇒n(A∪B)+n(A∩B)
On substituting the values, we get:
⇒7+2
On simplifying, we get:
⇒9, which is the value of the left-hand side.
Consider the right-hand side of the expression, we have:
⇒n(A)+n(B)
On substituting the values, we get:
⇒4+5
On simplifying, we get:
⇒9, which is the value of the right-hand side.
Since the values of both the sides are the same the expression is correct, hence proved.
Note: It is to be remembered that the n(P) represents the number of elements present in the set. A Venn diagram is a good method to simplify set problems such that they can be visualized. The various properties of Venn diagrams should be remembered to solve these types of questions.