Using properties of sets, show that
(i) \[A \cup (A \cap B) = A\]
(ii) \[A \cap (A \cup B) = A\]
Answer
639k+ views
Hint: According to the question, you can apply the following properties of set which are If \[X \subset Y\] and \[X \supset Y\] then \[X = Y\] and Distributive property i.e. \[A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\].
By using the properties, you can verify the above question.
Complete step-by-step answer:
(i) \[A \cup (A \cap B) = A\]
As we know that \[A \subset A\]
\[ \Rightarrow A \cap B \subset A\]
Therefore, \[A \cup (A \cap B)\] which is also a subset of \[A\] i.e.
\[A \cup (A \cap B) \subset A\] equation1
It is also obvious that \[A \subset A \cup (A \cap B)\] equation2
\[\therefore \]In equation 1 and 2 , we can apply property 1 as mentioned in the hint which is If \[X \subset Y\] and \[X \supset Y\] then \[X = Y\] to get the desired result.
Hence, \[A \cup (A \cap B) = A\] , which is the required result
(ii) \[A \cap (A \cup B) = A\]
Here, we can use the distributive property of set that is \[A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\] , Thus, we can write \[A \cap (A \cup B) = (A \cap A) \cup (A \cap B)\]
On Solving the Right hand side of above equation, we have, \[A \cup (A \cap B)\]
(As we know, \[(A \cap A) = A\] )
From the result in as shown in part (i) that is \[A \cup (A \cap B) = A\]
Therefore, \[A \cap (A \cup B) = A\] , which is the required result.
Hence both the parts have been proven.
Additional information: There are different types of properties of sets that are commutative property, associative property, distributive property and de-Morgan’s property. Based on these properties we have different types of questions and in some questions we have to prove these properties by making Venn’s diagram. Here, some of the operations are used that are intersection and union which are similar like addition and multiplication.
Note: To solve these types of questions, you should know all the properties of sets. You should also know how to use these properties according to the requirement of the question as above. Also, the properties make the question easier to solve.
By using the properties, you can verify the above question.
Complete step-by-step answer:
(i) \[A \cup (A \cap B) = A\]
As we know that \[A \subset A\]
\[ \Rightarrow A \cap B \subset A\]
Therefore, \[A \cup (A \cap B)\] which is also a subset of \[A\] i.e.
\[A \cup (A \cap B) \subset A\] equation1
It is also obvious that \[A \subset A \cup (A \cap B)\] equation2
\[\therefore \]In equation 1 and 2 , we can apply property 1 as mentioned in the hint which is If \[X \subset Y\] and \[X \supset Y\] then \[X = Y\] to get the desired result.
Hence, \[A \cup (A \cap B) = A\] , which is the required result
(ii) \[A \cap (A \cup B) = A\]
Here, we can use the distributive property of set that is \[A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\] , Thus, we can write \[A \cap (A \cup B) = (A \cap A) \cup (A \cap B)\]
On Solving the Right hand side of above equation, we have, \[A \cup (A \cap B)\]
(As we know, \[(A \cap A) = A\] )
From the result in as shown in part (i) that is \[A \cup (A \cap B) = A\]
Therefore, \[A \cap (A \cup B) = A\] , which is the required result.
Hence both the parts have been proven.
Additional information: There are different types of properties of sets that are commutative property, associative property, distributive property and de-Morgan’s property. Based on these properties we have different types of questions and in some questions we have to prove these properties by making Venn’s diagram. Here, some of the operations are used that are intersection and union which are similar like addition and multiplication.
Note: To solve these types of questions, you should know all the properties of sets. You should also know how to use these properties according to the requirement of the question as above. Also, the properties make the question easier to solve.
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Using properties of sets, show that
(i) \[A \cup (A \cap B) = A\]
(ii) \[A \cap (A \cup B) = A\]
(i) \[A \cup (A \cap B) = A\]
(ii) \[A \cap (A \cup B) = A\]

Class 11 MATHS Miscellaneous (Question - 7) | Sets Class 11 Chapter 1| NCERT | Ratan Kalra Sir
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