Question

How to prove this? 6) For sets A,B,C prove $ A \cup \left( {B \cap C} \right) = \left( {A \cup B} \right) \cap \left( {A \cup C} \right) $ by showing left side $ \subseteq $ Right side and Right Side $ \subseteq $ Left Side.

Answer 1

Hint: In order to prove the relation given in the question , first let the $ x $ belongs to the left-hand side of the equation and by explaining the meaning of all the operations and expanding, show left side $ \subseteq $ Right side and similarly by considering $ y $ belongs to the right-hand, show Right Side $ \subseteq $ Left Side.

Complete step by step solution:
To Proof: $ A \cup \left( {B \cap C} \right) = \left( {A \cup B} \right) \cap \left( {A \cup C} \right) $
Proof:
In order to prove the given relation, we first take the Left-hand side part and solve , to make it equivalent to the right-hand side and similarly we will take the Right-hand side part and solve, to make it equivalent to the left-hand side .
Let\[ x \in A \cup \left( {B \cap C} \right)\]
 \[
   \Rightarrow x \in A \vee x \in \left( {B \cap C} \right) \\
   \Rightarrow x \in A \vee \left( {x \in B \wedge x \in C} \right) \\
   \Rightarrow \left( {x \in A \vee x \in B} \right) \wedge \left( {x \in A \vee x \in C} \right) \\
   \Rightarrow x \in \left( {A \cup B} \right) \wedge x \in \left( {A \cup C} \right) \\
   \Rightarrow x \in \left( {A \cup B} \right) \cap \left( {A \cup C} \right) \;
 \]
As we have assumed \[ x \in A \cup \left( {B \cap C} \right)\] and in the above we obtained \[ x \in \left( {A \cup B} \right) \cap \left( {A \cup C} \right)\] .Hence we can state that
 $ A \cup \left( {B \cap C} \right) \subseteq \left( {A \cup B} \right) \cap \left( {A \cup C} \right) $ --------(1)
Now taking the Right-hand side
Let
\[ y \in \left( {A \cup B} \right) \cap \left( {A \cup C} \right)\]
\[
   \Rightarrow y \in \left( {A \cup B} \right) \wedge y \in \left( {A \cup C} \right) \\
   \Rightarrow \left( {y \in A \vee y \in B} \right) \wedge \left( {y \in A \vee y \in C} \right) \\
   \Rightarrow y \in A \vee \left( {y \in B \wedge y \in C} \right) \\
   \Rightarrow y \in A \vee y \in \left( {B \cap C} \right) \\
   \Rightarrow y \in A \cup \left( {B \cap C} \right) \;
 \]
As we have assumed \[ y \in \left( {A \cup B} \right) \cap \left( {A \cup C} \right)\] and in the above we obtained \[ y \in A \cup \left( {B \cap C} \right)\] .Hence we can state that
  $ \left( {A \cup B} \right) \cap \left( {A \cup C} \right) \subseteq A \cup \left( {B \cap C} \right) $ ----(2)
From both the equation (1) and (2) , we get by using the rule of equal set as
 $ A \cup \left( {B \cap C} \right) = \left( {A \cup B} \right) \cap \left( {A \cup C} \right) $
Hence, proved.
To get more clarity on the prove of $ A \cup \left( {B \cap C} \right) = \left( {A \cup B} \right) \cap \left( {A \cup C} \right) $ , let prove this formula with the help of an example.
So let $ A,B,C $ be sets as $ A = \left\{ {2,4,6,8,10} \right\},B = \left\{ {1,3,4,5,8} \right\}\,and\,C = \left\{ {5,8,10,12,15} \right\} $
Now to prove the statement , we are taking the LHS part of the equation
 $ LHS = A \cup \left( {B \cap C} \right) $
Intersection of sets B and C will be $ B \cap C = \left\{ {5,8} \right\} $ , putting this in above we have
 $
  LHS = \left\{ {2,4,6,8,10} \right\} \cup \left\{ {5,8} \right\} \\
  LHS = \left\{ {2,4,5,6,8,10} \right\} \;
  $
Therefore the result of LHS is $ \left\{ {2,4,5,6,8,10} \right\} $
Now let's take the RHS of $ A \cup \left( {B \cap C} \right) = \left( {A \cup B} \right) \cap \left( {A \cup C} \right) $ , we have
 $ RHS = \left( {A \cup B} \right) \cap \left( {A \cup C} \right) $
As we can see union of A and B will be $ A \cup B = \left\{ {2,4,6,8,10} \right\} \cup \left\{ {1,3,4,5,8} \right\}\, = \left\{ {1,2,3,4,5,6,8,10} \right\} $ and similarly the union of A with C will result in $ A \cup C = \left\{ {2,4,6,8,10} \right\} \cup \left\{ {5,8,10,12,15} \right\} = \left\{ {2,4,5,6,8,10,12,15} \right\} $
Now putting both the result in the RHS expression, we get
 $ RHS = \left\{ {1,2,3,4,5,6,8,10} \right\} \cap \left\{ {2,4,5,6,8,10,12,15} \right\} $
Now doing the intersection will lead to the result of RHS as
 $
  RHS = \left\{ {1,2,3,4,5,6,8,10} \right\} \cap \left\{ {2,4,5,6,8,10,12,15} \right\} \\
  RHS = \left\{ {2,4,5,6,8,10} \right\} \;
  $
As you can see Both LHS and RHS are equal
 $ \therefore LHS = RHS $
Hence, the relation $ A \cup \left( {B \cap C} \right) = \left( {A \cup B} \right) \cap \left( {A \cup C} \right) $ is verified.
So, the correct answer is “Option C”.

Note: 1. Rule of equal set is when $ A \subseteq B\,and\,B \subseteq A $ then both the sets $ A\,and\,B $ are equal i.e. $ A = B $ .
2.Here one can also solve the question by explaining using the Venn diagram.
3.Union of two sets(A and B) is basically a set which contains all the elements that are in A or in B and on the other hand intersection of two sets(A and B) is a set that contains only the elements that are common in A and B.
3. Verify all the relations used at the end of the result.
4.Students be careful while doing the intersection and union of sets, as there is a possibility that you can miss an element. So, verify twice the resultant set proceeding further.
5. Note that all the elements in a set are always distinct and no repetition is allowed inside the set.