Prove the following: A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C )

Given three sets A , B and C , prove that A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ) . Consider x ∈ A ∩ ( B ∪ C ) ⇔ x ∈ A and x ∈ ( B ∪ C ) ⇔ x ∈ A and ( x ∈ B or x ∈ C ) ⇔ ( x ∈ A and x ∈ B ) or ( x ∈ A and x ∈ C ) ⇔ x ∈ A ∩ B or x ∈ A ∩ C ⇔ x ∈ ( A ∩ B ) ∪ ( A ∩ C ) So, we have x ∈ A ∩ ( B ∪ C ) ⇒ x ∈ ( A ∩ B ) ∪ ( A ∩ C ) ⇒ A ∩ ( B ∪ C ) ⊂ ( A ∩ B ) ∪ ( A ∩ C ) . . . ( 1 ) Also, x ∈ ( A ∩ B ) ∪ ( A ∩ C ) ⇒ x ∈ A ∩ ( B ∪ C ) ⇒ ( A ∩ B ) ∪ ( A ∩ C ) ⊂ A ∩ ( B ∪ C ) . . . ( 2 ) Thus, A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ) (from ( 1 ) and ( 2 ) )

Prove the following: A ⊂ B ⇔ B ' ⊂ A ' ⇔ A ' ∪ B = U ⇔ B ' ∩ A = ϕ , where U is the universal set

Given two sets A and B , we have to prove that, A ⊂ B ⇔ B ' ⊂ A ' ⇔ A ' ∪ B = U ⇔ B ' ∩ A = ϕ , where U is the universal set. It is enough to prove A ⊂ B ⇔ B ' ⊂ A ' , B ' ⊂ A ' ⇔ A ' ∪ B = U and A ' ∪ B = U ⇔ B ' ∩ A = ϕ (Given the statements P , Q and R , if P ⇔ Q , Q ⇔ R then P ⇔ R ) (i) To prove: A ⊂ B ⇔ B ' ⊂ A ' Let A ⊂ B . Now consider, x ∈ B ' ⇒ x ∉ B ⇒ x ∉ A ( ∵ A ⊂ B , all the elements of A are present in B ) ⇒ x ∈ A ' ⇒ B ' ⊂ A ' (ii) To prove: B ' ⊂ A ' ⇔ A ' ∪ B = U Let B ' ⊂ A ' . Clearly A ' ∪ B ⊂ U . . . ( 1 ) . Now consider x ∈ U ⇒ x ∈ B or x ∈ B ' ⇒ x ∈ B or x ∈ A ' ⇒ x ∈ A ' ∪ B ⇒ U ⊂ A ' ∪ B ( 2 ) ⇒ A ' ∪ B = U (from ( 1 ) and ( 2 ) ) (iii) To prove: A ' ∪ B = U ⇔ B ' ∩ A = ϕ Let A ' ∪ B = U . ⇒ A ' ∪ ( B ' ) ' = U ⇒ A ∩ B ' ' = U (By De Morgan's law ( P ' ∪ Q ' ) = ( P ∩ Q ) ' ) ⇒ A ∩ B ' = ϕ (Complement of set A is U if and only if A = ϕ )

Prove the following: A ∪ B = U and A ∩ B = ϕ ⇒ B = A ' , where U is the universal set

Given two sets A and B such that A ∪ B = U and A ∩ B = ϕ , where U is the universal set A ∩ B = ϕ ⇒ There is no element of set B which is in set A ⇒ All the elements of set B are present in the set of A ' ⇒ B ⊂ A ' . . . ( 1 ) Now, A ∪ B = U ⇒ A ' ∩ ( A ∪ B ) = A ' ∩ U ⇒ ( A ' ∩ A ) ∪ ( A ' ∩ B ) = A ' ⇒ ϕ ∪ ( A ' ∩ B ) = A ' ⇒ ( A ' ∩ B ) = A ' ⇒ B = A ' ( ∵ B ⊂ A ' )

Prove the following: A ∪ B = A for all A ⇒ B = ϕ

Let A and B be the two sets such that A ∪ B = A for all A . A ∪ B is the set of the elements which are present in A or B Since A ∪ B = A for all A , it also should be true for A = ϕ . Then, we have A ∪ B = ϕ ⇒ ϕ ∪ B = ϕ ⇒ B = ϕ (Law of identity)

Prove that A - ∪ i = i n B i = ∩ i = 1 n ( A - B i ) and A - ∩ i = 1 n B i = ∪ i = 1 n ( A - B i ) .

To prove: A - ∪ i = i n B i = ∩ i = 1 n ( A - B i ) L.H.S = A - ∪ i = 1 n B i = A ∩ [ ∪ i = 1 n B i ] ' = A ∩ [ B 1 ∪ B 2 ∪ . . . ∪ B 2 ] ' = A ∩ [ B 1 ' ∩ B 2 ' ∩ . . . ∩ B 2 ' ] = [ ( A ∩ B 1 ' ) ∩ ( A ∩ B 2 ' ) ∩ . . . ( A ∩ B n ' ) ] ( ∵ P ∩ ( Q ∩ R ) = P ∩ Q ∩ R and P ∩ P ∩ Q = P ∩ Q ) = ∩ i = 1 n ( A ∩ B i ' ) = R.H.S Now, we have to prove that A - ∩ i = 1 n B i = ∪ i = 1 n ( A - B i ) Consider L.H.S = A - ∩ i = 1 n B i = A - ( B 1 ∩ B 2 ∩ . . . B n ) = A ∩ ( B 1 ∩ B 2 ∩ . . . B n ) ' = A ∩ ( B 1 ' ∪ B 2 ' ∪ . . . B n ' ) = ( A ∩ B 1 ' ) ∪ ( A ∩ B 2 ' ) ∪ . . . ( A ∩ B n ' ) ( ∵ P ∩ ( Q ∪ R ) = ( P ∩ Q ) ∪ ( P ∩ R ) and P ∩ Q = P ∩ P ∩ Q ) = ∪ i = 1 n ( A - B i ) = R.H.S

Prove that | A ∪ B ∪ C | = | A | + | B | + | C | + | A ∩ B ∩ C | - | A ∩ B | - | B ∩ C | - | C ∩ A ∣

Consider | A ∪ B ∪ C | = ( A ∪ B ) ∪ C = A ∪ B + C - ( A ∪ B ) ∩ C ( ∵ P ∪ Q = P + Q - P ∩ Q ) = A + B - A ∩ B + C - ( A ∩ C ) ∪ ( B ∩ C ) ( ∵ Intersection is distributive over union) = A + B - A ∩ B + C - ( A ∩ C + B ∩ C - ( A ∩ C ) ∩ ( B ∩ C ) ) = A + B - A ∩ B + C - ( A ∩ C + B ∩ C - A ∩ C ∩ B ∩ C ) = A + B + C - A ∩ C - B ∩ C - A ∩ B + A ∩ B ∩ C

E M B I B E

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 2 - Sets>Exercises- 2(b)>Q 5

EASY

11th Odisha Board

IMPORTANT

Earn 100

Prove the following:

$(A \ B) \ C = (A \ C) \ B = A \ (B \cup C)$

Important Questions on Sets

EASY

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 2 - Sets>Exercises- 2(b)>Q 5

Prove the following:

$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

MEDIUM

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 2 - Sets>Exercises- 2(b)>Q 5

Prove the following:

$A Δ (B Δ C) = (A Δ B) Δ C$

EASY

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 2 - Sets>Exercises- 2(b)>Q 5

Prove the following:

$A \subset B \Leftrightarrow B^{'} \subset A^{'} \Leftrightarrow A^{'} \cup B = U \Leftrightarrow B^{'} \cap A = ϕ$ , where $U$ is the universal set

EASY

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 2 - Sets>Exercises- 2(b)>Q 5

Prove the following:

$A \cup B = U$ and $A \cap B = ϕ \Rightarrow B = A^{'}$ , where $U$ is the universal set

EASY

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 2 - Sets>Exercises- 2(b)>Q 5

Prove the following:

$A \cup B = A$ for all $A \Rightarrow B = ϕ$

EASY

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 2 - Sets>Exercises- 2(b)>Q 7

Prove that $A - \cup_{i = i}^{n} B_{i} = \cap_{i = 1}^{n} (A - B_{i})$ and $A - \cap_{i = 1}^{n} B_{i} = \cup_{i = 1}^{n} (A - B_{i})$ .

EASY

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 2 - Sets>Exercises- 2(b)>Q 8

Prove that

$| A \cup B \cup C | = | A | + | B | + | C | + | A \cap B \cap C | - | A \cap B | - | B \cap C | - | C \cap A ∣$

EASY

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 2 - Sets>Exercises- 2(b)>Q 9

If $X$ and $Y$ are two sets such that $X \cup Y$ has $20$ objects, $X$ has $10$ objects and $Y$ has $15$ objects; how many objects does $X \cap Y$ have ?