Prove the following properties for any three sets A∩(B∪C)=(A∩B)∪(A∩C) andA∪(B∩C)=(A∪B)∩(A∪C)
Question
Prove the following properties for any three sets
A \cap (B \cup C) = (A \cap B) \cup (A \cap C)
and
A \cup (B \cap C) = (A \cup B) \cap (A \cup C)
Solution
Sure, let's prove these properties one by one.
- A∩(B∪C)=(A∩B)∪(A∩C)
Proof:
Let x be an arbitrary element of A∩(B∪C). Then x is in A and x is in B∪C. This means that x is in A and x is in B or x is in C. Therefore, x is in A∩B or x is in A∩C. So, x is in (A∩B)∪(A∩C). Hence, A∩(B∪C) is a subset of (A∩B)∪(A∩C).
Now, let x be an arbitrary element of (A∩B)∪(A∩C). Then x is in A∩B or x is in A∩C. This means that x is in A and x is in B or x is in A and x is in C. Therefore, x is in A and x is in B∪C. So, x is in A∩(B∪C). Hence, (A∩B)∪(A∩C) is a subset of A∩(B∪C).
Since A∩(B∪C) is a subset of (A∩B)∪(A∩C) and (A∩B)∪(A∩C) is a subset of A∩(B∪C), we conclude that A∩(B∪C)=(A∩B)∪(A∩C).
- A∪(B∩C)=(A∪B)∩(A∪C)
Proof:
Let x be an arbitrary element of A∪(B∩C). Then x is in A or x is in B∩C. This means that x is in A or x is in B and x is in C. Therefore, x is in A or x is in B and x is in A or x is in C. So, x is in (A∪B) and x is in (A∪C). Hence, A∪(B∩C) is a subset of (A∪B)∩(A∪C).
Now, let x be an arbitrary element of (A∪B)∩(A∪C). Then x is in A∪B and x is in A∪C. This means that x is in A or x is in B and x is in A or x is in C. Therefore, x is in A or x is in B and x is in C. So, x is in A∪(B∩C). Hence, (A∪B)∩(A∪C) is a subset of A∪(B∩C).
Since A∪(B∩C) is a subset of (A∪B)∩(A∪C) and (A∪B)∩(A∪C) is a subset of A∪(B∩C), we conclude that A∪(B∩C)=(A∪B)∩(A∪C).
Similar Questions
Prove the following properties for any three sets A∩(B∪C)=(A∩B)∪(A∩C) andA∪(B∩C)=(A∪B)∩(A∪C)
Let 𝐴 and 𝐵 be sets. Using set builder notation, show that:a) 𝐴 ∩ ∅ = ∅.b) 𝐴 ∪ 𝐴 = 𝑈.c) 𝐴 ∩ (𝐵 − 𝐴) = ∅
Show that the following four conditions are equivalent.(i) A ⊂ B (ii) A – B = Φ(iii) A ∪ B = B (iv) A ∩ B = A
If A is a subset of B and B is a subset of C, then cardinality of A ∪ B ∪ C is equal to
Let A𝐴 and B𝐵 are two events and P(A′)=0⋅3𝑃𝐴'=0·3, P(B)=0⋅4, P(A∩B′)=0⋅5𝑃𝐵=0·4, 𝑃𝐴∩𝐵'=0·5 then P(A∪B′)𝑃𝐴∪𝐵' is:
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.