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a complement intersect b complement
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Question:I cannot figure this out to save my life. Please help me!
Answers:(A B)' = A' U B' de Morgan's law
Answers:(A B)' = A' U B' de Morgan's law
Question:
Answers:Let A(Complement)=A' & B(Complement)=B' ^ means intersection U means union A and B are independent P(A^B)=P(A)(B) We have to prove that A' and B' are independent events ie., to prove that P(A'^B')=P(A')(B') Lets start with P(A'^B') P(A'^B')=P[(AUB)']By DeMorgan's law =1P(AUB) =1[P(A)+P(B)P(A^B)] =1P(A)P(B)+P(A^B) =1P(A)P(B)+P(A)P(B) =1[1P(A)]P(B)[1P(A)] =[1P(A)][1P(B)] =P(A')P(B') P(A'^B')=P(A')P(B') This shows that the events A' and B' are independent.
Answers:Let A(Complement)=A' & B(Complement)=B' ^ means intersection U means union A and B are independent P(A^B)=P(A)(B) We have to prove that A' and B' are independent events ie., to prove that P(A'^B')=P(A')(B') Lets start with P(A'^B') P(A'^B')=P[(AUB)']By DeMorgan's law =1P(AUB) =1[P(A)+P(B)P(A^B)] =1P(A)P(B)+P(A^B) =1P(A)P(B)+P(A)P(B) =1[1P(A)]P(B)[1P(A)] =[1P(A)][1P(B)] =P(A')P(B') P(A'^B')=P(A')P(B') This shows that the events A' and B' are independent.
Question:
Answers:When you have an "if and only if" condition, you must show that If A contained in B, then the intersection of A and B compliment is empty and If the intersection of A and B compliment is empty, then A is contained in B. Let x A. Since A is a subset of B, x A B Because every element contained in A is also contained in B so the intersection holds. x A B x A and x B x A and x B complement x A B compliment Since x is not a common element in A and B compliment, we then say that, A B complement = Now we must prove the opposite, which is a bit tricky. So, if we assume that A B complement = is true, then there exists an element x that belongs to one but not the other. So A B complement = implies that, x A B complement x A and x B complement Notice how I chose that there existed an x in A but not in B compliment. In abstract math, you choose how your proof is going to work (but in a reasonable manner). x A and x B x A B So A B I know this seems cheap and going in circles but this is one of the ways to prove the second half. I'm sure someone has a more elegant way. You can go to mathhelpforum.com and they have more mathematics on there that can help you.
Answers:When you have an "if and only if" condition, you must show that If A contained in B, then the intersection of A and B compliment is empty and If the intersection of A and B compliment is empty, then A is contained in B. Let x A. Since A is a subset of B, x A B Because every element contained in A is also contained in B so the intersection holds. x A B x A and x B x A and x B complement x A B compliment Since x is not a common element in A and B compliment, we then say that, A B complement = Now we must prove the opposite, which is a bit tricky. So, if we assume that A B complement = is true, then there exists an element x that belongs to one but not the other. So A B complement = implies that, x A B complement x A and x B complement Notice how I chose that there existed an x in A but not in B compliment. In abstract math, you choose how your proof is going to work (but in a reasonable manner). x A and x B x A B So A B I know this seems cheap and going in circles but this is one of the ways to prove the second half. I'm sure someone has a more elegant way. You can go to mathhelpforum.com and they have more mathematics on there that can help you.
Question:Ok, so I am doing my practice homework and come across a question with proving a set theory. I am a bit confused because it states that For all sets A and B, if B is a subset of A complement then A intersection of B would be an empty set. I guess i know it's true and I know that by definition if it is in A complement and B is a subset then the intersection has to be an empty set, I just don't know how to write it formally. I am just kind of worried about the form of writing the proof out. I hate proofs.
Thank you.
Answers:We prove by contradiction. Suppose A intersection B is not empty. Let x be an element of A intersection B. By definition of set intersection, x is in A and x is also in B. From the hypothesis, B is a subset of A' (A complement). This means that if x is in B, then x is also in A', or equivalently, x is not in A. So x is in A and x is not in A, which is a contradiction. Therefore, A intersection B is empty.
Answers:We prove by contradiction. Suppose A intersection B is not empty. Let x be an element of A intersection B. By definition of set intersection, x is in A and x is also in B. From the hypothesis, B is a subset of A' (A complement). This means that if x is in B, then x is also in A', or equivalently, x is not in A. So x is in A and x is not in A, which is a contradiction. Therefore, A intersection B is empty.