Discrete Math Proving some power sets

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To prove by contradiction that if the power set of A is a subset of the power set of B, then A must be a subset of B, one should assume that P(A) is a subset of P(B) while also assuming that A is not a subset of B. This leads to the conclusion that power set B must contain all sets in power set A, including the set A itself. Consequently, this implies that A is an element of P(B), which means A is a subset of B, contradicting the initial assumption. The correct approach involves clarifying the assumptions made during the proof. Understanding this logic is crucial for successfully completing the proof.
Servo888
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Ok so I need to prove (by contradiction) that... if the power set(A) is a subset of power set(B), then A is a subset of B.

I was given a hint to use proof by contradiction, but in general I'm lost as to what to do... I know the power set of (A) is {B|B subset A} and the powerset of (B) is {A|A subset B} but that's about it. As far as where to start; I'm drawing a blank. I need a point in the right direction of where to start, because as it stands now I've spent 2 hours looking at this problem.
 
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Ok I think I got it...

My proof is this; Assume that if P(A) is a subset of P(B), then A is not a subset of B (contradiction method). I then show that by definition of a power set; power set B must contain all sets that are in power set A. That would mean that powerset B must contain the set A, there for A is a subset of B (since P(B) contains the set A). Which contradicts the assumption.

Not sure if it's complete, but I think the logic is right (makes sense).
 
Unnecessary contradiction. A is an element of P(A), which is contained in P(B), ie A is an element of P(B), hence A is...
 
matt grime said:
Unnecessary contradiction. A is an element of P(A), which is contained in P(B), ie A is an element of P(B), hence A is...
a subset of B.
 
Assume that if P(A) is a subset of P(B), then A is not a subset of B
Also if you're doing it by contradiction, this is not the assumption you want to make. The negation of the statement "if P(A) is a subset of P(B), then A is not a subset of B" is "P(A) is a subset of P(B) and A is a subset of B," which you can't prove anyway. What you mean is "Assume P(A) is a subset of P(B) _and_ A is not a subset of B."
 
0rthodontist said:
Also if you're doing it by contradiction, this is not the assumption you want to make. The negation of the statement "if P(A) is a subset of P(B), then A is not a subset of B" is "P(A) is a subset of P(B) and A is a subset of B," which you can't prove anyway. What you mean is "Assume P(A) is a subset of P(B) _and_ A is not a subset of B."

Ah I understand what you are saying, so it's just assuming without implication (sort of). We are assuming P(A) is a subset of P(B) and we assume A is not a subset of B. That makes sense.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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