How Do You Prove Power Set Equations for Sets A and B?

AI Thread Summary
The discussion focuses on proving two power set equations involving sets A and B. The first equation, P(A U B) = P(A) U P(B), is shown to be false, with the subset relationship demonstrated from right to left. The second equation, P(A-B) = P(A) - P(B), is also proven false, as neither side is a subset of the other using specific counterexamples. Examples provided include sets A = {1,2} and B = {1,3}, illustrating the discrepancies in the power sets. The conclusion emphasizes that both equations do not hold true universally.
JasonJo
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How in the heck do i prove these:

Prove whether the following equations are true for all sets. For each one that's not always true, try to prove that one side is a subset of the other, and give a counterexample to the other direction. If neither side must be a subset of the other, give a counterexample to both directions:

let P(A) denote the power set of A

a) P(A U B) = P(A) U P(B)
b) P(A-B) = P(A) - P(B)
 
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a. is not true. Think about it, try an example or two. The subset goes from right to left. To prove things like this, you assume x is an element of P(A) U P(B) and show that x must be an element of P(A U B).

b. is also not true, the subset goes from left to right.
 
for part (b) i don't think either side is a subset of each other because:

let A = {1,2}
B = {1,3}
A-B = {2}

P(A) = {0, 1, 2, {1,2}}
P(B) = {0, 1, 3, {1,3}}
P(A)-P(B) = {2, {1,2}}
P(A-B) = {0, 2}

they aren't equal and one isn't a subset of another.
 
Ah, you're right.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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