Prove: P(A^B)=P(A)^P(B) where ^ is intersection?

Question

P(A) is the power set of A. If you don't know what that means, don't guess on the question, because you won't answer properly.

Answer 1

Let {x} element of P(A^B),
implies and implied 'x' is element of A^B
implies and implied 'x' is element of A and 'x' is element of B
implies and implied {x} is element of P(A) and {x} is element of P(B)
implies and implied {x} is element of P(A)^P(B)
impies and implied P(A^B) = P(A)^P(B)

Answer 2

An intersection element for those 2 curves potential x^3+4x = 3x^2+a million, this is comparable to x^3 - 3x^2 + 4x -a million =0. This function is monotonous! this is by way of the fact that is spinoff is 3x^2 - 6x + 4, that's constantly >0 (a normal examine, D= -12) So this function won't have greater effective than a million 0. even though it for sure has a million, via fact that could be a cubic function and because, as an occasion, for x=a million, that is equivalent to a million-3+4-a million=a million, for x=0, ---- 0-0+0-a million = -a million. So it has a nil someplace between 0 and a million. that's that is purely 0 - and the only intersection factor to the given curves.

Answer 3

In other words, a set C is a subset of A intersection B if and only if C is a subset of both A and B individually. This is an easy homework problem. The previous two posters have it wrong, BTW.

Answer 4

hey i have the answer but don't know how to put it in english termzz. ill try:

p(a^b)= p(aUb) - p(a) - p(b)
p(a)^p(b)= p(a) + P(b) - p(aUb). you add these two and u got ur answer. since we can't have - powers, it comes out P(a^b) = P(A)^P(B).