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Alex787 [66]
2 years ago
7

Determine whether each of these sets is the power set of a set, where a and b are distinct elements.

Mathematics
1 answer:
Schach [20]2 years ago
4 0
A. \varnothing cannot be the power set of any set. Consult Cantor's theorem, which says that the cardinality of the power set of any set (even the empty set) is strictly greater than the cardinality of the set.

(No part b?)

c. Also not the power set of any set, because any power set must have 2^n elements, where n is the cardinality of the original set. The cardinality of this set is 3, but there is no integer n such that 2^n=3. This set would be a power set if \{\varnothing\} (that is, the set containing the empty set) were a member of it.
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The first, second to last and last stements are true

Step-by-step explanation:

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e.g 8 x 1.50 = 12   is the same as marked up by 50%

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We have to identify the function which has the same set of potential rational roots as the function g(x)= 3x^5-2x^4+9x^3-x^2+12.

Firstly, we will find the rational roots of the given function.

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So, p= \pm 1, \pm 2, \pm 3, \pm 4, \pm 6

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So, q=\pm 1, \pm 3

So, the rational roots are given by \frac{p}{q} which are as:

\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}.

Consider the first function given in part A.

f(x) = 3x^5-2x^4-9x^3+x^2-12

Here also, Let 'p' be the factors of 12

So, p= \pm 1, \pm 2, \pm 3, \pm 4, \pm 6

Let 'q' be the factors of 3

So, q=\pm 1, \pm 3

So, the rational roots are given by \frac{p}{q} which are as:

\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}.

Therefore, this equation has same rational roots of the given function.

Option A is the correct answer.

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