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Dmitriy789 [7]
3 years ago
10

Set Question involving real and natural numbers

Mathematics
1 answer:
Allushta [10]3 years ago
3 0

Only two real numbers satisfy x² = 23, so A is the set {-√23, √23}. B is the set of all non-negative real numbers. Then you can write the intersection in various ways, like

(i) A ∩ B = {√23} = {x ∈ R | x = √23} = {x ∈ R | x² = 23 and x > 0}

√23 is positive and so is already contained in B, so the union with A adds -√23 to the set B. Then

(ii) A U B = {-√23} U B = {x ∈ R | (x² = 23 and x < 0) or x ≥ 0}

A - B is the complement of B in A; that is, all elements of A not belonging to B. This means we remove √23 from A, so that

(iii) A - B = {-√23} = {x ∈ R | x² = 23 and x < 0}

I'm not entirely sure what you mean by "for µ = R" - possibly µ is used to mean "universal set"? If so, then

(iv.a) Aᶜ = {x ∈ R | x² ≠ 23} and Bᶜ = {x ∈ R | x < 0}.

N is a subset of B, so

(iv.b) N - B = N = {1, 2, 3, ...}

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