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DerKrebs [107]
2 years ago
11

Find the truth set of each predicate.

Mathematics
1 answer:
Sedbober [7]2 years ago
5 0

The truth set for each predicate are:

a) d ∈ { -6, -3, -2, -1, 1, 2, 3, 6}

b) d ∈ {  1, 2, 3, 6}

c) x ∈ { [-2, -1] U [1, 2]}

d) x ∈ {-2, -1, 1, 2}

<h3>How to find the truth set of each predicate?</h3>

We only need to find the sets of values such that the statements are true.

a)

We want to find vales of d, such that d is an integer, and 6/d is an integer.

Here the possible values of d will be:

d ∈ { -6, -3, -2, -1, 1, 2, 3, 6}

Which are all the factors of 6, so all these integers divide 6.

b) Same as before, but this time the domain is a positive integer, so now the truth set will be:

d ∈ {  1, 2, 3, 6}

c) We want to find real values of x such that:

1 ≤ x² ≤ 4

If we apply the square root to the 3 sides, we get two inequalities:

-√1 ≥ x ≥ -√4

√1 ≤ x ≤ √4

Simplifying:

-1  ≥ x ≥ -2

1  ≤ x ≤ 2

So the truth set is:

x ∈ { [-2, -1] U [1, 2]}

c) Same as before, but now we only have integer solutions, so the truth set is:

x ∈ {-2, -1, 1, 2}

If you want to learn more about predicates:

brainly.com/question/18152046

#SPJ1

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Answer:

\displaystyle (5,5)

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