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If U = {all positive integers} and A = {x|x ∈ U and x is an odd positive integer}, which describes Ac?

Ac = {x|x ∈ U and is a negative integer}

Ac = {x|x ∈ U and is zero}

Ac = {x|x ∈ U and is not an integer}

Ac = {x|x ∈ U and is an even positive integer}

Verified Answer :

TaeKwonDoIsDead

Given:

Universal set is all positive integers {1,2,3...}

Set A is the set of all positive ODD integers {1,3,5,7...}


The question asks us to find \(A^{c}\), which is the complement of set A. The complement of a set refers to elements NOT in that set. Hence, complement of A should be all the elements NOT in A but in Universal Set, U.

It is clear from the question that set A houses all the odd positive numbers, so complement of A will have all the even positive numbers. Last choice is the correct one.


ANSWER: \(A^{c} =\) "{x|x ∈ U and is an even positive integer}"

You can use the definition of complement of a set to find out the value of \(A^c\)

The set that describes \(A^c\) is given by:
Option D:  \(A^c\) = {x|x ∈ U and is an even positive integer}

What is a complement of a set?

Firstly there is universal set specified in a given context which contains all the set discussed there in that context if not otherwise restricted.  It is generally indicated by U.

Let there is a set A in this universal set. Then its complement set would contain those elements of U which are not in U.

The complement of a set is written as  \(A^c\)

How to find the complement of set A here?

Since U is set of all positive integers, and A contains all odd positive integers, and since if an integer is not odd, it is surely even, thus the complement of A would contain all positive even integers since only those positive even integers are not in A which are in U.

Thus, we have

Option D:  \(A^c\) = {x|x ∈ U and is an even positive integer} as correct choice.

Learn more about complement of set here:

https://brainly.com/question/24403298

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