Question

Let U = {all integers}.
Consider the following sets:
A = {x | x ∈ U and x > 3}

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

C = {x | x ∈ U and 2x is an odd integer}

D = {x | x ∈ U and x is an odd integer}

Assuming 0 is an even integer, which set is the complement to set B?

Which set is an empty set?

Which set would contain the subset E = {1, 3, 5, 7}?

Answer 1

All integer numbers are either even or odd. So, the complement to the set B (to the set of all even integers) is the set of all odd integers, that is D.

The integer number of the form 2x is always even, because it is always divided by 2, so the set C is the empty set.

Numbers 1, 3, 5, 7 are odd, therefore they belong to the set D.

Answer 2

Answer:

D, C, D

Step-by-step explanation:

The complement of the set B is the set that contains all the elements that aren't in B. A number is either even or odd (it can't be both at the same time and it can't be neither of both). If an element is not in B, that means is not even. So, it's odd and it belongs to D. Therefore, D is the complement of B.

Let's notice that 4 belongs to U and is greater than 3, so it belongs to A. A is not empty. It also belongs to B because is even. So, B isn't empty. And 5 belongs to D because is an odd integer, D isn't empty as well.  

Let's see what happens with C: every number multiply by 2 is even, so 2x can't be an odd integer. Therefore, C is an empty set.

1, 3, 5 and 7 are all odd integers, so E is a subset of D.  

A can't contain it because 1 < 3.

B can't contain it because 1, 3, 5 and 7 aren't even.

C can't contain it because we said is empty.