(The set of all such that)
A={n in Z| 2 divides n}
B={n in Z| 3 divides n}
Both are contained in the universal set
U=Z
a/ What is A union B?
b/ What is the complement of A?
a/ Sorry, it was what's A intersection B?
A={n ∈ Z| 2 divides n}
B={n ∈ Z| 3 divides n}
A∩B
={n in Z| 2|n and 3|n}
={n in Z| 6|n}
That is
A∩B is the set of all integers divisible by 6.
A'={n in Z| 2∤n}
= all odd integers.
To find the solutions to these set operations, we need to understand what each set represents and then apply the corresponding operations.
a/ A union B (A ∪ B) is the set that contains all elements that are in either set A or set B, or in both.
In set A, we have A = {n in Z | 2 divides n}. This means that A is the set of all integers (Z) where each number is divisible by 2. So, the elements of A are ..., -4, -2, 0, 2, 4, 6, ... (every even number).
In set B, we have B = {n in Z | 3 divides n}. This means that B is the set of all integers (Z) where each number is divisible by 3. So, the elements of B are ..., -6, -3, 0, 3, 6, 9, ... (every multiple of 3).
To find A union B, we need to combine the elements of both sets, removing any duplicates. So, the elements of A union B are ..., -6, -4, -3, -2, 0, 2, 3, 4, 6, 9, ...
b/ The complement of set A (A') is the set that contains all elements of the universal set U (Z in this case), but are not part of set A. In other words, A' consists of all integers that are not divisible by 2.
To find the complement of A, we need to consider the elements of U (Z) that do not belong to A. Since A is the set of all even integers, the complement of A consists of all odd integers.
So, the complement of A (A') is the set of all odd integers: ..., -5, -3, -1, 1, 3, 5, ....