The set A can be rewritten using the definition of multiple as follows:
"A = \\left\\{ 5x \\mid x\\in\\mathbb{Z}\\right\\}."Set B consists of 6 elements because 20 has 6 divisors and "20=2^2*5". Because of that:
Since the universe in our case is integer numbers, then "B^{c}=\\mathbb{Z}\\setminus B =\\mathbb{Z}\\setminus \\left\\{1,2,4,5,10,20\\right\\}", it means that "B^c" contains all integer numbers beside 1, 2, 4, 5, 10, 20. Then the intersection of these sets will be the following set:
The second condition occurs from the fact that 5=5*1, 10=5*2, 20=5*4 are not included in the set "B^c" , thus cannot be in the intersection.
Venn diagram:
Comments
Leave a comment