Let U be the universal set and A, B, C, the subsets of U such that
U= {x|x £ Z; 0<x<=30}
A= {U{all primes}}, B= {all multiples of 3}, C= {1, 2, 5, 8, 9, 12, 15} . . .
Using the above information, verify Democrat's Law
Demorgan's Law states that the complement of the union of two sets is equal to the intersection of their complements.
In this case, we need to verify if (A ∪ B)' = A' ∩ B'.
First, let's find A ∪ B and its complement:
A ∪ B is the set of all integers that are either primes or multiples of 3.
Primes in the universal set U = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}. Therefore, A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}.
Multiples of 3 within the universal set are {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}. Therefore, B = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}.
So, A ∪ B = {2, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 18, 19, 21, 23, 24, 27, 29, 30}.
Then, (A ∪ B)' would be the set of all elements in the universal set that are not in (A ∪ B).
Calculating (A ∪ B)', we get (A ∪ B)' = {1, 4, 8, 10, 14, 16, 20, 22, 25, 26, 28}.
Now, let's find A' and B' and their intersection:
A' would be the set of all integers that are not primes.
So, A' = {1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30}.
B' represents the set of all integers that are not multiples of 3.
Therefore, B' = {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26, 28, 29}.
Finally, A' ∩ B' represents the intersection of A' and B', which is {1, 8, 14, 25}.
This verifies that (A ∪ B)' = A' ∩ B', which satisfies DeMorgan's Law in this context.