Answer:
See below
Step-by-step explanation:
[tex]A = \{2, 4, 6\}[/tex]
[tex]A^c =\{1, 3, 5\}\\B = \{4, 5, 6\}\\B^c = \{ 1,2, 3\}[/tex]
To prove 1 we have
[tex](A \bigcup B) = \{2, 4, 6\} \bigcup \{4, 5, 6\} = \{2, 4, 5, 6\}\\(A \bigcup B) ^c =\bold{\{1,3\}}\\A^c \bigcap B^c = {\{1, 3, 5\} \bigcap \{1, 2, 3\} =\bold{\{1,3\}}\\[/tex]
Hence proved
To prove 2
[tex](A \bigcap B) = \{2, 4, 6\} \bigcap \{4, 5, 6\} = \{4, 6\}\\\\(A \bigcap B)^c = \bold{\{1,2, 3, 5\}}\\A^c \bigcup B^c = \{1, 3, 5\} \bigcup \{1, 2, 3\} = \bold{\{1,2,3,5\}}\[/tex]
Hence proved
Notes
The complement of a set is the set of all elements not in the set
Here the set of all outcomes is {1, 2, 3,4,5, 6}
So if A = {2, 4, 6} then the complement of A s=is the set of outcomes not in A ie the set {1, 3, 5} which is the set of odd numbers on a die throw
The union of two sets is the set of all elements in both sets without duplication
The intersection of two sets is the set of all elements common to both sets
