If <em>U</em> = {1, 2, 3, …, 12} is the universal set, and
<em>A</em> = {1, 2, 6, 10}
<em>B</em> = {3, 6, 9, 10, 11}
<em>C</em> = {1, 2, 4, 7, 11}
then
(1) <em>A</em> U <em>B</em> is the set containing all elements from <em>A</em> and <em>B</em>,
<em>A</em> U <em>B</em> = {1, 2, 3, 6, 9, 10, 11}
(2) <em>A</em> ∩ <em>B</em> is the set of elements that are contained in both <em>A</em> and <em>B</em>,
<em>A</em> ∩ <em>B</em> = {6, 10}
(3) Unfortunately, <em>A</em> ∩ <em>B</em> U <em>C</em> is somewhat ambiguous. It could mean (<em>A</em> ∩ <em>B</em>) U <em>C</em> or <em>A</em> ∩ (<em>B</em> U <em>C </em>). Then either
(<em>A</em> ∩ <em>B</em>) U <em>C</em> = {6, 10} U {1, 2, 4, 7, 11} = {1, 2, 4, 6, 7, 10, 11}
or
<em>A</em> ∩ (<em>B</em> U <em>C </em>) = {1, 2, 6, 10} ∩ {1, 2, 3, 4, 6, 7, 9, 10, 11} = {1, 2, 6, 10}
The first interpretation is probably the intended one, since that essentially reads the set operations from left to right.
(4) <em>A'</em> U <em>B</em> is the union of <em>A'</em> and <em>B</em>, where <em>A'</em> is the complement of <em>A</em>, or all elements in <em>U</em> that are not in <em>A</em>. We have
<em>A'</em> = <em>U</em> - <em>A</em> = {3, 4, 5, 7, 8, 9, 11, 12}
and so
<em>A'</em> U <em>B</em> = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
(5) We have
<em>A</em> U <em>C</em> = {1, 2, 4, 6, 7, 10, 11}
so that
(<em>A</em> U <em>C </em>)<em>'</em> = <em>U</em> - (<em>A</em> U <em>C</em> ) = {3, 5, 8, 9, 12}
(6) We have
<em>B'</em> = <em>U</em> - <em>B</em> = {1, 2, 4, 5, 7, 8, 12}
and so
<em>A</em> ∩ <em>B'</em> = {1, 2}
(7) Using the complements found in (4) and (6), we have
<em>A'</em> U <em>B'</em> = {1, 2, 3, 4, 5, 7, 8, 9, 11, 12}
Alternatively, we can use the fact that
<em>A'</em> U <em>B'</em> = (<em>A</em> ∩ <em>B</em>)<em>'</em>
and since we know from (2) that <em>A</em> ∩ <em>B</em> = {6, 10}, we end up with the same result,
(<em>A</em> ∩ <em>B</em>)<em>'</em> = <em>U</em> - (<em>A</em> ∩ <em>B</em>) = {1, 2, 3, 4, 5, 7, 8, 9, 11, 12}
(8) We have
<em>A</em> U <em>B</em> U <em>C</em> = {1, 2, 3, 4, 6, 7, 9, 10, 11}
so that
(<em>A</em> U <em>B</em> U <em>C</em> )<em>'</em> = {5, 8, 12}
