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kramer
3 years ago
10

Find set A={1, 2, 6, 10} B={3, 6, 9, 10, 11} C = {1, 2, 4, 7, 11}

Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
6 0

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}

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