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Akimi4 [234]
3 years ago
14

Find a counterexample to show that the following statement is false. For all sets A and B, (A ∪ B)c = Ac ∪ Bc. Assume that all s

ets are subsets of a universal set U = {1, 2, 3, 4, 5}. (Enter your answer as a comma-separated list in the form A, B where both A and B are written in set-roster notation. Enter EMPTY or ∅ for the empty set.)
Mathematics
1 answer:
olganol [36]3 years ago
7 0

Answer:

Given,

The universal set,

U = {1, 2, 3, 4, 5},

Let A = {1, 2},

B = {2, 3, 4}

A\cup B=\text{ all elements of A and B }=\{1, 2, 3, 4\}

(A\cup B)^c=U-(A\cup B) = \{1, 2, 3, 4, 5\} - \{1, 2, 3, 4\} = \{5\}

Now,

A^c = U - A = \{3, 4, 5\}

B^c=U-B=\{1, 5\}

A^c\cup B^c = \{1, 3, 4, 5\}

\implies (A\cup B)^c\neq A^c\cup B^c

Hence, proved......

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