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Consider rolling a six-sided die. Let A be the set of outcomes where the roll is an even number. Let B be the set of outcomes where the roll is greater than 3. Calculate and compare the sets on both sides of De Morgan’s law.
1. [tex](A \cup B)^c = A^c \cap B^c[/tex]
2. [tex](A \cap B)^c = A^c \cup B^c[/tex]

Sagot :

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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

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