Question

Let A and B be two sets. (a) Show that Ac = (Ac ∩ B) ∪ (Ac ∩ Bc ), Bc = (A ∩ Bc ) ∪ (Ac ∩ Bc ).

(b) Show that (A ∩ B) c = (Ac ∩ B) ∪ (Ac ∩ Bc ) ∪ (A ∩ Bc ).

(c) Consider rolling a fair six-sided die. Let A be the set of outcomes where the roll is an odd number. Let B be the set of outcomes where the roll is less than 4. Calculate the sets on both sides of the equality in part (b), and verify that the equality holds.

Answer 1

(a) The distributive property is $\dpi{80} X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)$ .

Hence, we have

$\dpi{80} (A^c \cap B) \cup (A^c \cap B^c) = A^c \cap (B \cup B^c)$

or $\dpi{80} (A^c \cap B) \cup (A^c \cap B^c) = A^c \cap U$ (for $\dpi{80} U = \phi^c$ be the universal set)

or $\dpi{80} (A^c \cap B) \cup (A^c \cap B^c) = A^c$ .

Again, we have

$\dpi{80} (A \cap B^c) \cup (A^c \cap B^c) = (A \cup A^c) \cap B^c$ (the distributive property)

or $\dpi{80} (A \cap B^c) \cup (A^c \cap B^c) = U \cap B^c$

or $\dpi{80} (A \cap B^c) \cup (A^c \cap B^c) = B^c$ .

(b) We have, $\dpi{80} (A^c \cap B^c) = (A \cup B)^c$ by the De Morgan's law.

Also, we have $\dpi{80} (A^c \cap B) \cup (A^c \cap B^c) = A^c$ , and hence,

$\dpi{80} (A^c \cap B) \cup (A^c \cap B^c) \cup (A \cap B^c) = A^c \cup (A \cap B^c)$

or $\dpi{80} (A^c \cap B) \cup (A^c \cap B^c) \cup (A \cap B^c) = (A^c \cup A) \cap (A^c \cup B^c)$

or $\dpi{80} (A^c \cap B) \cup (A^c \cap B^c) \cup (A \cap B^c) = U \cap (A^c \cup B^c)$

or $\dpi{80} (A^c \cap B) \cup (A^c \cap B^c) \cup (A \cap B^c) = (A^c \cup B^c)$

or $\dpi{80} (A^c \cap B) \cup (A^c \cap B^c) \cup (A \cap B^c) = (A \cap B)^c$ (De Morgan's law).

(c) We have, $\dpi{80} A = \left \{ 1,3,5 \right \}$ and $\dpi{80} B = \left \{ 1,2,3 \right \}$ .

$\dpi{80} (A \cap B)^c = (\left \{ 1,3,5 \right \} \cap \left \{ 1,2,3 \right \})^c$

or $\dpi{80} (A \cap B)^c = (\left \{ 1,3 \right \})^c$

or $\dpi{80} (A \cap B)^c = \left \{ 2,4,5,6 \right \}$ .

Also,

$\dpi{80} (A^c \cap B) = (\left \{ 1,3,5 \right \}^c \cap \left \{ 1,2,3 \right \})$ or $\dpi{80} (A^c \cap B) = (\left \{ 2,4,6 \right \} \cap \left \{ 1,2,3 \right \})$ or $\dpi{80} (A^c \cap B) = \left \{ 2 \right \}$ ,

$\dpi{80} (A^c \cap B^c) = (\left \{ 1,3,5 \right \}^c \cap \left \{ 1,2,3 \right \}^c)$ or $\dpi{80} (A^c \cap B^c) = (\left \{ 2,4,6 \right \} \cap \left \{ 4,5,6 \right \})$ or $\dpi{80} (A^c \cap B^c) = \left \{ 4,6 \right \}$ ,

and $\dpi{80} (A \cap B^c) = (\left \{ 1,3,5 \right \} \cap \left \{ 1,2,3 \right \}^c)$ or $\dpi{80} (A \cap B^c) = (\left \{ 1,3,5 \right \} \cap \left \{ 4,5,6 \right \})$ or $\dpi{80} (A \cap B^c) = \left \{ 5 \right \}$ .

Hence, we have $\dpi{80} (A^c \cap B) \cup (A^c \cap B^c) \cup (A \cap B^c) = \left \{ 2 \right \} \cup \left \{ 4,6 \right \} \cup \left \{ 5 \right \}$ or $\dpi{80} (A^c \cap B) \cup (A^c \cap B^c) \cup (A \cap B^c) = \left \{ 2,4,5,6 \right \}$ .

Hence, we may see that $\dpi{80} (A^c \cap B) \cup (A^c \cap B^c) \cup (A \cap B^c)$ is indeed equal to $\dpi{80} (A \cap B)^c$ .