1. ## Set Questions

1. Show
i. $(A' \cup B') \cap (A' \cup B) \cap (A \cup B)=A' \cap B$

ii. $(A \cup B') \cap (A \cup B)=A$

iii. $(A \cup B' \cup C') \cap [A \cup (B \cap C)]=A$

iv. $A \cup (A' \cup B)'=A$

2. A certain sixth form science class consists of 90 pupils, all of whom take mathematics, 45 study chemistry, 42 study biology, and 52 study physics. Of those taking three subjects, 18 study physics and chemistry, 22 study physics and biology, and 16 study chemistry and biology while 5 take all four subjects. Find the number of pupils who study mathematics only.

Note: I'd like a Venn Diagram and the formula for question 2.

2. Originally Posted by cloud5

1. Show
i. $(A' \cup B') \cap (A' \cup B) \cap (A \cup B)=A' \cap B$

(A' u B') n [( A' n A) u B ]

=( A' u B' ) n ( nullset u B )

=(A' u B' ) n B

= (A' n B ) u ( B' n B )

= ( A' n B ) u null set

= you got ur prove .

3. Originally Posted by cloud5

1. Show

ii. $(A \cup B') \cap (A \cup B)=A$

iii. $(A \cup B' \cup C') \cap [A \cup (B \cap C)]=A$
(2) A u ( B' n B ) = A u null set - distributive

(3) A u [( B' u C' ) n (B n C )] - distributive

= A u [( B n C )' n ( B n C )] -- de morgans

= A u null set

= A

(4) This is not clear . Its A u (A' u B)' or [A u (A' u B)]'

(2) A u ( B' n B ) = A u null set - distributive

(4) This is not clear . Its A u (A' u B)' or [A u (A' u B)]'
Opss... made a mistake there... it is

$A \cup (A' \cup B)'=A$

5. Originally Posted by cloud5
Opss... made a mistake there... it is

$A \cup (A' \cup B)'=A$

( A n universal set ) u ( A n B' ) = A n ( universal set u B )

= A n universal set

=A

6. Hello, cloud5!

I'll do the first two.

We need a couple of Theorems:

. . $\begin{array}{cc}
[1] & P \cap P' \:=\:\emptyset \\ \\[-3mm]
[2] & P \cup \emptyset \:=\:P \end{array}$

$(ii)\;\;(A \cup B') \cap (A \cup B) \;=\;A$

$(A \cup B') \cap (A \cup B)$

. . $\begin{array}{ccc}= & A \cup (B' \cap B) & \text{Distr.} \\ \\
= &A \cup \emptyset & [1] \\ \\
=& A & [2] \end{array}$

$(i)\;\;(A' \cup B') \cap (A' \cup B) \cap (A \cup B) \:=\:A' \cap B$

$(A' \cup B') \cap (A' \cup B) \cap (A \cup B)$

. . $\begin{array}{cccc}= & \bigg[A' \cup (B' \cap B)\bigg] \cap (A \cup B) & \text{Distr.} \\ \\
= & \bigg[A' \cup \emptyset\bigg] \cap (A \cup B) & [1] \\ \\
= & A' \cap (A \cup B) & [2] \\ \\
= & (A' \cap A) \cup (A' \cap B) & \text{Distr.} \\ \\
= & \emptyset \cup (A' \cap B) & [1] \end{array}$

. . $\begin{array}{cccc}
\end{array}$