$A \cap B = A$
Also $A \subset A \cup B$
$\therefore A \cap (A \cup B) = A$
50 questions · timed · auto-graded




Let A = {x : x is an odd natural number}
$A' = U - A = \{ x:x \in N\} -${x : x is an odd natural number}
= {x : x is an even natural number}
C - B = {2,4,6,8,10,12,14,16} - {4,8,12,16,20}
= {2,6,10,14}
C - A = {2,4,6,8,10,12,14,16} - {3,6,9,12,15,18,21}
= {2,4,8,10,14,16}
C - D = {2,4,6,8,10,12,14,16} - {5,10,15,20}
= {2,4,6,8,12,14,16}
D - B = {5,10,15,20} - {4,8,12,16,20}
= {5,10,15}
$C \cap D$ = {x : x is an odd natural number} $ \cap ${x : x is a prime number}
= {x : x is an odd prime number}
$B \cap D$ = {x : x is an even natural number) $ \cap ${x : x is a prime number}
= {2}
$B \cap C$= {x : x is an even natural number} $ \cap${x : x is an odd natural number}
$ = \phi $
$A \cap D$= {x : x is a natural number) $ \cap${x : x is a prime number}
= {x : x is a prime number}
= D
$A \cap C$ = {x : x is a natural number} $ \cap ${x : x is an odd natural number}
= { x : x is an odd natural number }
= C