Let A = {–3, –1}, B = {1, 3) and C = {3, 5). Find:
(i) $A \times B$
(ii) $(A \times B) \times C$
(iii) $B \times C$
(iv) $A \times(B \times C)$
(i) Given: $A=\{-3,-1\}$ and $B=\{1,3\}$
To find: A × B
By the definition of the Cartesian product
Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e.
$P \times Q=\{(p, q): p \in P, q \in Q\}$
Here, $A=\{-3,-1\}$ and $B=\{1,3\} .$ So,
$A \times B=\{-3,-1\} \times\{1,3\}$
$=\{(-3,1),(-3,3),(-1,1),(-1,3)\}$
(ii) Given: $C=\{3,5\}$
From part (i), we get $A \times B=\{(-3,1),(-3,3),(-1,1),(-1,3)\}$
So
$(A \times B) \times C=\{(-3,1),(-3,3),(-1,1),(-1,3)\} \times(3,5)$
$=(-3,1,3),(-3,1,5),(-3,3,3),(-3,3,5),(-1,1,3),(-1,1,5),(-1,3,3),(-1,3,5)\}$
(iii) Given: $B=\{1,3\}$ and $C=\{3,5\}$
To find: B $\times \mathrm{C}$
By the definition of the Cartesian product
Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e.
$P \times Q=\{(p, q): p \in P, q \in Q\}$
Here, $B=\{1,3\}$ and $C=\{3,5\} .$ So,
$B \times C=(1,3) \times(3,5)$
$=\{(1,3),(1,5),(3,3),(3,5)\}$
(iv) Given: $A=\{-3,-1\}$
From part (iii), we get $B \times C=\{(1,3),(1,5),(3,3),(3,5)\}$
So,
$A \times(B \times C)=\{-3,-1\} \times\{(1,3),(1,5),(3,3),(3,5)\}$
$=(-3,1,3),(-3,1,5),(-3,3,3),(-3,3,5),(-1,1,3),(-1,1,5),(-1,3,3),(-1,3,5)\}$