Exercise 1.1
1. Find $A \times B$, $A \times A$ and $B \times A$
(i) $A = \{2, -2, 3\}$ and $B = \{1, -4\}$
(ii) $A = B = \{p, q\}$
(iii) $A = \{m, n\}$ ; $B = \phi$
(i) $A = \{2, -2, 3\}$ and $B = \{1, -4\}$
(ii) $A = B = \{p, q\}$
(iii) $A = \{m, n\}$ ; $B = \phi$
Solutions:
(i) Given: $A = \{2, -2, 3\}$ and $B = \{1, -4\}$
To find $A \times B$:
$A \times B = \{2, -2, 3\} \times \{1, -4\}$
$A \times B = \{(2, 1), (2, -4), (-2, 1), (-2, -4), (3, 1), (3, -4)\}$
$A \times B = \{2, -2, 3\} \times \{1, -4\}$
$A \times B = \{(2, 1), (2, -4), (-2, 1), (-2, -4), (3, 1), (3, -4)\}$
To find $A \times A$:
$A \times A = \{2, -2, 3\} \times \{2, -2, 3\}$
$A \times A = \{(2, 2), (2, -2), (2, 3), (-2, 2), (-2, -2), (-2, 3), (3, 2), (3, -2), (3, 3)\}$
$A \times A = \{2, -2, 3\} \times \{2, -2, 3\}$
$A \times A = \{(2, 2), (2, -2), (2, 3), (-2, 2), (-2, -2), (-2, 3), (3, 2), (3, -2), (3, 3)\}$
To find $B \times A$:
$B \times A = \{1, -4\} \times \{2, -2, 3\}$
$B \times A = \{(1, 2), (1, -2), (1, 3), (-4, 2), (-4, -2), (-4, 3)\}$
$B \times A = \{1, -4\} \times \{2, -2, 3\}$
$B \times A = \{(1, 2), (1, -2), (1, 3), (-4, 2), (-4, -2), (-4, 3)\}$
(ii) Given: $A = B = \{p, q\}$
Since $A$ and $B$ are the same sets, their cross products $A \times B$, $A \times A$, and $B \times A$ will all yield the identical result.
To find $A \times B$:
$A \times B = \{p, q\} \times \{p, q\}$
$A \times B = \{(p, p), (p, q), (q, p), (q, q)\}$
$A \times B = \{p, q\} \times \{p, q\}$
$A \times B = \{(p, p), (p, q), (q, p), (q, q)\}$
To find $A \times A$:
$A \times A = \{(p, p), (p, q), (q, p), (q, q)\}$
$A \times A = \{(p, p), (p, q), (q, p), (q, q)\}$
To find $B \times A$:
$B \times A = \{(p, p), (p, q), (q, p), (q, q)\}$
$B \times A = \{(p, p), (p, q), (q, p), (q, q)\}$
(iii) Given: $A = \{m, n\}$ ; $B = \phi$
Recall that $\phi$ represents an empty set $\{\}$. The Cartesian product of any set with an empty set is always an empty set.
To find $A \times B$:
$A \times B = \{m, n\} \times \phi$
$A \times B = \phi$
$A \times B = \{m, n\} \times \phi$
$A \times B = \phi$
To find $A \times A$:
$A \times A = \{m, n\} \times \{m, n\}$
$A \times A = \{(m, m), (m, n), (n, m), (n, n)\}$
$A \times A = \{m, n\} \times \{m, n\}$
$A \times A = \{(m, m), (m, n), (n, m), (n, n)\}$
To find $B \times A$:
$B \times A = \phi \times \{m, n\}$
$B \times A = \phi$
$B \times A = \phi \times \{m, n\}$
$B \times A = \phi$