If \(A = \{5,9,10\}\) and \(B = \{0,6\}\) then
(i) finds \(A \times B\) and \(B \times A\).
(ii) Is \(A \times B = B \times A\)? If not, why?
(iii) Show that \(n(A \times B) = n(B \times A) = n(A) \times n(B)\)
Answer:
(i) \(A \times B =\)
\(B \times A =\)
(ii) Since all the ordered pairs of \(A \times B\) and \(B \times A\) are , then \(A \times B\) \(B \times A\).
(iii) \(n(A) =\)
\(n(B) =\)
\(n(A \times B) =\)
\(n(B \times A) =\)
Therefore, \(n(A \times B)\) \(n(B \times A)\) \(n(A) \times n(B)\).
Answer variants:
not equal
\(\neq\)
\(2\)
\(6\)
\(3\)
\(=\)