Verify the property: $x \times y = y \times x$ by taking:
(i) $ x=-\frac{1}{3}, y=\frac{2}{7} $
(ii) $ x=\frac{-3}{5}, y=\frac{-11}{13} $
(iii) $ x=2, y=\frac{7}{-8} $
(iv) $ x=0, y=\frac{-15}{8} $
To do:
We have to verify $x \times y = y \times x$.
Solution:
(i) LHS $=x \times y$
$=-\frac{1}{3} \times \frac{2}{7}$
$=\frac{-1 \times 2}{3 \times 7}$
$=\frac{-2}{21}$
RHS $=y \times x$
$=\frac{2}{7} \times \frac{-1}{3}$
$=\frac{2 \times(-1)}{7 \times 3}$
$=\frac{-2}{21}$
LHS $=$ RHS
Therefore,
$x \times y = y \times x$.
(ii) LHS $=x \times y$
$=\frac{-3}{5} \times \frac{-11}{13}$
$=\frac{-3 \times -11}{5 \times 13}$
$=\frac{33}{65}$
RHS $=y \times x$
$=\frac{-11}{13} \times \frac{-3}{5}$
$=\frac{-11 \times(-3)}{13 \times 5}$
$=\frac{33}{65}$
LHS $=$ RHS
Therefore,
$x \times y = y \times x$.
(iii) LHS $=x \times y$
$=2 \times \frac{7}{-8}$
$=\frac{2 \times -7}{8}$
$=\frac{-7}{4}$
RHS $=y \times x$
$=\frac{7}{-8} \times 2$
$=\frac{-7 \times2}{8}$
$=\frac{-7}{4}$
LHS $=$ RHS
Therefore,
$x \times y = y \times x$.
(iv) LHS $=x \times y$
$=0 \times \frac{-15}{8}$
$=\frac{0 \times -15}{8}$
$=0$
RHS $=y \times x$
$=\frac{-15}{8} \times 0$
$=\frac{-15 \times0}{8}$
$=0$
LHS $=$ RHS
Therefore,
$x \times y = y \times x$.
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