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,

$x \times y = y \times x$.