Verify the property: x × y = y × x by taking:

Solution:

We have to verify that $x \times y=y \times x$.

(i) $x=\frac{-1}{3}, y=\frac{2}{7}$

$x \times y=\frac{-1}{3} \times \frac{2}{7}=\frac{-2}{21}$

$y \times x=\frac{2}{7} \times \frac{-1}{3}=\frac{-2}{21}$

$\therefore \frac{-1}{3} \times \frac{2}{7}=\frac{2}{7} \times \frac{-1}{3}$

Hence verified.

(ii) $x=\frac{-3}{5}, y=\frac{-11}{13}$

$\mathrm{x} \times \mathrm{y}=\frac{-3}{5} \times \frac{-11}{13}=\frac{33}{65}$

$y \times x=\frac{-11}{13} \times \frac{-3}{5}=\frac{33}{65}$

$\therefore \frac{-3}{5} \times \frac{-11}{13}=\frac{-11}{13} \times \frac{-3}{5}$

Hence verified.

(iii) $x=2, y=\frac{7}{-8}$

$x \times y=2 \times \frac{7}{-8}=\frac{7}{-4}$

$y \times x=\frac{7}{-8} \times 2=\frac{7}{-4}$

$\therefore 2 \times \frac{7}{-8}=\frac{7}{-8} \times 2$

Hence verified.

$($ iv $) x=0, y=\frac{-15}{8}$

$x \times y=0 \times \frac{-15}{8}=0$

$y \times x=\frac{-15}{8} \times 0=0$

$\therefore 0 \times \frac{-15}{8}=\frac{-15}{8} \times 0$

Hence verified.