Verify the property: x × (y × z) = (x × y) × z by taking:

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

(i) $x=\frac{-7}{3}, y=\frac{12}{5}, z=\frac{4}{9}$

$\mathrm{x} \times(\mathrm{y} \times \mathrm{z})=\frac{-7}{3} \times\left(\frac{12}{5} \times \frac{4}{9}\right)=\frac{-7}{3} \times \frac{16}{15}=\frac{-112}{45}$

$(x \times y) \times z=\left(\frac{-7}{3} \times \frac{12}{5}\right) \times \frac{4}{9}=\frac{-28}{5} \times \frac{4}{9}=\frac{-112}{45}$

$\therefore \frac{-7}{8} \times\left(\frac{15}{5} \times \frac{4}{9}\right)=\left(\frac{-7}{8} \times \frac{15}{5}\right) \times \frac{4}{9}$

Hence verified.

(ii) $x=0, y=\frac{-3}{5}, z=\frac{-9}{4}$

$x \times(y \times z)=0 \times\left(\frac{-3}{5} \times \frac{-9}{4}\right)=0 \times \frac{27}{20}=0$

$(x \times y) \times z=\left(0 \times \frac{-3}{5}\right) \times \frac{-9}{4}=0 \times \frac{-9}{4}=0$

$\therefore 0 \times\left(\frac{-3}{5} \times \frac{-9}{4}\right)=\left(0 \times \frac{-3}{5}\right) \times \frac{-9}{4}$

Hence verified.

(iii) $x=\frac{1}{2}, y=\frac{5}{-4}, z=\frac{-7}{4}$

$x \times(y \times z)=\frac{1}{2} \times\left(\frac{5}{-4} \times \frac{-7}{4}\right)=\frac{1}{2} \times \frac{35}{16}=\frac{35}{32}$'

$(x \times y) \times z=\left(\frac{1}{2} \times \frac{5}{-4}\right) \times \frac{-7}{4}=\frac{5}{-8} \times \frac{-7}{4}=\frac{35}{32}$

$\therefore \frac{1}{2} \times\left(\frac{5}{-4} \times \frac{-7}{4}\right)=\left(\frac{1}{2} \times \frac{5}{-4}\right) \times \frac{-7}{4}$

Hence verified.

$($ iv $) x=\frac{5}{7}, y=\frac{-12}{13}, z=\frac{-7}{18}$

$x \times(y \times z)=\frac{5}{7} \times\left(\frac{-12}{13} \times \frac{-7}{18}\right)=\frac{5}{7} \times \frac{14}{39}=\frac{10}{39}$

$\left.(x \times y) \times z=\frac{5}{7} \times \frac{-12}{13}\right) \times \frac{-7}{18}=\frac{-60}{91} \times \frac{-7}{18}=\frac{10}{39}$

$\therefore \frac{5}{7} \times\left(\frac{-12}{13} \times \frac{-7}{18}\right)=\left(\frac{5}{7} \times \frac{-12}{13}\right) \times \frac{-7}{18}$

Hence verified.