Verify the property: $x \times (y \times z) = (x \times y) \times z$ by taking:
(i) $ x=\frac{-7}{3}, y=\frac{12}{5}, z=\frac{4}{9} $
(ii) $ x=0, y=\frac{-3}{5}, z=\frac{-9}{4} $
(iii) $ x=\frac{1}{2}, y=\frac{5}{-4}, z=\frac{-7}{5} $
(iv) $ x=\frac{5}{7}, y=\frac{-12}{13}, z=\frac{-7}{18} $
To do:
We have to verify $x \times (y \times z) = (x \times y) \times z$.
Solution:
(i) LHS $=x \times (y \times z)$
$=\frac{-7}{3} \times(\frac{12}{5} \times \frac{4}{9})$
$=\frac{-7}{3} \times(\frac{4 \times 4}{5 \times 3})$
$=\frac{-7}{3} \times \frac{16}{15}$
$=\frac{-7 \times 16}{3 \times 15}$
$=\frac{-112}{45}$
RHS $=(x \times y) \times z$
$=(\frac{-7}{3} \times \frac{12}{5}) \times \frac{4}{9}$
$=\frac{-7 \times 4}{1 \times 5} \times \frac{4}{9}$
$=\frac{-28}{5} \times \frac{4}{9}$
$=\frac{-28\times4}{5\times9}$
$=\frac{-112}{45}$
LHS $=$ RHS
Therefore,
$x \times (y \times z) = (x \times y) \times z$.
(ii) LHS $=x \times (y \times z)$
$=0 \times(\frac{-3}{5} \times \frac{-9}{4})$
$=0 \times \frac{(-3) \times(-9)}{5 \times 4}$
$=0 \times \frac{27}{20}$
$=0$
RHS $=(x \times y) \times z$
$=(0 \times \frac{-3}{5}) \times \frac{-9}{4}$
$=0 \times \frac{-9}{4}$
$=0$
LHS $=$ RHS
Therefore,
$x \times (y \times z) = (x \times y) \times z$.
(iii) LHS $=x \times (y \times z)$
$=\frac{1}{2} \times(\frac{5}{-4} \times \frac{-7}{5})$
$=\frac{1}{2} \times(\frac{5 \times(-7)}{-4 \times 5})$
$=\frac{1}{2} \times(\frac{-7}{-4})$
$=\frac{1}{2} \times \frac{7}{4}$
$=\frac{1 \times 7}{2 \times 4}$
$=\frac{7}{8}$
RHS $=(x \times y) \times z$
$=(\frac{1}{2} \times \frac{5}{-4}) \times \frac{-7}{5}$
$=\frac{1 \times 5}{2 \times(-4)} \times \frac{-7}{5}$
$=\frac{5}{-8} \times \frac{-7}{5}$
$=\frac{5\times-7}{-8\times5}$
$=\frac{7}{8}$
LHS $=$ RHS
Therefore,
$x \times (y \times z) = (x \times y) \times z$.
(iv) LHS $=x \times (y \times z)$
$=\frac{5}{7} \times(\frac{-12}{13} \times \frac{-7}{18})$
$=\frac{5}{7} \times(\frac{-12 \times(-7)}{13 \times 18})$
$=\frac{5}{7} \times(\frac{14}{39})$
$=\frac{5}{7} \times \frac{14}{39}$
$=\frac{5 \times 14}{7 \times 39}$
$=\frac{10}{39}$
RHS $=(x \times y) \times z$
$=(\frac{5}{7} \times \frac{-12}{13}) \times \frac{-7}{18}$
$=\frac{5 \times -12}{7 \times13} \times \frac{-7}{18}$
$=\frac{-60}{91} \times \frac{-7}{18}$
$=\frac{-60\times(-7)}{91\times18}$
$=\frac{70}{273}$
$=\frac{10}{39}$
LHS $=$ RHS
Therefore,
$x \times (y \times z) = (x \times y) \times z$.
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