# 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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Updated on: 10-Oct-2022

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