Question

Verify the property: $x\times (y\times z)=(x\times y)\times zbytaking:$

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

Answer 1

(i)$x\times (y\times z)=(x\times y)\times zx=\frac{-7}{3},y=\frac{12}{5},z=\frac{4}{9}L.H.S.=x\times (y\times z)=\frac{-7}{3}\times (\frac{12}{5}\times \frac{4}{9})=\frac{-7}{3}\times \left(\frac{4\times 4}{5\times 3}\right)=\frac{-7}{3}\times \frac{16}{15}=\frac{-7\times 16}{3\times 15}=\frac{-112}{45}R.H.S.=(x\times y)\times z=(\frac{-7}{3}\times \frac{12}{5})\frac{4}{9}=\frac{-7\times 4}{1\times 5}\times \frac{4}{9}=\frac{-28}{5}\times \frac{4}{9}=\frac{-28\times 4}{5\times 9}=\frac{-112}{45}\therefore L.H.S.=R.H.S.$

(ii) $x\times (y\times z)=(x\times y)\times zx=0,y=\frac{-3}{5},z=\frac{-9}{4}L.H.S.=x\times (y\times z)=0\times (\frac{-3}{5}\times \frac{-9}{4})=0\times \frac{(-3)\times (-9)}{5\times 4}=0\times \frac{27}{20}=0R.H.S.=(x\times y)\times z=(0\times \frac{-3}{5})\times \frac{-9}{4}=0\times \frac{-9}{4}=0\therefore L.H.S.=R.H.S.$

(iii) $x\times (y\times z)=(x\times y)\times zx=\frac{1}{2},y=\frac{5}{-4},z=\frac{-7}{5}L.H.S.=x\times (y\times z)=\frac{1}{2}\times \left(\frac{5}{-4}\frac{-7}{5}\right)=\frac{1}{2}\left(\frac{5\times (-7)}{-4\times 5}\right)=\frac{1}{2}\times \left(\frac{-7}{4}\right)=\frac{1}{2}\times \frac{7}{4}=\frac{1\times 7}{2\times 4}=\frac{7}{8}R.H.S.=(x\times y)\times z=(\frac{1}{2}\times \frac{5}{-4})\times \frac{-7}{5}=\frac{1\times 5}{2\times (-4)}\times \frac{-7}{5}=\frac{5}{-8}\times \frac{-7}{5}=\frac{5\times (-7)}{-8\times 5}=\frac{1\times (-7)}{-8\times 1}=\frac{-7}{-8}=\frac{7}{8}\therefore L.H.S.=R.H.S.$

(iv)$x\times (y\times z)=(x\times y)\times zx=\frac{5}{7},y=\frac{-12}{13},z=\frac{-7}{18}L.H.S.=x\times (y\times z)=\frac{5}{7}\times (\frac{-12}{13}\times \frac{-7}{18})=\frac{5}{7}\times \left(\frac{-12\times (-7)}{13\times 18}\right)=\frac{5}{7}\times \frac{-2\times (-7)}{13\times 3}=\frac{5}{7}\times \frac{14}{39}=\frac{5\times 14}{7\times 39}=\frac{5\times 2}{1\times 39}=\frac{10}{39}R.H.S.=(x\times y)\times z=(\frac{5}{7}\times \frac{-12}{13})\times \frac{-7}{18}=\frac{5\times (-12)}{7\times 13}\times \frac{-7}{18}=\frac{-60}{91}\times \frac{-7}{18}=\frac{-60\times (-7)}{91\times 18}=\frac{-10\times (-1)}{13\times 3}=\frac{10}{39}\therefore L.H.S.=R.H.S.$