Question

Prove (i) If x $\alpha$ y and y $\alpha$ z then x $\alpha$ z (ii) x $\alpha$$\frac{1}{y}$ and y $\alpha$$\frac{1}{z}$ then $x\alpha z$.

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)x\alpha y \\ \therefore x=ky\left(ki\mathrm{s}\mathrm{constant}.\right)...\left(\mathrm{i}\right) \\ And, \\ y\alpha z \\ \therefore y=lz\left(l\mathrm{is}\mathrm{another}\mathrm{constant}.\right)...\left(\mathrm{ii}\right) \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}\left(\mathrm{ii}\right)\mathrm{in}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ x=k\left(lz\right) \\ \Rightarrow x=kl\left(z\right) \\ \Rightarrow x=mz\left(m=kl\mathrm{is}\mathrm{any}\mathrm{constant}.\right) \\ \therefore x\alpha z \end{gathered}$$

$$\begin{gathered} \left(ii\right)x\alpha \frac{1}{y} \\ \therefore xy=u\left(u\mathrm{is}\mathrm{constant}.\right)...\left(\mathrm{i}\right) \\ \mathrm{And}, \\ y\alpha \frac{1}{z} \\ \therefore y=\frac{v}{z}\left(v\mathrm{is}\mathrm{any}\mathrm{constant}.\right)...\left(\mathrm{ii}\right) \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}\left(\mathrm{ii}\right)\mathrm{in}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ x\times \frac{v}{z}=u \\ x=\frac{u}{v}z \\ x=wz\left(w=\frac{u}{v}\mathrm{is}\mathrm{any}\mathrm{constant}.\right) \\ \therefore x\alpha z \end{gathered}$$