(a) Using Proposition 1.3.2 and/or the definition of absolute value, prove that if b ‡ 0, 1 1|2|= |b|` then (b) Prove that for all a E R and b = 0, we have |a| |b| a .

The proposition is given by :  For any two real numbers  x and y  we have 

     $\sqrt{x^2}=\left|x\right| , {\left|x\right|}^2=x^2 , x\le \left|x\right| and \left|x·y\right|=\left|x\right|·\left|y\right|$

We have to prove that ,

$$\begin{gathered} \left(a\right) if b\ne 0, then \left|\frac{1}{b}\right|=\frac{1}{\left|b\right|} \\ \left(b\right) for all a\in \mathrm{ℝ} and b\ne 0 , we have \left|\frac{a}{b}\right|=\frac{\left|a\right|}{\left|b\right|} \end{gathered}$$