# Proving an absolute value inequality.

I just started studying the first lessons of Analysis for my first year of college and I met this inequality in the absolute value lesson, but i'm unable to prove it, need some help..

Question is :

For $a\in R$ $and$ $a \neq 0$ and $x \in R$  we have $\lvert x-a\rvert$ $\lt$ $\lvert a\rvert$, show that $x$ $\neq 0$ and that $x$ and $a$ have the same sign $(-)$ or $(+)$.

For the first one I simply did : if $x = 0$, we get $\lvert -a\rvert$ $<$ $\lvert a\rvert$ which is not true so $x$ $\neq$ $0$.

Second one i'm kind of confused, i tried $x=6$ , $a=2$ and we get $\lvert 4\rvert$ $<$ $\lvert 2\rvert$, which is not true either.. so what am i not understanding here?.

Thanks

The question is: if(!) $|x-a|<|a|$, then $x$ and $a$ have the same sign.
We have $|x-a|<a \iff -|a|<x-a<|a|$.
Case 1: $a>0$. Then we have $0<x<2a$ and we are done.
Case 2: $a<0$. Then we have $2a<x<0$ and we are done.