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.