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?.



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.

  • $\begingroup$ oh okay, so i simply misunderstood the whole question, o' well. thank you for your answer, have a good day! $\endgroup$ – Mario SOUPER Aug 16 '17 at 12:31

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