1
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.