# Dealing with modulus inequality

How to prove the following?

1)$|f(x)|< g(x) \iff -g(x)< f(x) < g(x)$

2)$|f(x)|> g(x) \iff f(x)<-g(x) \mbox{ or } f(x)>g(x)$

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Seperate for the cases when $f(x) > 0$, $f(x)=0$ and $f(x) < 0$ and note that $|f(x)| \ge 0$.

For example, if $f(x) > 0$ then $f(x) < g(x) \iff |f(x)| < g(x)$. If $f(x) = 0$ then $0 < g(x) \iff -g(x) < 0 \mbox{ and } 0 < g(x)$. If $f(x) < 0$ then $-f(x) > 0$ and then $|f(x)|<g(x) \iff -f(x) < g(x) \iff -g(x) < f(x)$ (follow the algebraic passages I did, such as adding $f(x)$ to both sides and substracting $g(x)$ from both sides).

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To solve |f(x)|>g(x), – newbowl Mar 11 '13 at 13:19
To solve |f(x)|>g(x), when f(x)>0, then |f(x)|>g(x) <=> f(x)>g(x). Why don't we have to take the intersection of f(x)>0 and f(x)>g(x)? Why can we straight away say f(x)>g(x)? – newbowl Mar 11 '13 at 13:29

Recall the definition of the modulus function:

$|f(x)|= $$\begin{cases} &f(x) \space \space\space\space\space f(x)\geq0\\ &-f(x) \space\space\space\space f(x)<0 \end{cases}$$$.

Then, saying, for example, that $|f(x)|<g(x)$ means that $f(x)<g(x)$ and $-f(x)<g(x)$. Multiplying the latter by $(-1)$ we have $f(x)>-g(x)$, and combining the two we have: $-g(x)<f(x)<g(x)$.

For the second case:

$|f(x)|>g(x)$ means that either $f(x)>g(x)$ or $-f(x)>g(x)$ (the latter means $f(x)<-g(x)$).

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To solve |f(x)|>g(x), when f(x)>0, then |f(x)|>g(x) <=> f(x)>g(x). Why don't we have to take the intersection of f(x)>0 and f(x)>g(x)? Why can we straight away say f(x)>g(x)? – newbowl Mar 11 '13 at 13:33
Not sure I understand the question. And, anyway, it is not true that $|f(x)|>g(x)$ iff $f(x)>g(x)$. I'll edit my answer. – Ludolila Mar 11 '13 at 13:39