Is it safe to use $|f(x)|

Question

Some peoples often use the following.

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

The weird part occurs when $g(x)<0$, for example:

$$10<f(x)<-10\to |f(x)|<-10$$

where $x\in\{\}$ for both sides of implication which is true.

Is it safe to use $|f(x)|<g(x)\iff -g(x)<f(x)<g(x)$?

Yeah, this looks ok to me. But using $\leq$ instead of $<$ would make this a little sounder. — K.defaoite, Sep 19 '21 at 10:02

score 3 · Accepted Answer · answered Sep 19 '21 at 10:05

3

It is correct. If you have $|f(x)|<g(x)$, then $g$ must be positive everywhere since $0 \le |f(x)|$. Conversely, if $-g(x)<f(x)<g(x)$, the again $g$ must be positive everywhere because we have $-g(x)<g(x)$ which is impossible if $g(x) \le 0$.

answered Sep 19 '21 at 10:05

Paul Frost

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For the sake of completeness, how about the following, is it still safe? $$|f(x)|>g(x) \iff f(x)<-g(x) \quad \text{or}\quad f(x)>g(x)$$ – Display Name Sep 19 '21 at 10:10
Yes, this is also correct. – Paul Frost Sep 19 '21 at 10:12

score 2 · Answer 2 · answered Sep 20 '21 at 09:16

For each $f(x)$ and negative $g(x),$ $$|f(x)|<g(x)$$ and $$-g(x)<f(x)<g(x)$$ are both false. Hence, by deductive explosion, $$\forall f \;\;\forall g<0 \bigg(|f(x)|<g(x)\; \iff \; -g(x)<f(x)<g(x)\bigg).$$

Also for each $f(x)$ and negative $g(x),$ $$|f(x)|> g(x)$$ and $$f(x)<-g(x) \;\text{ or }\; f(x)> g(x)$$ are both true. Hence, since a true consequent automatically means a true implication, $$\forall f \;\;\forall g<0 \bigg(|f(x)|> g(x)\;\iff \; f(x)<-g(x) \;\text{ or }\; f(x)> g(x) \bigg).$$

Is it safe to use $|f(x)|

2 Answers2