Mathematical Logic: Having trouble understanding this inference So here I wanted for fun to infer that: $$(\lvert a\rvert<\varepsilon) \rightarrow (a > -\varepsilon)
\land (a < \varepsilon)$$
In other words, if it is true that the absolute value of $a$ is less than $\varepsilon$, then $a$ is in between $-\varepsilon$ and $\varepsilon$.
So I started by the definition of $\lvert a \rvert$. Which says that:
$$if\; a > 0 \; then \; \lvert a\rvert=a$$
$$if\; a < 0 \; then \; \lvert a\rvert=-a$$
Which I know them to be true (by definition). So then I do the following logical steps:
$$suppose\;\lvert a\rvert<\varepsilon$$
$$a>0 \rightarrow \;a<\varepsilon$$
$$a<0 \rightarrow \;-a<\varepsilon$$
from here I am literally stuck. I do not know how to infer/conclude that $a > -\varepsilon 
\land a < \varepsilon $.
Where I get confused is in here: If I suppose that $a$ is greater than $0$, then I get that $a<\varepsilon$. If I also suppose that $a$ is less than $0$, then I get $-a<\varepsilon$ or $a>-\varepsilon$.
But how can $a$ be both positive and negative? In other words how can this statement: $a>0\,\land\;a<0$ be true?
I want them to be both true so that through modus ponens I infer $a > -\varepsilon 
\land a < \varepsilon $.
So long story short, I'd like to see this proof done by only involving logical steps.
 A: If $a < b$ and $r > 0$, then $ar < br.$
If $a < b$ and $r < 0$, then $ar > br$.
In other words, the rule is that when you multiply the inequality by a negative number, you must reverse the direction of the inequality.
So, $-a < \epsilon \implies (a) = [(-a) \times (-1)] > [\epsilon \times (-1)] = -\epsilon.$
Therefore, under the premises that

*

*$a < 0$


*$|a| < \epsilon$,
you can conclude that $a > -\epsilon.$  Combining this with the premise that $a < 0$, you have that then $-\epsilon < a < 0.$
A: HINT
As long as $\varepsilon \geq 0$, we can square both sides of the proposed inequality which holds an equivalent transformation. Having said that, we can reach the desired result:
\begin{align*}
|a| < \varepsilon & \Longleftrightarrow |a|^{2} \leq \varepsilon^{2}\\\\
& \Longleftrightarrow |a|^{2} - \varepsilon^{2} \leq 0\\\\
& \Longleftrightarrow a^{2} - \varepsilon^{2} \leq 0\\\\
& \Longleftrightarrow (a - \varepsilon)(a + \varepsilon)\leq 0
\end{align*}
The last inequality is equivalent to the following collection of systems of inequalities:
\begin{align*}
\begin{cases}
a - \varepsilon \geq 0\\\\
a + \varepsilon \leq 0
\end{cases}\quad\vee\quad
\begin{cases}
a - \varepsilon \leq 0\\\\
a + \varepsilon \geq 0
\end{cases}
\end{align*}
Can you take it from here?
A: As you've observed, if $a > 0$, $a < \epsilon$ and if $a < 0$, $a > -\epsilon$. Now $\epsilon > 0$ (you haven't said this explicitly, but it's implicitly stated in the inequality $|a| < \epsilon$), so the following statements are also true:

*

*If $a < 0$, $a < \epsilon$


*If $a > 0$, $a > -\epsilon$


*If $a = 0$, $-\epsilon < a < \epsilon$
So, regardless of whether $a$ is positive, negative, or $0$, $-\epsilon < a < \epsilon$.
