# Solve absolute value inequality

I have to show the inequality

$$\left|\frac{1}{2 + a}\right| < 1.$$

How do I do this?

I know that a fraction is less than 1 when the denominator is greater than the numerator, but I cannot just check if $2 + a > 1$ because of the absolute value sign.

# Edit

If I use

$$\left|\frac{1}{2 + a}\right| < 1 \Leftrightarrow -1 < \frac{1}{2 + a} < 1.$$

I have to look at the inequalities separately, i.e. $\frac{1}{2+a} > -1 \Leftrightarrow 1 > -2 - a \Leftrightarrow a > -3$ and $\frac{1}{2+a} < 1 \Leftrightarrow 1 < 2+a \Leftrightarrow a > -1$.

Since $a > -1 > -3$, $a$ must just be greater than $-1$. But what about $a = -2$ which yields a zero in the denominator?

• Hint: $$|x|<1\iff -1<x<1$$ – Did Oct 2 '15 at 6:53
• If $a$ is a real number, the inequality might or might not be true, depending on exactly which real number $a$ is. Do you have any further information about the value of $a$? – David K Oct 2 '15 at 7:01
• I have to show for which values $a \in \mathbb{R}$ the inequality is true. – Jamgreen Oct 2 '15 at 7:04
• I have edited my question – Jamgreen Oct 2 '15 at 7:08
• Hint: Take two cases : . $|2+a|<1$ and $|2+a| \geq 1$ . – Nizar Oct 2 '15 at 7:10

Both sides are positive, so you can take their reciprocals (of course the 'less than' flips to 'greater than'): $$\left|\frac 1{2+a}\right| < 1 \iff \frac 1{\left|\frac 1{2+a}\right|} = \frac {|2+a|}{|1|}= |2+a| > 1$$ That is equivalent to an alternative: $$(2+a) < -1 \lor (2+a) > 1$$ which resolves to: $$a < -3 \lor a > -1$$ Equivalently $$a\in (-\infty, -3)\cup (-1,\infty)$$

EDIT in reply to the comment

No, $1/(2+a)>−1$ does not imply $a>−3$.

When you multiply both sides by $(2+a)$ you must consider the sign of the multiplicand term. If the term is negative, the direction of an inequality gets reversed. So you have two possible cases here:

$$\color{red}{1/(2+a) > -1} \quad |\,\times(2+a)$$ $$\begin{cases}1 > -1\times(2+a) & \text{ if}\ (2+a) > 0 \\ \qquad \text{or} \\ 1 < -1\times(2+a) & \text{ if}\ (2+a) < 0 \end{cases}$$ This is equivalent to $$1 > -2-a \ \text{and}\ 2+a > 0 \ \text{or} \ 1 < -2-a\ \text{and}\ 2+a < 0$$ $$a > -3 \ \text{and}\ a > -2 \ \text{or} \ a < -3\ \text{and}\ a <-2$$ Finally $$\color{red}{a > -2 \ \text{or} \ a < -3}$$

Similary from the other inequality we get

$$\color{green}{1/(2+a)<1} \quad |\,\times(2+a)$$ $$\begin{cases}1 < 1\times(2+a) & \text{ if}\ (2+a) > 0 \\ \qquad \text{or} \\ 1 > 1\times(2+a) & \text{ if}\ (2+a) < 0 \end{cases}$$ $$1 < 2+a \ \text{and}\ 2+a > 0 \ \text{or} \ 1 > 2+a\ \text{and}\ 2+a < 0$$ $$a > -1 \ \text{and}\ a > -2 \ \text{or} \ a < -1\ \text{and}\ a <-2$$ $$\color{green}{a > -1 \ \text{or} \ a < -2}$$

Together they make $$(\color{red}{a > -2 \ \text{or} \ a < -3}) \ \text{and} \ (\color{green}{a > -1 \ \text{or} \ a < -2})$$ so $$a < -3 \ \text{or} \ a > -1$$

• Your result makes sense!, but If $|x| < 1 \Leftrightarrow -1 < x < 1$, then I also have $|1/(2+a)| < 1 \Leftrightarrow -1 < 1/(2+a) < 1$ which yields two inequalities: $1/(2+a) > -1$ and $1/(2+a) < 1$, and from the first inequality I get $a > -3$ instead of $a < -3$. How come? – Jamgreen Oct 2 '15 at 14:55
• But that is not true... If you say $a>-3$ then let's verify $a=-3+0.001=-2.999$. We then get $2+a=-0.999$, next $|2+a|=0.999$ and finally $\left|\frac 1{2+a}\right|= 1.\overline{001}$ which is greater than $1$. That contradicts the given inequality. – CiaPan Oct 2 '15 at 19:06
• @Jamgreen See the edited answer. – CiaPan Oct 2 '15 at 22:04