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I'm trying to nail down the logic behind absolute value inequalities. Given an inequality like $$\Bigg\lvert\frac{2}{x+4}\Bigg\rvert>2,$$ the normal process is to use cases and the definition of absolute value to determine the solution set.

However, upon thinking about the process a little more I've confused myself. When faced with this problem the goal is to find the values of $x$ that make the relation true. But by using cases I am assuming that $\frac{2}{x+4}$ is nonnegative in one case and negative in another and then I proceed to find the solutions that satisfy each case. Is this logic circular since the sign of $\frac{2}{x+4}$ depends on $x$? It seems like I'm assuming something about $x$ before I find it. Can someone explain the logic here? Thanks for the help.

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  • $\begingroup$ Are you familiar with predicate logic? $\endgroup$ – Ovi Jun 28 '17 at 2:22
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    $\begingroup$ If $S\subset\mathbb{R}$, the assumptions consist in dividing $\mathbb{R}$ in sets $A_1,A_2,A_3,.., A_n$. Those are your cases. Then by finding the solutions inside each case you are computing $S\cap A_1$, $S\cap A_2$, ... etc. Since $\mathbb{R}=\bigcup_i A_i$, then $S=\bigcup_i (S\cap A_i)$. $\endgroup$ – OR. Jun 28 '17 at 2:25
  • $\begingroup$ @Ovi Yes, I am familiar. $\endgroup$ – Newman Jun 28 '17 at 2:30
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There are two ways for the given inequality to be true. Either $\left(\frac{2}{x+4}\geq 0\right) \land \left(\frac{2}{x+4}>2\right)$, or else $\left(\frac{2}{x+4} < 0\right) \land \left(-\frac{2}{x+4}>2\right)$.

That's a total of $4$ inequalities, joined together with "and"s and "or"s. The solution to each one is some subset of the number line, which can be boiled down to something about $x$. Because of the "and"s and "or"s involved, you want the union of two sets, one of which is the intersection of the solutions of the first two inequalities, and the other is the intersection of the solutions of the last two inequalities.

There's no circular reasoning, because we're not assuming anything about $x$ to get solutions. We're just writing down all of the conditions that have to be satisfied, and finding out where that happens.

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$$\left|\frac{2}{x+4}\right|>2$$

is equivalent to $$\left|\frac{1}{x+4}\right|>1$$

We need to avoid $x+4 = 0$, so we must strive to keep $x+4$ strictly positive or strictly negative.

Which leads to the equivalent statements

step 1

$\dfrac{1}{x+4} < -1 \quad \text{or} \quad \dfrac{1}{x+4} > 1$

step 2

$-1 < x+4 <0 \quad \text{or} \quad 0 < x+4 < 1$

step 3

$x \in (-5, -4) \cup (-4, -3)$

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Let the solution set be $S$. The thing is that $x$ is not just any number; you assume that $x$ satisfies $\left|\dfrac {2}{x+4} \right|>2 $, and now you are trying to find other neccesary and sufficient conditions to judge if a number$s$ is in $S$ or not (other than the condition $\left|\dfrac {2}{s+4}\right|>2 )$.

In other words, we can deduce

$s \in S \iff \left(\dfrac {2}{s+4}>2 \right)\vee \left(\dfrac {2}{s+4}<-2 \right) \iff ... \iff x \in (-5, -4) \cup ( -4, -3)$

So therefore $\{x \in \mathbb{R} : \left|\dfrac {2}{x+4} \right|>2 \} = \{ x \in \mathbb{R} : x \in (-5, -4) \cup ( -4, -3) \}$

Hope this helps.

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  • $\begingroup$ What does $|a>b|$ mean? $\endgroup$ – G Tony Jacobs Jun 28 '17 at 2:52
  • $\begingroup$ @GTonyJacobs Haha thank you for pointing that out I made the mistake in the beginning and copy and pasted it everywhere else. $\endgroup$ – Ovi Jun 28 '17 at 2:53
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For solving of your inequality there are two cases:

  1. $$\frac{1}{x+4}>1,$$ which is $$\frac{1}{x+4}-1>0,$$ which is $$\frac{-3-x}{x+4}>0,$$ which is $$-4<x<-3;$$ 2.$$\frac{1}{x+4}<-1,$$ which is $$\frac{x+5}{x+4}<0,$$ which is $$-5<x<-4$$ and we get the answer: $$(-5,-3)\setminus\{-4\}$$

I think the cases $x+4>0$ and $x+4<0$ are not necessary here.

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" Is this logic circular since the sign of depends on ? "

no. Why would that be circular? It's because the value of $|\frac 2 {x+4}|$ does depend on $x $ that doing cases tells us what $x $ has to be to make $|\frac 2 {x+4}|$ positive or negative.

Consider Sam's religion determines his diet and we know he drank milk on friday in 1954 and he either ate ham, fish, chicken or vegetables with it. If he ate ham or chicken he is not Jewish. If he ate didn't eat fish he isn't catholic. whatever he ate he isn't Jain. That's not circular at all.

If $x >-4$ then that forces $\frac 2 {x+4}$ to be positive. So $x >-4$ and $\frac 2 {x+4}>0$ is possible. That's acceptable. If $x <-4$ that forces $\frac 2 {x+4}$ to be negative . So $x <-4$ and $\frac 2 {x+4}<0$ is possible. That's acceptable. But nothing else is as $x=-4$ is impossible.

So we have two options: $x >-4$ and $\frac 2 {x+4}>0$ or $x <-4$ and $\frac 2 {x+4}<0$.

There is nothing circular about that.

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The source task is falling apart in two :

$$\left(\frac{1}{x+4}\ >1\right) \vee \left(\frac{1}{x+4}\ <-1\right) $$

Solve the both and the answer is the union of the result sets. Both are trivial, for you know the sign of x+4 in every one.

$$\left(\frac{1}{x+4}\ >1 \vee x+4>0 \right) \vee \left(\frac{1}{x+4}\ <-1 \vee x+4<0 \right) $$

$$\left(x+4<1 \land x+4>0 \right) \vee \left(x+4 >-1 \land x+4<0 \right) $$

$$ \left(-5<x<-4 \right) \vee \left(-4<x<-3\right) $$

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