# Solving $|a| < |b|$

I apologize if this question in general, but I've been having trouble finding solutions as Google discards absolute value signs and inequality symbols.

I am looking for a way to eliminate absolute value functions in $|a| < |b|$.

I can solve $|a| < b$ and $|a| > b$, but I am unsure what method / combination of methods to use to eliminate absolute value signs from both sides.

Thank you!

An example problem:

$$|x + 2| < |x - 4|$$

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perhaps you could give some specific problems that trouble you. – Will Jagy Aug 11 '12 at 5:19
Good suggestion. I've added one above. – Mark V. Aug 11 '12 at 5:41
It's useful to review the definition of the absolute value function: $$\text{abs}(a) = \begin{cases} a & 0 \leq a \\ -a & a < 0 \end{cases}$$ In order to eliminate one absolute value function from an expression, you will need to consider two cases: the case where its argument is non-negative, and the case where its argument is negative. This will give you an expression with two cases with one less absolute value function. Iterating this process will give an expression without the absolute value sign. – danportin Aug 11 '12 at 7:13
Since you mentioned trying Google, maybe you should try Wolfram Alpha instead – Ben Millwood Aug 11 '12 at 11:45

We have $|a| \lt |b|\,$ if any of these is true:

(i) $\,a$ and $b$ are $\gt 0$ and $a \lt b$

(ii) $a\lt 0$ and $b\ge 0$ and $-a \lt b$

(iii) $b \lt 0$ and $a \gt 0$ and $a \lt -b\,$

(iv) $\,a\lt 0$ and $b\lt 0$ and $-a\lt -b$. We can rewrite this as $b \lt a$.

Four cases! Not surprising, since eliminating a single absolute value sign often involves breaking up the problem into $2$ cases.

Sometimes, one can exploit the simpler $|a| \lt| b|\,$ iff $\,a^2\lt b^2$. But squaring expressions generally makes them substantially messier.

Added: With your new sample problem, squaring happens to work nicely. We have $|x+2| \lt |x-4|$ iff $(x+2)^2 \lt (x-4)^2$. Expand. We are looking at the inequality

$$x^2+4x+4 \lt x^2-8x+16.$$

The $x^2$ cancel, and after minor algebra we get the equivalent inequality $12x \lt 12$, or equivalently $x\lt 1$. The squaring strategy works well for any inequality of the form $|ax+b| \lt |cx+d|$.

But the best approach for this particular problem is geometric. Draw a number line, with $-2$ and $4$ on it. Our inequality says that we are closer to $-2$ than we are to $4$. The number $1$ is halfway between $-2$ and $4$, so we must be to the left of $1$.

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Ha! I never would of thought of this. This is brilliant! Thank you so much for your help! – Mark V. Aug 11 '12 at 18:10

There are different approaches; one is to look at the zeroes of the expressions inside the absolute values, and split up $\mathbb R$ into intervals accordingly. In your example $|x + 2| < |x - 4|$, the points of interest are at $x=-2$ and $x=4$. You can therefore consider three cases:

1. If $x \in (-\infty,-2)$, then $x+2 < 0$ and $x-4 < 0$, so the absolute values will reverse the signs of both. This gives: \begin{align} -(x+2) &< -(x-4) \\ x+2 &> x-4 \\ 2 &> -4 \end{align} This is true for all $x$ in the interval.

2. If $x \in [-2,4)$, then $x+2 \geq 0$ so its sign is unaffected by the absolute value, but $x-4 <0$ so its sign will be reversed: \begin{align} x+2 &< -(x-4) \\ 2x+2 &< 4 \\ x &< 1 \end{align} Combining this last inequality with the assumption that $x \in [-2,4)$, we see that any $x$ in $[-2,1)$ is valid.

3. Finally, if $x \in [4,\infty)$, neither expression's sign is reversed: \begin{align} x+2 &< x-4 \\ 2 &< -4 \end{align} This is false for all $x$ in the interval.

Putting all the information together from the above three cases, we have $x \in (-\infty, 1)$.

Note: as some other contributors have mentioned, there are simpler ways to deal with your problem, such as viewing it geometrically. The method that I have shown above is more useful when the expressions are more complicated or when you have several absolute values; for example, an inequality like $3 |x^2-1|+|x-2|+|x^2-3x| > 5$.

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Pedantry: you probably want $x\in[-2,4)$ for step 2 and $x\in[4,\infty)$ for step 3 (or change where you say $x-4<0$, but I think my way is more natural/consistent) – Ben Millwood Aug 11 '12 at 11:03
@BenMillwood: Fair enough! I've changed it. – Théophile Aug 11 '12 at 18:13

We can see $|x-a|$ as a distance point $x$ from point $a$.

Now, the question with the above "definition" would be like:

For which $x$ values distance point $x$ from $-2$ be less than distance point $x$ from $4$?

Clearly, by drawing it maybe, you can observe that the answer is for all $x<1$.

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We can divide by $|b|$ to get $|a/b|<1$. Let $x=a/b$ then $|x|<1$ so $-1<x<1$.

Now multiply through by $b$.

If $b>0$ then $-b<a<b$.

If $b<0$ then $-b>a>b$

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