Find all $x$ to satisfy $|x-a|<|x-b|$.

Question

Exercise:

Using signs of inequality alone (not using signs of absolute value) specify the values of $x$ which satisfy the following relation. Discuss all cases. $$|x-a|<|x-b|$$

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Solution:

$|x-a|<|x-b| \tag{1}$

Since $RHS \geq 0$ in (1): $-|x-b|<x-a<|x-b| \tag{2}$

Split (2): $-|x-b|<x-a \tag{3}$ $x-a<|x-b| \tag{4}$

Working on (3): $|x-b|<a-x \tag{5}$

For (5) to be valid: $RHS \geq 0 \implies a-x\geq 0 \implies x \leq a \tag{6}$

Since $RHS \geq 0$ in (5): $-(a-x)<x-b<a-x \implies x-a<x-b<a-x \tag{7}$

Split (7): $x-a<x-b \implies -a<-b \implies a>b \tag{8}$

$x-b<a-x \implies x<\frac{a+b}{2} \tag{9}$

Working on (4); due to (6): $LHS \leq 0 \tag{10}$

Due to (10), $0 < |x-b| \implies x \neq b \tag{11}$

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Answer:

$$a > b \tag{a}$$ $$x < \frac{a+b}{2} \tag{b}$$ $$x \neq b \tag{c}$$

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Request:

Is my answer correct? If so, can my answer be any more specific?

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Update:

As per @Bernard's hint, here is my new attempt.

$|x-a|<|x-b|$

$(x-a)^2<(x-b)^2$

$x^2-2ax+a^2<x^2-2bx+b^2$

$-2ax+a^2<-2bx+b^2$

$a^2-b^2<2ax-2bx$

$(a-b)(a+b)<2(a-b)x$

$a+b<2x$

$$\frac{a+b}{2}<x$$

I feel as if I "lost information" when interchanging the absolute values with squares. Is this so?

Answer 1

You can have it in a much simpler way if you eliminate the absolute values.

Hint:

$$\lvert A\rvert <\lvert B\rvert \iff A^2 <B^2.$$

Answer 2

Interpreting the expressions inside the absolute values as distances is very helpful. More precisely, $|x-a|$ is the distance from $x$ to $a$, and $|x-b|$ is the distance from $x$ to $b$. Therefore you are looking for the real numbers $x$ that are closer to $a$ (than to $b$).

For the three possible cases, you get immediatly:

(1) $a<b$: the interval $(-\infty,\frac{a+b} 2)$.

(2) $a=b$: the empty set.

(3) $a>b$: the interval $(\frac{a+b} 2,\infty)$.