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Earlier, I posted this question, and received great answers. Now, I'm attempting another, similar problem.


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


Attempt:

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

RE (1): $RHS \geq 0$ to be defined; so: $x \geq b \tag{2}$

RE (1): $RHS \geq 0$ to be defined; so: $|x-a|<|x-b| \tag{3}$

RE (3): According to previous exercise (see link above):

$x > \frac{a+b}{2} \space\text{when}\space a>b \tag{4}$

$x < \frac{a+b}{2} \space\text{when}\space a<b \tag{5}$

$x \space\text{undefined when}\space a=b \tag{6}$

RE (2), (4) & (6): $x \geq b > \frac{a+b}{2} \space\text{when}\space a>b \tag{7}$

RE (7): Simplified: $x > b \space\text{when}\space a>b \tag{8}$


Answer:

RE (5), (6), & (8):

$$x < \frac{a+b}{2} \space\text{when}\space a<b$$ $$x \space\text{undefined when}\space a=b$$ $$x > b \space\text{when}\space a>b$$


Request:

Is my answer correct? If not, where'd I go wrong?

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  • $\begingroup$ @Macavity -- I'm not quite understanding you. Don't I already cover all relationships of $a$ and $b$ ($=$, $>$, $<$)? $\endgroup$
    – Fine Man
    Dec 13 '16 at 5:58
  • $\begingroup$ @mathlove -- Oh, I see. But, where'd I go wrong? $\endgroup$
    – Fine Man
    Dec 13 '16 at 6:04
  • $\begingroup$ Statement (7) is where you went wrong. When $a> b$, $\frac{a+b}2 > b$. $\endgroup$
    – Macavity
    Dec 13 '16 at 6:06
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    $\begingroup$ @SirJony Note that $-(x-b) < x-a < x-b$ in turn is equivalent to BOTH of the following inequalities being true: $-(x-b) < x-a$ and $x - a < x - b$. The second of these in particular shows that there can't be a solution if $a \leq b$. $\endgroup$
    – user169852
    Dec 13 '16 at 6:13
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    $\begingroup$ @Bungo -- OK, I'll go do that. On a note to all: Thanks to all three answer! They were all helpful, using unique methods. However, I've decided to check-mark Macavity's because his was the closest to my attempt, correcting only where I had gone wrong. $\endgroup$
    – Fine Man
    Dec 13 '16 at 22:02
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Expanding my comment into an answer:

We can eliminate the absolute value as follows. Observe that for any real numbers $c$ and $d$, the inequality $|c| < d$ means precisely that both $c < d$ and $-c < d$, or equivalently, that $$-d < c < d$$ Applying this to your problem, with $c = x-a$ and $d = x-b$, we see that $|x-a| < x-b$ is equivalent to $$-(x-b) < x-a < x-b$$ This in turn is equivalent to both of the following inequalities being true: $$-(x-b) < x-a$$ and $$x-a < x-b$$ We can simplify the first inequality as follows: $$-(x-b) < x-a \quad \iff \quad x > (a+b)/2$$ and the second inequality as: $$x-a < x-b \quad \iff \quad a > b$$ This means that there is no solution if $a \leq b$, whereas if $a > b$, any $x$ satisfying $x > (a+b)/2$ is a solution.

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As there were a lot of steps, showing the right expression in green below, and showing unnecessary or wrong steps in red:

Your Attempt:

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

RE (1): $\color{green}{RHS > 0}$ to be defined; so: $\color{green}{x > b} \tag{2}$

RE (1): $RHS \geq 0$ to be defined $\color{red}{\text{unnecessary step}}$; so: $|x-a|<|x-b| \tag{3}$

RE (3): According to previous exercise (see link above):

$x > \frac{a+b}{2} \space\text{when}\space a>b \tag{4}$

$x < \frac{a+b}{2} \space\text{when}\space a<b \tag{5}$ $\color{green}{\text{But this (5) cannot give solutions as }\frac{a+b}2 < b \text{ in this case..}}$

$x \space \color{red}{\text{undefined when}}\color{green}{\text{ has no solutions when}}\space a=b \tag{6}$

RE (2), (4) & (6): $x \color{red}{\geq b} > \frac{a+b}{2} \color{green}{> b} \space\text{when}\space a>b \tag{7}$ $\color{green}{\text{Note }\frac{a+b}2 > b \text{ in this case..}}$

RE (7): Simplified: $\color{red}{x > b}\color{green}{ \quad x > \frac{a+b}2} \space\text{when}\space a>b \tag{8}$


Hope that helps!

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    $\begingroup$ Oh! My only was that I (stupidly) claimed $b >\frac{a+b}{2}$ when $a > b$. I think you guys overcomplicated things in the comments under the question. :) Thanks, though! $\endgroup$
    – Fine Man
    Dec 13 '16 at 21:55
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I would always recommend drawing a picture first:

  • Case 1: $a>b$

enter image description here You can see that $|x-a|<x-b$ for values of $x$ to the right of the intersection of the lines $y=x-b$ and $y=a-x$.

Equating the two gives: $x-b=a-x \Rightarrow 2x=a+b \Rightarrow x=\frac {a+b}2$.

So $|x-a|<x-b$ for $x>\frac {a+b}2$

  • Case 2: $a=b$

enter image description here You can see that there are no instances where $|x-a|<x-b$

  • Case 3: $a<b$

enter image description here You can see that there are no instances where $|x-a|<x-b$

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