0
$\begingroup$

Suppose I have any three numbers $a,b,$ and $c$. I'm wondering if it's true in general that

$$||a - c| - |b - c || = |a - b|$$

I've been testing this out with numbers and thinking about it in terms of length of lines and it seems to hold true. But I haven't been able to prove it using properties of absolute value.

Edit: Assuming $a$,$b$, and $c$ are all constant

$\endgroup$
  • 1
    $\begingroup$ Try $a=1$, $b=-1$, $c=0$. $\endgroup$ – Bungo Nov 22 '17 at 0:23
  • 2
    $\begingroup$ No, but you can say $||a-c| - |b-c|| \le |a-b|$ This is a variant of the triangle inequality. $\endgroup$ – Doug M Nov 22 '17 at 0:26
  • $\begingroup$ @Bungo what if a,b, and c are restricted to being positive numbers? $\endgroup$ – Patty Nov 22 '17 at 0:30
  • $\begingroup$ For a positive counterexample, take $a=2$, $b=1$, $c=3/2$. $\endgroup$ – Bungo Nov 22 '17 at 0:31
  • $\begingroup$ I see, thanks for the counter examples. $\endgroup$ – Patty Nov 22 '17 at 0:32
1
$\begingroup$

What is true, in fact, is that $||a-c| - |b-c|| \leq |a-b|$. This follows from the fact that $|x| \leq |y| + |x-y|$, so $|x-y| \geq |x|-|y|$, and similarly switch $x$ and $y$ to get $|y-x| \geq|y| - |x|$. Since th LHS on both sides are equal, we get $|y-x| \geq ||y|-|x||$. Put $y = a-c,x=b-c$, then $y-x = a-b$.

The other side would not hold trivially i.e. if we choose inequality in the triangle inequality for example. Taking $x=1,y=-2$ would do, so $c = 0,a=-2,b=1$ would not work, since $|a-c| = 2,|b-c| = 1$ but $|a-b| = 3$, so the inequality is strict.

If $a,b,c$ are positive, then with the same $x=1,y=-2$ we could get various values of $a,b,c$ e.g. $a=1,b=4,c=3$, then $|a-c| = 2,|b-c| = 1$ but $|a-b| = 3$.

$\endgroup$
  • $\begingroup$ Thanks a lot for the explanation. The inequality $||a−c|−|b−c||≤|a−b|$ is just as useful for what I'm trying to do, so I really appreciate the answer. $\endgroup$ – Patty Nov 22 '17 at 0:49
  • $\begingroup$ You are welcome! $\endgroup$ – астон вілла олоф мэллбэрг Nov 22 '17 at 0:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.