Can we conclude that if $|x|-|y|=|x-a|-|y-a|$ then $x=y$?

Question

I want to prove a secondary result by induction, but to complete it relies on the fact that

$$|x|-|y|=|x-a|-|y-a| \implies x = y$$

It seems intuitively true to me, since the $-a$ excludes the case $x=-y$ and vise versa. But at the same time I would be surprised if some kind of special case construction could violate this.

Is this result true? And if so, is there an easy way I can show this?

edit:

Forgot to add important details:

1. $a \neq 0$ by construction

2. We can extend this to hold for any $a_1,\dots,a_n$ so long as $n$ is finite.

Answer 1

There are in general 16 cases that may be reduced with the specified $a_1$ and $a_2$.

$$|x - a_1| - |y - a_1| = |x - a_2| - | y - a_2| $$

Here for the right hand side $a_2 = a$ (that can be either positive or negative) is the critical point where we need to split $x$-versus-$a$ and $y$-versus-$a$ so that's $4$ from 2-by-2.

For the left hand side $a_1 = 0$ is a critical point, with also $4$ combo of $x$-versus-$0$ times $y$-versus-$0$.

Given $a_1 = 0$, if you want to exclude $a_2 = a = 0$, that will have no effect (at most this can effect two cases).

Case++++

If $x>0,y>0,x>a,y>a$, then $$x - y = (x-a)-(y-a) \implies \text{trivial identity, we learn nothing}$$ Numerical example, $x = 2,y=1, a = -\pi \implies 1 = (2-\pi)-(1-\pi)$

Case+++-

If $x>0,y>0,x>a,y\color{magenta}<a$, thus necessarily $a$ is positive as well since $a > y >0$ $$x - y = (x-a)-(a-y) \implies -y = -2a + y \implies y = a$$ Numerical example, $x = 4, a = \pi \implies 4-y = (4-\pi)-(\pi-y)$

Case-+++

If $x \color{magenta}\leq 0,y>0,x>a,y>a$, where the last condition is sort of redundant since $y > 0 \leq x > a$ by the previous three. $$-x - y = (x-a)-(y-a) \implies -x-y = x - y \implies x = 0$$ Numerical example $y = 1, a = -\pi \implies -x-1 = (x-\pi)-(1-\pi)$

Shall I go on?

nah, you get the point