$\frac{|x|-|y|}{|x-y|}$

Question

Let $x\neq y$. What values $\frac{|x|-|y|}{|x-y|}$ can take?

I tried: If $|x|>|y|$, it is $1$, if $|x|<|y|$, it is $-1$ and if $|x|=|y|$, it is $0$. Have I missed anything?

Answer 1

Let $y=0$ and $x=1$.

Thus, we get a value $1$.

We'll prove that it's a maximal value.

Indeed, we need to prove that $$\frac{|x|-|y|}{|x-y|}\leq1$$ or $$|x|-|y|\leq|x-y|$$ or $$|x-y|+|y|\geq|x|,$$ which is true by the triangle inequality: $$|x-y|+|y|\geq|x-y+y|=|x|.$$

For $x=0$ and $y=-1$ we get a value $-1$.

We'll prove that it'a minimal value.

Indeed, we need to prove that $$\frac{|x|-|y|}{|x-y|}\geq-1$$ or $$|x|-|y|\geq-|x-y|$$ or $$|x-y|+|x|\geq|y|,$$ which is true by the triangle inequality again: $$|x-y|+|x|=|x-y|+|-x|\geq|x-y-x|=|y|.$$ Id est, $$-1\leq \frac{|x|-|y|}{|x-y|}\leq1.$$ Now, let $x>0$, $k\geq0$ and $y=-kx$.

Thus, $$\frac{|x|-|y|}{|x-y|}=\frac{1-k}{1+k}$$ and since for all $l\in(-1,1]$ there is $k\geq0$ for which $-1<\frac{1-k}{1+k}\leq1$,

we obtain that $\frac{|x|-|y|}{|x-y|}$ can get all value from $[-1,1]$.

Answer 2

We have $|x-y| \geq ||x| - |y||$ so $$ \frac{|x|-|y|}{|x-y|} \in [-1,1] $$ for all $x,y \in \Bbb R$, $x \neq y$.

Moreover, every element $a \in [-1,1]$ can be obtained. You already found $1$, $-1$ and $0$.

For $a \in (0,1)$, setting $x = -\frac{1}{2}(\frac{1}{a} + 1)$ and $y = \frac{1}{2}(\frac{1}{a} - 1)$ gives $$|x-y| = \frac{1}{a}$$ and $$|x| - |y| = 1.$$ For $a \in (-1,0)$ you can take $x = \frac{1}{2}(\frac{1}{|a|} - 1)$ and $y = -\frac{1}{2}(\frac{1}{|a|} + 1)$ to obtain $$ |x-y| = \frac{1}{|a|} $$ and $$|x| - |y| = -1.$$

In any case, you get $\frac{|x|-|y|}{|x-y|} = a$ as desired.