$x \leq y \leq z \iff |x-y|+|y-z|=|x-z|$

Question

I want to prove one statement.

if $x,y,z \in \mathbb R$

$x \leq y \leq z \iff |x-y|+|y-z|=|x-z|$

So with $\Rightarrow$ I don't have a problem. But with another direciton... The solution says: To establish the converse, show that $y<x$ and $y>z$ are impossible. For example,if $y<x \leq z$, it follows from what we have shown and the given relationshipthat$|x−y|= 0$, so that $y=x$, a contradiction. I didn't understand anything. How did we get this contradiction? Why are $y<x$ and $y>z$ impossible, if it's possible. Because if $y<x$ and $y>z$, then $z<y<x$ and we have

$|x-y|+|y-z|=|x-z|$

$x-y+y-z=x-z$

$x-z=x-z$ and it seems ok. Where did I make a mistake or where my missunderstanding? Thank you for help!

Answer 1

Suppose $y<x\leq z$, it follows $$|x-y|-(y-z)=-(x-z) \Leftrightarrow |x-y|+x-y=0 \Leftrightarrow 2|x-y|=0.$$ This is a contradiction to the assumption $y<x$. I think your attempt is also correct since $|x-y|+|y-z|=|x-z|$ is symmetric in x and z.

Answer 2

As noticed the implication doesn't hold and we can find many counter examples.

Let for example $x-y=a>0$ and $y-z=b<0$ then

$$|x-y|+|y-z|= |x-z| \iff a -b= |a+b| \iff ab=0$$