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I'm doing this exercise from Robert Bartle's Introduction to Analysis, it's a if only if excersise and I've done the half part, but I can't figure this part of the proof: $x,y,z \in \Re$ and $|x-y| + |y-z| = |x-z|$ then $x \le y \le z$

Thanks for any hint.

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    The equality holds iff $x -y$ and $y-z$ have the same sign (or are zero), so a correct statement would have "$x \leq y \leq z$ or $x \geq y \geq z$" instead of $x \leq y \leq z$.2011-09-23

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I think the best way is just to work it out with brute force. Then, having $|x-y| + |y-z| = |x-z|$,assume $y$ is not between $z$ and $x$; without loss of generality (because the other cases will be proved analogously), we suppose $y < x$ and $y \leq z$. Also, without loss of generality, we assume $x \leq z$ (as already pointed out in the comments above, we can just prove that $y$ is between $x$ and $z$, without assumption on who's greater between them). Then, by the implication you've already proved, since $y < x \leq z$ we have $|y-z|=|y-x|+|x-z|$; substituting in the original identity $|x-y| + |y-z| = |x-z| \iff |x-y| + |y-x|+|x-z| = |x-z| \iff 2|x-y|=0$ contradiction.

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    WLOG you may assume that $y=0$ since distances are preserved by translation. It makes the reasoning easier.2011-09-23