Proving distance defined is a metric

Question

$\forall x,y \in \mathbb R$ if distance $d(x,y) = { (x - y)^2}$,

I want to show that ${(x - y)^2} \leq {(x - z)^2} + {(z - y)^2} $. Basically, I am trying to prove that this distance satisfy triangular inequality.

I have established

$ |p \cdot q | \leq |p| \cdot |q| $

$ |p + q | \leq |p| + |q| $

$ |p - q | \leq |p| - |q| $

Problem seems to introduce $z$ in the inequality. Any hints?

I started with $(x-y)^2 = (x+z-z-y)^2$ But this doesn't lead to where I wanted to go.

A related question Euclidean distance proof had confusing answer to me.

Isn't this just absolute value? (upvoted so you can comment) — IAmNoOne, Aug 06 '16 at 22:56
in other terms $\sqrt{(x - y)^2} \leq \sqrt{(x - z)^2} + \sqrt{(z - y)^2}$ can written $|x - y| \leq |x - z|+|z - y|$... — Jean Marie, Aug 06 '16 at 22:58
If $ x<z<y$, RHD $=|x-y|$. Then this is obvious except this. — Takahiro Waki, Aug 06 '16 at 23:03
I am sorry, but wanted to prove that without square root. Corrected typo. Any hints? — C34nm, Aug 06 '16 at 23:07
It is intuitive but I'm seeking hint for proof in introducing another variable in inequality. — C34nm, Aug 06 '16 at 23:25
@Exodd I think you missed the idea of what C34nm was saying. Go ahead and try his/her points. — rnrstopstraffic, Aug 06 '16 at 23:45
@rnrstopstraffic As already stated in the answer, it is not a distance, and I proposed a counterexample — Exodd, Aug 07 '16 at 11:01
@Exodd I got my tags mixed up and meant to point OP back to your counterexample. Sorry about that! — rnrstopstraffic, Aug 07 '16 at 14:33

score 2 · Answer 1 · answered Aug 06 '16 at 23:39

2

This is not a distance. Prove with $$(x,y,z)=(3,5,4)$$ In this case you have $${(x - y)^2} \gt {(x - z)^2} + {(z - y)^2}$$

answered Aug 06 '16 at 23:39

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1 Answers1