Metric defined by $d(x,y)=\frac{||x|-|y||}{1+|x||y|}$

Question

Q. Let $d :\mathbb R \times \mathbb R →[0,\infty)$ be defined by $$d(x,y)=\frac{||x|-|y||}{1+|x||y|}$$ then is it a metric on $\mathbb R $?

Positive definitness is clear but i couldn't clear the triangular inequality, can we do with $d(x,y)=\frac{||x|-|y||}{1+|x||y|}\leq |x-y|$?

As José Carlos Santos pointed out in his answer, identity of indiscernibles fails to hold. Perhaps you meant $$d(x,y) = \frac{|x-y|}{1+|x||y|}$$ instead? I suspect this would be a metric. — Luke Collins, Jun 08 '19 at 15:55

score 4 · Accepted Answer · answered Jun 08 '19 at 15:52

4

No, because $d(1,-1)=0$, in spite of the fact that $1\neq-1$.

answered Jun 08 '19 at 15:52

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