Do these define metrics on $\mathbb{R}$?

Question

Which of the following define a metric on $\mathbb{R}$?

1. $$d_1(x,y) = \frac{\bigl||x|-|y|\bigr|} {1+|x||y|},$$

2. $$d_2(x,y) = \frac{\bigl||x|+|y|\bigr|} {1+|x||y|}.$$

I think that both option 1 and option 2 are true as both satisfy the triangle inequality and symmetry property. Am I right?

Answer 1

None of them is a metric:

1. $d(1,-1)=0$, but $1\neq-1$;
2. $d(1,1)\neq0$.

Answer 2

We'll try to prove that $d_1$ is a metric.

Let $|x|=\tan\alpha$, $|y|=\tan\beta$ and $|z|=\tan\gamma$, where $\{\alpha\beta,\gamma\}\subset\left[0,\frac{\pi}{2}\right)$ and $\alpha\geq\beta\geq \gamma$.

Thus, $$\tan(\alpha-\gamma)=\max\{\tan(\alpha-\gamma),\tan(\beta-\gamma),\tan(\alpha-\beta)\}$$ and it's enough to prove that $$\tan(\alpha-\beta)+\tan(\beta-\gamma)\geq\tan(\alpha-\gamma)$$ or $$\frac{\sin(\alpha-\gamma)}{\cos(\alpha-\beta)\cos(\beta-\gamma)}\geq\frac{\sin(\alpha-\gamma)}{\cos(\alpha-\gamma)}$$ or $$\cos(\alpha-\beta+\beta-\gamma)\geq\cos(\alpha-\beta)\cos(\beta-\gamma)$$ or $$-\sin(\alpha-\beta)\sin(\beta-\gamma)\geq0,$$ which is wrong.

Answer 3

The first one $$d_1(x,y) = \frac{||x|-|y||} {1+|x||y|}$$is a metric. The second one$$ d_2(x,y) = \frac{||x|+|y||} {1+|x||y|}$$is not a metric because $ d(x,x)=0$ fails to be true for all x in $R$ for example d(1,1) = 1.