Has any one ever tried *this* to define a new version of metric? (division instead of subtraction) Let $x \in \Bbb{R}^+$ be a positive real number.  Define $|x|_{\bullet} = \max(x,\dfrac{1}{x})$.
Then define for $a, b \in \Bbb{R}^+, \ \ d(a,b) := |\dfrac{a}{b}|_{\bullet}$
$d$ satisfies:
$$
(1) \ d(a,b) \geq 1 \\
(2) \ d(a,b) = d(b,a) \\
(3) \ d(a,b) = 1 \iff a = b \\
(4) \ d(a,b) \leq d(a,c) \cdot d(b,c)
$$
Let's take a look at $(4)$.  


*

*Suppose that $a \leq b \leq c$: $d(a,b) = \dfrac{b}{a} \leq \dfrac{c}{b}\cdot \dfrac{c}{a}$ since we're working with positive numbers here and can cancel $a$, and bring $b$ to the upper LHS to give equivalently $b^2 \leq c^2$.  But the RHS equals $d(b,c)\cdot d(a,c)$.

*The condition $a\leq b$ is WLOG for our purposes which isn't necc. easy to see, so give me just this one to make the proof simpler.  Now all we need to check are the cases $a \leq c \leq b$ and $c \leq a \leq b$.

*Suppose that $a \leq c \leq b$: $d(a,b) = \dfrac{b}{a} \leq \dfrac{b}{c}\cdot\dfrac{c}{a} = \dfrac{b}{a}$ the RHS being $d(b,c)\cdot d(a,c)$.

*Suppose that $c \leq a \leq b$:  $d(a,b) = \dfrac{b}{a}\leq \dfrac{a}{c}\cdot \dfrac{b}{c} \implies \dots$ (similar to bullet one).


Thus $d$ satisfies $(1)$ - $(4)$ and is "metric-like" in these properties.  So what now can we do with $d$.  What about the topology generated by the balls $B_{\epsilon}(x) = \{ y \in \Bbb{R}^+ : d(x,y) \lt \epsilon \}$ where $\epsilon \gt 1$?  Specifically I'm interested in $\Bbb{Z}^+$ so define a metric by restricting to this domain, and so on...
Any ideas what could be done with this "metric", or have you seen it before?
 A: Eric Wofsey's comments are very much to the point: your notion of "metric" is just the thing you get when you apply the $\exp$ function to a standard metric. However, there are some things to be said about the specific example you have given.
The metric you have defined (or, more precisely the metric $d(a,b)=\lvert \log a- \log b \rvert$) is an invariant metric on the multiplicative group of reals. This is easy to check: if you multiply $a$ and $b$ by the same number, the distance remains unchanged (because numerator and denominator in your quotient change in the same way).
The general fact there is that whenever you have a topological group whose topology is metrisable, then you can find a left-invariant metric which induces the topology. Those metrics are the ``nice'' ones in such a group, and it is the case here: when you think about positive reals as a group with multiplication, the standard metric inherited from reals (which is invariant under addition) is far less natural than the one you have described.
Moreover, we can also consider the positive reals as a Lie group. In a Lie group, as soon as we fix an inner product in the tangent space at the identity, we can use translation to obtain a so-called metric tensor on the group, which allows us to measure the lengths of (piecewise smooth) curves. If the group is connected, this also gives rise to a metric (where the distance is simply the infimum of lengths of curves connecting the two points). The metric that this process gives in case of positive reals (with the standard inner product in the tangent space at $1$) is just $d(a,b)=\lvert \log a-\log b\rvert$.

Edit:
I feel silly for not thinking about it sooner, but there is also a notion of a valuation. Given a ring $R$ and an ordered abelian group $\Gamma$, a valuation is a function from $R$ to $\Gamma\cup \{\infty\}$ which satisfies certain axioms, and this gives a notion of a distance in the ring -- the oddity here is that an element is close to $0$ if its valuation is very large: zero has valuation $\infty$. A valuation ring is a ring in which all elements have nonnegative valuation.
But there's a twist: sometimes, valuations are written multiplicatively. In this case, we think about $\Gamma$ as a multiplicative group, and instead look at homomorphisms $R\to \Gamma\cup \{0\}$. Under this convention, a valuation ring is a ring in which all elements have valuation $\leq 1$ (notice the inequality), and the "distance" between two elements is given by $v(a-b)=v(a)/v(b)$. The nice thing here is that under this convention, elements are close when the valuation is small.
If $\Gamma$ is actually a subgroup of (positive) reals, we can get from "multiplicative" world to "additive" world by taking each $\gamma$ to $-\log(\gamma)$, and in the other direction by taking $\gamma$ to $\exp(-\gamma)$. Even if $\Gamma$ is not a subgroup of reals, we can still think about the abstract exponential/logarithm maps.
A very specific example of this are the $p$-adic valuations, which (in additive form) take an integer to $n$, where $n$ is the largest natural number such that $p^n$ divides the integer. Thus the elements close to $0$ are those which are divisible by large powers of $p$. It extends to the rational numbers in the obvious way, and a completion with respect to the notion of distance that arises gives us the field of $p$-adic numbers. A multiplicative $p$-adic valuation is just $2^{-n}$ (or $\exp(-n)$, or $p^{-n}$, it doesn't really matter what base for the exponent you choose, as long as it's greater than $1$).
A: This is completely equivalent to the normal notion of a metric.  Specifically, a function $d(a,b)$ satisfies your axioms (1-4) iff the function $\log d(a,b)$ satisfies the usual definition of a metric.  It is also easy to see that such a $d$ generates the same topology as the metric $\log d$ (since $\log$ is monotone and $\log x$ goes to $0$ as $x$ goes to $1$).
In the case of your particular "metric" on $\mathbb{R}^+$, it generates a topology equivalent to the usual topology because the function $\log:\mathbb{R}^+\to\mathbb{R}$ is a homeomorphism for both the usual topology on $\mathbb{R}^+$ and for the topology given by your $d$ (since it is in fact an isometry from the metric $\log d$ to the usual metric on $\mathbb{R}$).
