Operation that makes the negative real numbers, $\mathbb{R}_{<0}$, a group The real numbers with addition, $\left( \mathbb{R}, + \right)$, and the positive-real numbers with multiplication, $\left( \mathbb{R}_{>0}, \cdot \right)$, both are Abelian groups. For reasons of symmetry, I think that it should be also possible to construct the group of negative-real numbers.
How does the operation
$$
\ast : \mathbb{R}_{<0} \to \mathbb{R}_{<0}
$$
have to look like to make
$$
\left( \mathbb{R}_{<0}, \ast \right)
$$
an Abelian group?
 A: If you must, define $a*b = -(ab)$.
A: The simplest I have in my mind is this:
$$x * y = -xy$$
Actually, if you have any abelian group $(G,+)$ and any bijection $f: \mathbb{R}_{<0} \to G$, then
$$x*y = f^{-1}(f(x)+f(y))$$
endows $\mathbb{R}_{<0}$ with a group structure isomorphic to $G$, and $f$ becomes an isomorphism of groups. In my first case, if you take the map $x \mapsto -x$, you can recognize that $(\Bbb{R}_{< 0}, *)$ is just isomorphic to $(\Bbb{R}_{> 0}, \cdot)$
A: If $G$ and $H$ are sets with equal cardinality, and $(G,+_G)$ is a group, then $H$ can be turned into a group by choosing a bijection $\phi:G\to H$. The idea is to define $+_{H}$ so that
$$
(\forall x,y,z)(x+_Gy=z \iff \phi(x)+_H\phi(y)=\phi(z)) \, , \tag{*}\label{*}
$$
i.e. for any equation $x+_Gy=z$ in $G$, there is a corresponding equation $\phi(x)+_H\phi(y)=\phi(z)$ in $H$ (and vice versa), meaning that $(G,+_G)$ and $(H,+_H)$ behave identically as groups (in more technical language, they are isomorphic).
Thus for any $u,v\in H$, we define
$$
u+_{H}v=\phi\left(\phi^{-1}(u)+_G\phi^{-1}(v)\right)
$$
so that $\eqref{*}$ holds. In the case of $G=\mathbb R$, $H=\mathbb R_{<0}$, we can choose the map $\phi(x)=-e^x$, so
$$
u+_Hv=-\exp(-\log u-\log v)=-\frac{1}{uv} \, .
$$
We also note that $\mathbb R_+$ is also a group under the usual multiplication, and since the map $\phi:\mathbb R_{+}\to\mathbb R_{-}$ given by $\phi(x)=-x$ is a bijection, we can define
$$
u+_{H}v=-(-u\cdot -v)=-uv \, .
$$
