the groups of real numbers

Proposition 1.

The additive group of real number $\mathrm{⟨}\mathrm{R}\mathrm{,}\mathrm{+}\mathrm{⟩}$ is isomorphicto the multiplicative group of positive real numbers $\mathrm{⟨}{\mathrm{R}}^{\mathrm{+}}\mathrm{,}\mathrm{\times }\mathrm{⟩}$.

Proof.

Let $f(x)=e^x$. This maps the group $⟨ℝ,+⟩$ to the group$⟨{ℝ}^{+},\times ⟩$. As $f$ has an inverse $f^{-1}(x)=\ln x$we observe $f$ is invertible. Furthermore, $f(x+y)=e^{x+y}=e^x{e}^y=f(x)f(y)$ so$f$ is a homomorphism. Thus $f$ is an isomorphism.∎

Corollary 2.

The multiplicative group of non-zero real number ${\mathrm{R}}^{\mathrm{\times }}$ is isomorphic to ${\mathrm{Z}}_{\mathrm{2}}\mathrm{\oplus }\mathrm{⟨}\mathrm{R}\mathrm{,}\mathrm{+}\mathrm{⟩}$.

Proof.

Use the map $f:{\mathbb{Z}}_2\oplus ⟨ℝ,+⟩\to {ℝ}^{\times }$ defined by $f(s,r)={(-1)}^s{e}^r$. Then

$$f((s_1,r_1)+(s_2,r_2))=f(s_1+s_2,r_1+r_2)={(-1)}^{s_1+s_2}{e}^{r_1+r_2}={(-1)}^{s_1}{e}^{r_1}{(-1)}^{s_2}{e}^{r_2}=f(s_1,r_1)f(s_2,r_2)$$

so that $f$ is a homomorphism. Furthermore, $f^{-1}(r)=(\mathrm{sign}r,\ln |r|)$is the inverse of $f$ so that $f$ is bijective and thus an isomorphism of groups.∎