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Homethe groups of real numbers

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# the groups of real numbers

###### Proposition 1.

The additive group of real number $\langle\mathbb{R},+\rangle$ is isomorphic to the multiplicative group of positive real numbers $\langle\mathbb{R}^{+},\times\rangle$.

###### Proof.

Let $f(x)=e^{x}$. This maps the group $\langle\mathbb{R},+\rangle$ to the group $\langle\mathbb{R}^{+},\times\rangle$. 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 $\mathbb{R}^{\times}$ is isomorphic to $\mathbb{Z}_{2}\oplus\langle\mathbb{R},+\rangle$.

###### Proof.

Use the map $f:\mathbb{Z}_{2}\oplus\langle\mathbb{R},+\rangle\rightarrow\mathbb{R}^{\times}$ defined by $f(s,r)=(-1)^{s}e^{r}$.^{1}^{1}We write $(-1)^{s}$ to mean $(-1)^{{s^{{\prime}}}}$ for any integer $s^{{\prime}}$ representative of the equivalence class of $s$ in
$\mathbb{Z}_{2}$. 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, $f^{{-1}}(r)=(\operatorname{sign}r,\ln|r|)$ is the inverse of $f$ so that $f$ is bijective and thus an isomorphism of groups. ∎

## Mathematics Subject Classification

54C30*no label found*26-00

*no label found*12D99

*no label found*

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