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 $f(s,r)=(-1)^{s}e^{r}$ .¹¹We 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. ∎

Title	the groups of real numbers
Canonical name	TheGroupsOfRealNumbers
Date of creation	2013-03-22 16:08:50
Last modified on	2013-03-22 16:08:50
Owner	Algeboy (12884)
Last modified by	Algeboy (12884)
Numerical id	5
Author	Algeboy (12884)
Entry type	Result
Classification	msc 54C30
Classification	msc 26-00
Classification	msc 12D99
Related topic	ExponentialFunction
Related topic	GroupHomomorphism
Related topic	GroupsInField