Strictly Positive Real Numbers under Multiplication form Subgroup of Non-Zero Real Numbers
Theorem
Let $\R_{>0}$ be the set of strictly positive real numbers, that is:
- $\R_{>0} = \set {x \in \R: x > 0}$
The structure $\struct {\R_{>0}, \times}$ forms a subgroup of $\struct {\R_{\ne 0}, \times}$, where $\R_{\ne 0}$ is the set of real numbers without zero, that is:
- $\R_{\ne 0} = \R \setminus \set 0$
Proof
From Non-Zero Real Numbers under Multiplication form Abelian Group we have that $\struct {\R_{\ne 0}, \times}$ is a group.
We know that $\R_{>0} \ne \O$, as (for example) $1 \in \R_{>0}$.
Now, verify that the conditions for Two-Step Subgroup Test are satisfied:
Closure under $\times$
Let $a, b \in \R_{>0}$.
We take on board the fact that the Real Numbers form Ordered Integral Domain.
Then:
- $a b \in \R_{\ne 0}$
From Positive Elements of Ordered Ring :
- $a \times b > 0$
so $a b \in \R_{>0}$.
$\Box$
Closure under Inverse
Let $a \in \R_{>0}$.
Then $a^{-1} = \dfrac 1 a \in \R_{>0}$.
$\Box$
Hence, by the Two-Step Subgroup Test, $\struct {\R_{>0}, \times}$ is a subgroup of $\struct {\R_{\ne 0}, \times}$.
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 5.2$. Subgroups: Example $91$
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Subgroups