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 $a b \in \R_{>0}$.

Closure under Inverse
Let $a \in \R_{>0}$.

Then $a^{-1} = \dfrac 1 a \in \R_{>0}$.

Hence, by the Two-Step Subgroup Test, $\struct {\R_{>0}, \times}$ is a subgroup of $\struct {\R_{\ne 0}, \times}$.