Strictly Positive Real Numbers under Multiplication form Subgroup of Non-Zero Real Numbers

From ProofWiki
Jump to navigation Jump to search

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

Retrieved from "https://proofwiki.org/w/index.php?title=Strictly_Positive_Real_Numbers_under_Multiplication_form_Subgroup_of_Non-Zero_Real_Numbers&oldid=522156"