Multiplicative Group of Rationals is Normal Subgroup of Reals

Theorem
Let $\left({\Q_{\ne 0}, \times}\right)$ be the multiplicative group of rational numbers.

Let $\left({\R_{\ne 0}, \times}\right)$ be the multiplicative group of real numbers.

Then $\left({\Q_{\ne 0}, \times}\right)$ is a normal subgroup of $\left({\R_{\ne 0}, \times}\right)$.

Proof
From the definition of real numbers, it is clear that $\Q_{\ne 0}$ is a subset of $\R_{\ne 0}$.

As $\left({\R_{\ne 0}, \times}\right)$ is a group, and $\left({\Q_{\ne 0}, \times}\right)$ is a group, it follows from the definition of subgroup that $\left({\Q_{\ne 0}, \times}\right)$ is a subgroup of $\left({\R_{\ne 0}, \times}\right)$.

As $\left({\R_{\ne 0}, \times}\right)$ is abelian, it follows from Subgroup of Abelian Group is Normal that $\left({\Q_{\ne 0}, \times}\right)$ is normal in $\left({\R_{\ne 0}, \times}\right)$.