Multiplicative Group of Rationals is Subgroup of Reals

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Theorem

Let $\struct {\Q_{\ne 0}, \times}$ be the multiplicative group of rational numbers.

Let $\struct {\R_{\ne 0}, \times}$ be the multiplicative group of real numbers.


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


Proof

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

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

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

$\blacksquare$


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