Cardinality of rational exponentiation orbit space

Question

Let $X=(0,\infty)$ be the set of positive real numbers. Let $G=\mathbb{Q}\backslash\{0\}$ be the multiplicative group of rational numbers. $G$ acts freely on $X$ by exponentiation: $r\cdot x=x^r$ for $x\in X$ and $r\in G$. The orbits are clearly countable, but what about the orbit space $X/G$? Is it true that the cardinality $|X/G|$ is the cardinality of the continuum, i.e. $|X/G|=|X|$?

Answer 1

Assuming the axiom of choice, every partition of a set of size continuum into sets of size $\aleph_0$ must have continuum many parts.

To see this, note that if $\{P_i\mid i\in I\}$ is such a partition then $2^{\aleph_0}=|I|\cdot\sup_{i\in I}|P_i|$. In this case, $|P_i|\leq\aleph_0$ for all $i$, so it has to be the case that $|I|=2^{\aleph_0}$.