# Structure of $(\mathbb{R}, +)$ as an abelian divisible group

It’s clear that $$(\mathbb{R}, +)$$ is a divisible group. By the structure theorem of abelian divisible groups, it follows that it’s isomorphic to a direct sum of copies of $$\mathbb{Q}$$, i.e., $$\mathbb{R} \cong \mathbb{Q}^{(\kappa)}$$, since $$\mathbb{R}$$ is torsion-free (all of its elements except the identity $$0$$ are of infinite order), so there are no Prüfer $$p$$-groups in its decomposition. My question is about the value of $$\kappa$$. I assume it can’t be $$\aleph_{0}$$, since the elements of a countable direct sum of copies of $$\mathbb{Q}$$ have finite support, so $$|\mathbb{Q}^{(\aleph_{0})}| = \aleph_{0}$$. Therefore, I guess that $$\kappa = 2^{\aleph_{0}}$$, but just because of cardinality reasons, so I’m not sure about it. Another speculation is that this could be a pure algebraic construction of the “real numbers”, but just seen as an additive group, so I don’t now if it could inherit the order of the reals or the product in an algebraic natural way.

• Are you asking for the cardinality of a Hamel basis? The cardinality of a Hamel basis is the continuum. Commented Sep 4 at 23:39
• @Lucenaposition I guess, so then $\kappa = 2^{\aleph_{0}}$ Commented Sep 5 at 0:02
• Yes, and that's true ($\kappa = 2^{\aleph_0}$) whether or not $\mathsf {CH}$ holds. You are in fact looking for the cardinality of a (Hamel) basis for $\Bbb R$ considered as a vector space over $\Bbb Q$. Commented Sep 5 at 0:50
• @RobertShore I see. Thank you. Commented Sep 5 at 1:07