Can $\mathbb{Q×Q}$ be embedded in $\mathbb{R}$ as group?

Question

I think ans is NO : if possible let that is true hence there is a monomorphism from $H= \mathbb{Q×Q}$ to $\mathbb{R}$. as $\mathbb{R} $ has only subgroups which is cyclic or dense and $H$ is not cyclic hence dense but it's proper subgroup $\mathbb{Z×Z}$ is neither cyclic nor dense in $\mathbb{R}$ hence contradiction. Hence the claim.

Is my proof correct??
Thanks.

Answer 1

The map $(a,b)\mapsto a+b\sqrt2$ is an injection $\mathbb{Q×Q} \to \mathbb{R}$ because $\sqrt2$ is irrational.

$\sqrt2$ is not special here; any irrational number works, for instance $\pi$.

Answer 2

Let $\{e_i: i \in I\}$ be a basis for the vector space of $\Bbb R$ over the field $\Bbb Q$. It is clear from cardinality considerations that $I$ is uncountable.

Pick any two distinct elements from the base, say $e_{i_1}$ and $e_{i_2}$. Map $(q,q') \in \Bbb Q^2$ to $qe_{i_1} + q'e_{i_2} \in \Bbb R$ and note that we have group embedding.

This of course works for any finite power of the rationals.