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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.

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    $\begingroup$ How do you know $\mathbb{Z}\times\mathbb{Z}$ is not dense in $\mathbb{R}$? $\endgroup$
    – Eric Wofsey
    Jul 6, 2019 at 3:54
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    $\begingroup$ Try $(a,b)\mapsto a+b\sqrt2$. $\endgroup$
    – Angina Seng
    Jul 6, 2019 at 3:56
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    $\begingroup$ Your proof is incorrect. While you claim to have derived a contradiction, you did not actually arrive at any contradiction. Note that while $\Bbb Z\times\Bbb Z$ may not be dense within $\Bbb Q\times\Bbb Q$ with the product topology, that doesn't mean its isomorphic copy in $\Bbb R$ isn't dense. $\endgroup$
    – runway44
    Jul 6, 2019 at 4:22
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    $\begingroup$ If you know that $\mathbb{R}$ is an infinite-dimensional vector space over $\mathbb{Q}$ then the answer is trivially yes. $\endgroup$
    – trisct
    Jul 6, 2019 at 4:35
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    $\begingroup$ The question has nothing to do with topology—only the algebraic structures of the groups—so arguments about density are unlikely to be relevant. $\endgroup$
    – Greg Martin
    Jul 6, 2019 at 4:44

2 Answers 2

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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.

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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 $\pi$.

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  • $\begingroup$ Based on a comment by Lord Shark the Unknown. $\endgroup$
    – lhf
    Dec 23, 2019 at 18:50

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