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Prove that as vector spaces over $\mathbb{Q}, \mathbb{R}^n \cong \mathbb{R},$ for all $n \in \mathbb{Z}^+.$

My idea is to create a $\mathbb{Q}$-module homomorphism and show that is bijective but I can't seem to come up with the correct map. Any ideas?

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    $\begingroup$ It's easier to use that every vector space has a basis, and that $\mathbb R$ must thus have a $\mathbb Q$-basis, which must be infinite. Then show that $\mathbb R^n$ must have a basis of the same cardinality. $\endgroup$ – Thomas Andrews Jan 31 '18 at 3:36
  • $\begingroup$ Basically, there isn't any "good" $\mathbb Q$-module homomorphism that you can just define; you can only prove it exists. $\endgroup$ – Thomas Andrews Jan 31 '18 at 3:37
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Consider a basis $(e_i)_{i\in I}$ of the $\mathbb{Q}$ vector space $\mathbb{R}$, you can define the basis of $n$-uple $(e_{i_1},...,e_{i_n})$ of $\mathbb{R}^n$, these basis have the same cardinal and a bijection between them induces an isomorphism between the $\mathbb{Q}$ vector space $\mathbb{R}$ and the $\mathbb{Q}$ vector space $\mathbb{R}^n$.

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  • $\begingroup$ Doesn't this have the same issue as trying to create a isomorphism between $\mathbb{R}^n$ and $\mathbb{R}$? I don't see how there's going to be a bijection between the bases. $\endgroup$ – Smash Jan 31 '18 at 14:21
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It's easy to notice that $\mathbb{R^n}$ must be infinite-dimensional over $\mathbb{Q}$. Otherwise it would have been isomorphic to some $\mathbb{Q^m}$ but that's impossible since this sets have different cardinality.

In this blog post it's proven that infinite-dimensional vector space over a countable field has dimension equal to cardinality of the space.

The criteria that two vector spaces are isomorphic iff their algebraic dimensions are the same holds not only for finite- but also for infinite-dimensional vector spaces. So since $R$ and $R^n$ have the same cardinality then they have the same dimension and then they are isomorphic.

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