showing that $\frac{\mathbb R}{\mathbb Q}\cong\mathbb R$, both over $\mathbb Q$

In my last question The following groups are the same., three groups were given and I wanted to verify that they are isomorphic to each other. Derek suggested some points and I got my answer about one of them completely. But one was remained. I am trying to simplify my question about it again.

There; I had to show that $\frac{\mathbb R}{\mathbb Q}\cong\mathbb R$ since I need it. A theorem tells me that $\frac{\mathbb R}{\mathbb Q}$ is a vector space over $\mathbb Q$ because it is abelian divisible torsion-free and the same is true for $\mathbb R$. So I consider $\mathcal{B}$ be a basis for vector space $\mathbb R$ over $\mathbb Q$.

Now can I have $\mathcal{B}+\mathbb Q$ as a basis for vector space $\frac{\mathbb R}{\mathbb Q}$ over $\mathbb Q$ and concluding that $\#\mathcal{B}=\#(\mathcal{B}+\mathbb Q)$? I wanted to use what Derek pointed there for these two vector space again. Thanks.

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if $\mathcal B$ is a basis for $\mathbb R/ \mathbb Q$, you will get a basis for $\mathbb R$ by picking one representative in each element of the basis (which are equivalence classes), and then adjoin the element $1$ to this family. This will get you a basis for $\mathbb R$ – mercio Nov 16 '12 at 19:39
$\ddot\smile\quad$ – amWhy Apr 17 '13 at 1:02
@amWhy: Thanks Amy for your supports. Up to this time, the line has been remained corrupted. No access, just a little. :-( – Babak S. Apr 17 '13 at 4:46

Let $\cal B$ be a basis of $\Bbb R$ over $\Bbb Q$ (as a vector space, that is). The cardinality of $\cal B$ is $\frak c$. Note that $\Bbb{R/Q}$ is like removing one basis element, or one copy of $\mathbb Q$.

Therefore the kernel of the quotient map $\Bbb{R\to R/Q}$ has dimension $1$, so the image has to have dimension $\mathfrak c-1=\frak c$. Therefore $\Bbb R$ and $\Bbb{R/Q}$ are isomorphic as vector spaces over $\Bbb Q$. But what are isomorphic vector spaces? To begin with, those are isomorphic groups! Which is what you want to begin with.

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Is this true that the base B for R has just one rational number in shared with Q? – Babak S. Nov 16 '12 at 19:41
@BabakSorouh: Every two rational numbers are linearly dependent over $\Bbb Q$ in quite an obvious way. – Asaf Karagila Nov 16 '12 at 19:42
Thanks for your time. – Babak S. Nov 16 '12 at 19:43

Note that $\mathbb{R}/\mathbb{Q}$ and $\mathbb{R}$ are both size continuum.

So as vector spaces over the $\mathbb{Q}$ they must have continuum size bases.

And if two vector spaces have bases of the same size then they are isomorphic, so we're done.

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Direct proof: if $\,\{r_i\}_{i\in I}\,$ is a basis for $\,\Bbb R_\Bbb Q\,$ , then $\,\{r_i+\Bbb Q\}_{i\in I}\,$ is a basis for $\,\left(\Bbb R/\Bbb Q\right)_\Bbb Q\,$ , since:

$$(1)\;\;\;\;\;\;\;\;\;\forall r\in\Bbb R\,\,\exists\,J\subset I\,\,,\,|J|<\infty\,,\,a_j\in\Bbb Q\,\,,\,\,\,s.t.\;\; r=\sum_{j\in J}a_jr_j\Longrightarrow r+\Bbb Q=\sum_{j\in J}a_j\left(r_j+\Bbb Q\right)$$

$$(2)\;\;\;\;\;\;\;\;\;\text{If}\;\;\exists\,J\subset I\,\,,\,|J|<\infty\,\,,\,s.t.\;\;\sum_{j\in J}a_j\left(r_j+\Bbb Q\right)=\overline 0=\Bbb Q\in\Bbb R/\Bbb Q\,\,,\,\text{then}$$

$$\,\,\sum_{j\in J}a_jr_j\in\Bbb Q\Longrightarrow a_j=0\,\,\forall j\in J$$

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$\mathbb Q$ is also in the span of the original basis, so the last implication isn't correct. You could modify this by taking $\{1\}\cup\{r_i\}_{i\in I}$ to be the original basis and leaving the rest the same. – Jonas Meyer Nov 16 '12 at 20:06
Point taken, thanks. – DonAntonio Nov 16 '12 at 20:08