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I wonder if the set of real numbers $\mathbb{R}$ could be a vector space over $\mathbb{R}$ with dimension 2. Which then would be the proper "sum" and "product"? Thanks.

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    $\begingroup$ Since $\Bbb R^2$ has the same cardinality as $\Bbb R$, pick a bijection and use it to transport the operationsfrom $\Bbb R ^2$ to $\Bbb R$ $\endgroup$ – Hagen von Eitzen Sep 18 '17 at 18:46
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Let $f : \mathbb{R} \to \mathbb{R}^2$ be any bijection.

Define the operations $\oplus : \mathbb{R}\times\mathbb{R}\to \mathbb{R}$ and $\odot : \mathbb{R}\times\mathbb{R}\to \mathbb{R}$ like this:

$$x \oplus y = f^{-1}(f(x) + f(y)), \quad\forall x,y \in \mathbb{R}$$

$$\lambda \odot x = f^{-1}(\lambda \cdot f(x)), \quad\forall \lambda,x \in \mathbb{R}$$

You can check that $\left(\mathbb{R}, \oplus, \odot\right)$ is indeed a vector space over $\mathbb{R}$.

Futhermore, notice that $f$ is an isomorphism of $(\mathbb{R}, \oplus, \odot)$ and $(\mathbb{R}^2, +, \cdot)$, so $$\dim \,(\mathbb{R}, \oplus, \odot) = \dim\,(\mathbb{R}^2, +, \cdot) = 2$$

Indeed:

$$f(x \oplus y) = f\big(f^{-1}(f(x) + f(y))\big) = f(x) + f(y), \quad\forall x,y \in \mathbb{R}$$ $$f(\lambda \odot y) = f\big(f^{-1}(\lambda \cdot f(x))\big) = \lambda \cdot f(x), \quad\forall \lambda,x \in \mathbb{R}$$

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  • $\begingroup$ Great!! Thanks a lot! $\endgroup$ – dmtri Sep 18 '17 at 19:56
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Yes, there is a vector space structure on $(\mathbb R,{+})$ that has dimension $2$ over $\mathbb R$. But of course the scalar multiplication of this vector space structure will not agree with ordinary multiplication of real numbers.

First, $\mathbb R$ and $\mathbb R^2$ both have dimension $2^{\aleph_0}$ as vector spaces over $\mathbb Q$. (Simply considering cardinalities tells us it cannot be otherwise). This means they are isomorphic as $\mathbb Q$-vector spaces, and in particular the additive groups $(\mathbb R,{+})$ and $(\mathbb R^2,{+})$ are isomorphic.

We can use this latter isomorphism to pull the scalar multiplication of $\mathbb R^2$ back to $(\mathbb R,{+})$ to get a vector space that is isomorphic to $\mathbb R^2$, and therefore has dimension $2$.

Unfortunately, this construction does not produce a description of how the scalar multiplication would work. It depends on the axiom of choice, and this dependence feels essential (if you want the addition to agree with ordinary real addition), so it is likely that there is no solution with an explicit description.

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    $\begingroup$ Surely all you need is a bijection between $\Bbb{R}$ and $\Bbb{R}^2$ (without regard to any algebraic structure). And that doesn't need AC. (Maybe you are trying to preserve the underlying additive group structure, but the OP isn't asking for that.) $\endgroup$ – Rob Arthan Sep 18 '17 at 19:14
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    $\begingroup$ @RobArthan: My goal here is to preserve the additive group. I thought I was pretty explicit about that. $\endgroup$ – Henning Makholm Sep 18 '17 at 19:28
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    $\begingroup$ It's clear, but I just wanted to confirm that it is more than is being asked for: the original question has an affirmative answer in ZF without choice. $\endgroup$ – Rob Arthan Sep 18 '17 at 19:32
  • $\begingroup$ Just something more guys...In case, I do not want to tell my students about space filling curves ( this is needed propably for the bijection between $R$ and $R^2$ ), what else can I do? $\endgroup$ – dmtri Sep 19 '17 at 19:35
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    $\begingroup$ @dmtri: Space-filling curves do not produce bijections anyway (they can't be injective). The interleaving-digits technique should be enough, though. $\endgroup$ – Henning Makholm Sep 19 '17 at 19:47

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