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The problem is,

Could $\mathbb{N}$ form a 2-dimensional linear space over $\mathbb{Q}$?

My attempt,

First, I thought this proposition is wrong because when natural numbers are multiplied by rational numbers (e.g., multiplying the natural number $1$ by $1/2$), the result typically yields a fraction, not necessarily a natural number.

But $\mathbb{N}$ and $\mathbb{Q}$ have the same cardinality, thus there exists a bijection between them,so do $\mathbb{N}$ and $\mathbb{Q}\times \mathbb{Q}$. And I want to use an isomorphism approach, but I'm not sure how to do it.

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    $\begingroup$ It's not entirely clear to me whether you want to retain any of the usual arithmetic structure on $\mathbb N$. The first of the two longer paragraphs seems to imply that you do, and the second one seems to imply that you don't. Once you clarify that, you've already answered the question: If you retain the usual arithmetic structure, $\mathbb N$ is not a vector space over $\mathbb Q$. If you don't, you can put it in bijection with $\mathbb Q^2$ and thus induce the structure of that two-dimensional vector space over $\mathbb Q$. $\endgroup$
    – joriki
    Commented Jan 6 at 3:21
  • $\begingroup$ @joriki Yes I want to put it in the bijection with $\mathbb{Q}^2$, but how to define the arithmetic in this bijection? $\endgroup$
    – Zhiwei
    Commented Jan 6 at 3:37
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    $\begingroup$ Take your favorite bijection $f: \mathbb{N} \rightarrow \mathbb{Q}^2$. Being a bijection, $f$ has an inverse map $f^{-1}: \mathbb{Q}^2 \rightarrow \mathbb{N}$. You have the usual addition operation $+$ on $\mathbb{Q}^2$, so you can use $f$ and its inverse to define an addition operation $\oplus$ on $\mathbb{N}$ simply as $x \oplus y = f^{-1}(f(x) + f(y))$ - you first use $f$ to take your inputs into $\mathbb{Q}^2$, perform the addition there, then bring the output back to $\mathbb{N}$ using $f^{-1}$. Scalar multiplication works similarly. $\endgroup$
    – Z. A. K.
    Commented Jan 6 at 3:54
  • $\begingroup$ @Z.A.K. Thank you, I understand now! $\endgroup$
    – Zhiwei
    Commented Jan 6 at 4:17
  • $\begingroup$ This is an answer to the question, not just a comment. Per site policy, comments should only be used to clarify, not answer the question. See How do comments work for more information. $\endgroup$
    – Martin Brandenburg
    Commented Jan 6 at 10:29

1 Answer 1

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Community wiki answer so the question doesn't remain unanswered:

As noted in a comment, the arithmetic structure of $\mathbb Q^2$ as a vector space over $\mathbb Q$ can be imposed on $\mathbb N$ by using a bijection $f:\mathbb N\to\mathbb Q^2$ and defining

\begin{eqnarray} m\oplus n&=&f^{-1}(f(m)+f(n))\;,\\ q\odot n&=&f^{-1}(q\cdot f(n))\;. \end{eqnarray}

Then $(\mathbb N,\oplus,\odot)$ is a $2$-dimensional vector space over $\mathbb Q$.

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