It might be an interesting question before studying the concept of orientation on $\mathbb{R}$ as it is studied on $\mathbb{R^2}$ & $\mathbb{R^3}$

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    $\begingroup$ yes, it's one-dimensional; addition and scalar multiplication satisfy the required properties $\endgroup$ May 27 '20 at 22:28
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    $\begingroup$ To check if something is vector space, you first need to define addition and scalar multiplication operations on it, then verify each of the vector space operations – if they all hold, then you have a vector space, and otherwise you don't. So, step 1: define the addition and scalar multiplication operations. Step 2: check the axioms. Where along this path are you first having trouble? $\endgroup$ May 27 '20 at 22:37
  • $\begingroup$ There are many similar questions which have already been asked and answered on Math SE. A sampling of more-or-less relevant questions (found via the Google search "is R a vector space over itself") includes: [1], [2], [3] $\endgroup$
    – Xander Henderson
    May 27 '20 at 22:48
  • $\begingroup$ @XanderHenderson So sorry you thought my question was inappropriate. First of all, it's my first question on Math Stack. I have suspected that the answer to my question would be yes through the demonstration of vector space. But not seeing the answer explicitly in examples on linear algebra books, or google, or even on this website, has made me insecure. I thought it would be a simple answer (to formulate and to answer) and here I am. Greetings and thanks to all of you for your answers, even if it seemed like a trivial question. Sorry again. $\endgroup$ May 27 '20 at 23:02
  • $\begingroup$ @JesusBlanco You did nothing wrong. The question is not a bad question (though it needs some context; e.g. when asking a question on MSE, provide your basic definitions, a link to the book you are working out of, etc; but the question itself is natural). As a newbie, you are not expected to know these things---you have to live and learn. My vote-to-close is not an indication that you did anything wrong, only that the question isn't really appropriate for MSE (it lacks context, and also happens to be a duplicate). For future reference: math.meta.stackexchange.com/questions/9959 . $\endgroup$
    – Xander Henderson
    May 27 '20 at 23:14

Any ring $R$ is a an $R$-module via its intrinsic multiplication. So in the case when $R$ is a field, $R$ is a vector space over itself. In particular, $\mathbb{R}$ is a vector space over itself that is one-dimensional generated, for example, by $1$. Similarly, $\mathbb{C}$ is a $\mathbb{C}$ vector space of dimension one and $\mathbb{F}_2$ is a vector space over $\mathbb{F}_2$ of dimension one.

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    $\begingroup$ Of course you're right, but I can't imagine that this will be of any help to the OP – if they didn't already know off the top of their head that any field is a vector space over itself (let alone the more general result for rings as modules over themselves), why would they ask this question? Your answer leaves that result as an unexplained black box, which doesn't help the OP understand why it's true. $\endgroup$ May 27 '20 at 22:35
  • $\begingroup$ perhaps "is one-dimensional generated, for example by $1$" is the useful part of this $\endgroup$
    – Henry
    May 27 '20 at 22:40
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    $\begingroup$ I have downvoted this answer for three reasons. The first is highlighted by @diracdeltafunk, above. Secondly, I do not think that it is good practice to answer poorly posed questions where the asker does not provide any context. It often creates a disconnect between the level of the question and the answer, and encourages the propagation of low quality questions. Finally, I strongly suspect that this question is a duplicate (just looking at the sidebar, I see several potential dupe targets). It is better to link to dupes than to answer them. $\endgroup$
    – Xander Henderson
    May 27 '20 at 22:42

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