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As far as I understand, a vector space is a set + two operations (addition and multiplication by a number) + a set of coefficients. The set is closed with respect to operations, operations satisfy axioms, the set of coefficients is a field.

Well, for example, a set of directed line segments on a plane with the traditional operations of addition and multiplication by a number, and the coefficients from $\mathbb{Q}$ are a vector space, isn't it?

If this is a vector space, then the question about the basis arises. After all, an ordinary pair of non-parallel vectors will not be a basis here. It turns out that we got some kind of infinite-dimensional space for nothing?

In general, debunk my delusions, please...

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    $\begingroup$ Just to be clear, are you describing $\mathbb{Q}^2$? $\endgroup$
    – angryavian
    Commented Nov 23, 2021 at 19:06
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    $\begingroup$ I'm okay with you using whatever approach you want, but then you need to be precise. I gave you two directed line segments. How do you add them? How do you define "equality of vectors"? The government doesn't like it when I read minds without a warrant, so I need you to say it explicitly and not just say "It's what I'm thinking". $\endgroup$ Commented Nov 23, 2021 at 19:55
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    $\begingroup$ Note that if you consider, for example, the line segment from $(1,1)$ to $(1,2)$ to be "the same vector" as the line segment from $(0,1)$ to $(0,2)$, then you are not considering the set of directed line segments: in that set, the two line segments are different. You are actually considering a quotient set, the partition induced by an equivalence relation on the set of line segments. That's a very different object from "the set of directed line segments." That's not a figure of speech, but a rather important mathematical distinction. $\endgroup$ Commented Nov 23, 2021 at 19:59
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    $\begingroup$ So then your vector space does not consist of the set of all directed line segements, because the line segment from $(0,0)$ to $(1,1)$ is "the same" as the line segment from $(1,1)$ to $(2,2)$. And so, you are taking a quotient set modulo an equivalence relation. And thus, you can take a single representative from each equivalence class instead. And the standard representative is the vector that begins at $(0,0)$, and is described exclusively by their terminal points... and so you end up with the usual description. $\endgroup$ Commented Nov 23, 2021 at 20:13
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    $\begingroup$ The sum is not "addition of coordinates" until you translate them so that they all begin at the origin. Yet you objected vocally to my talking about the vectors all beginning at the origin. I don't want you to copy pages from the book, I just wanted you to be clear and explicit. Note: Your vector space does not consist of all directed line segments, it consists of equivalence classes of directed line segments. And so they can be represented as the set of vectors starting at the origin, exactly as I said and what you objected to. $\endgroup$ Commented Nov 23, 2021 at 20:15

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The vector space $\mathbb{R}^2$ of ordered pairs of real numbers, considered as a vector space over $\mathbb{Q}$, is infinite dimensional. You do not have any misunderstandings or delusions there. It is not difficult to exhibit an infinite linearly independent subset, since in fact we can do that for $\mathbb{R}$ considered as a vector space over $\mathbb{Q}$; so just take the ordered pairs of the form $(\sqrt{p},0)$ with $p$ ranging over all primes to exhibit an infinite $\mathbb{Q}$-linearly independent set of vectors. This suffices to show that this vector space does not have a finite basis.

Whether this vector space has a basis at all, though, depends on your set theory. In standard set theory, the Axiom of Choice guarantees that every vector space has a basis. In fact, the Axiom if Choice is equivalent to the statement that "Every vector space has a basis".

But there are models of set theory without the Axiom of Choice in which $\mathbb{R}$ does not have a basis over $\mathbb{Q}$. In those models, $\mathbb{R}^2$ does not have a basis over $\mathbb{Q}$ either, even though you can exhibit infinitely many linearly independent vectors.

Again, there is nothing to debunk: one can prove in Zermel-Fraenkel Set Theory that $\mathbb{R}^2$ is not finite dimensional over $\mathbb{Q}$.

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Let $\mathbb{K}$ a general field (such as $\mathbb{R}$ or $\mathbb{Q}$). We define vector space $\mathbb{V}$ and we write $(\mathbb{K}, +,\cdot)$ a structure where are defined the following operations:

$+:\, \mathbb{V}\cdot\mathbb{V}\to\mathbb{V}$ that is known as vector addition: $\forall v, w \in \mathbb{V}\,\, v+w$.

$\cdot\,: \mathbb{K}\cdot\mathbb{V}\to \mathbb{V}$, known as scalar multiplication: $\forall v\in \mathbb{V} \forall k\in \mathbb{K}\,\, k\cdot v$.

$\mathbb{Q}$ is a a field, but $\mathbb{N}$ is not. (because it's not close with sum). Thus, $(\mathbb{Q}, +,\cdot)$ is a vector space.

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