# Problems with the definition of vectors as directed line segments in $\mathbb{R}^3$

First I'll say where I'm working: The vectorial spaces $\mathbb{R}^2$ and $\mathbb{R}^3$.

Then I'll define a vector of this spaces as the following:

$\textbf{Definition. }$ A vector $\vec{v}$ is the set of all equal directed line segments.

Now suppose that $$\underbrace{\overrightarrow{AB}}_{\mbox{directed line segment}} \in \vec{v},$$ which is a correct notation, by definition. So why do we write: $$\overrightarrow{AB} = \vec{v}$$

So I'm a little bit confused. If we understand this two objects as sets, from my point of view:

• $\overrightarrow{AB}$ as a directed line segment, it's a set of points in space(plane).
• $\vec{v}$ as a vector, it's a set of directed line segments, with infinite elements.

$\textbf{Question. }$How can these two objects coincide as sets? Which implies $\overrightarrow{AB} = \vec{v}$.

• So it seems in that textbook, the notation $\overrightarrow{AB}$ does not mean "the directed line segmant from $A$ to $B$", but instead it means "the vector represented by the directed line segmant from $A$ to $B$". – GEdgar Jun 10 '16 at 16:23

This is a reasonable question.

The point is that "equality of directed line segments" is an equivalence relation, so every directed line segment is art of some "vector" (where a vector is an equivalence class). There's even a function --- let's call it $p$ --- that sends a directed line segment to its equivalence class (but there's no function in the other direction! Too many equivalent directed-segments for each vector!). So properly speaking, we should write:

"Let $\vec{v} = p(AB)$"

to indicate the association of a directed segment to a vector. But no one ever does this. I suppose the reason is that the association is pretty much a natural one, and carrying around an extra name for it would be a pain in the neck.

We do the same thing in other contexts, too. For instance, one can define rational numbers as equivalence classes of pairs of integers under a certain equivalence relation. An integer $n$ is then represented, as a rational number, as the equivalence class of the pair $(n, 1)$. But we simply write things like "$n \in \mathbb Q$" rather than distinguishing $n$ from the equivalence class of $(n, 1)$.

• A question: Why you said "a vector is an equivalence class"? In linear algebra, say $\overrightarrow v=(1,5)$ in $\mathbb{R^2}$, this vector $\overrightarrow v$ is just a vector. – Eric Oct 26 '16 at 2:28
• There are many different (and in various ways, equivalent) definitions of "a vector" or even of the vector space we call $\mathbb R^2$. The one that the OP asked about is one of them, and I was answering a question about that. You, apparently, are used to a different one, which some folks might call a "coordinate vector" version. But that doesn't mean I can't answer a question about a different formulation. – John Hughes Oct 26 '16 at 11:40
• English is not my native language, I just reviewed my words and I found it might be a little offensive. I just wanted to know your definition, so I asked it. :) And in fact, I just asked a question about this before seeing this post. :) – Eric Oct 26 '16 at 15:01
• I took no offense (I try to be aware of the not-native-language issue, esp. since my only semi-competent other language is French, and I can't even say "Good Morning" without it potentially offending someone!) I'll add an answer to your linked question. – John Hughes Oct 26 '16 at 15:27

$\overrightarrow v$ is a class of equivalence, so you take one element there to represent all the others, that way there is no problem in stating $\overrightarrow v = \overrightarrow {AB}.$

If you take as the first point $A$ the origin $O$ of $\mathbb R^n$ you have a canonical representant of the vector $\overset{\to}{v}=p(AB)$, according to the notation of John Hughes: $$\overset{\to}{v}=p(AB)\cong OB'$$ for a suitable $B'$ (which is denoted, by the way, as $B'=B-A$)