why does there's need of axiom that vector spaces must be closed under scalar multiplication? I am learning linear algebra and I just learnt about vector spaces. There're axioms that vector spaces must be closed under addition , vector spaces must be closed under scalar multiplication.
So I am thinking that if the vector space is closed under addition the it must also be closed under scalar multiplication.
Reason:
Suppose set $ \mathbf{X} $ is closed under addition.
if $ \mathbf{v} \in \mathbf{X} $ then $ k\mathbf{v} \in \mathbf{X}$
$ \because k\mathbf{v} = \mathbf{v}_{1} + \mathbf{v}_{2} + \mathbf{v}_{3} + ... + \mathbf{v}_{k}$
while i am writing this question i thought if $ k \in \mathbb{Q} $
then $ k = \frac{m}{n} $
then $\frac{m}{n}\mathbf{v} = \frac{(\mathbf{v}_{1} + \mathbf{v}_{2} + ... + \mathbf{v}_{m})}{n} $
= $ (\mathbf{v}_{1} + \mathbf{v}_{2} + ... + \mathbf{v}_{m}) -  (n\mathbf{u}_{1} + n\mathbf{u}_{2} + ... + n\mathbf{u}_{v})$ where $ \mathbf{u} \in \mathbf{X} $
if $ \mathbf{X} $ is $ \mathbb{R}^{n} $ then $\mathbb{u}$ will be vector of one's $ \begin{bmatrix} 1 & 1 & 1 ... 1_{n} \end{bmatrix} $
But if $ \mathbf{X} $ is not $ \mathbb{R}^{n} $ then it doesn't need to have such vector. Right?
basically, i am trying to think multiplication as addition and division as subtraction and subtraction is a kind of addition. That's why i am not getting the need of axiom vector spaces must be closed under scalar multiplication.
What am i doing wrong here ? Am i missing something ?
Many Many thanks :)
 A: Your reasoning is correct to some extent: addition can mimick multiplication by an integer. I am not getting why you can mimick rational numbers: this is false. For example, take the points in the plane that have just integer coordinates: this is closed for addition,  but not for multiplication by rational numbers.
In general, vector spaces are closed by multiplication for "numbers" in a field, where the most used field is the one of real numbers. This is probably the only field your textbook introduce; maybe it will also introduce the one of complex numbers.
The fundamental idea is that a field contains a lot of things more than the integers, and in general a lot more. Technically, your argument does not workby the following counterexample: just take $\mathbb{Z}$. This is closed by addition but not for scalar multiplication!!
A: If $v$ is a vector, by your argument so is $kv$ for integer $k$. To do the same for rational $k=p/q$ with $p,\,q\in\Bbb Z$, we'd need to prove or assume some $w$ satisfies $qw=pv$. The simplest approach is to demand closure under scalar multiplication as an axiom, which also handles $k$ over a field of our choice, such as $\Bbb R$ or $\Bbb C$. That's an added bonus: if you pick such a field, you get a larger continuity than from $\Bbb Q$ alone.
