Is $F(c\vec{v})$ = $cF(\vec{v})$ condition of linear transformation the same thing as $F(\vec{a})+F(\vec{b})=F(\vec{a}+\vec{b})$? Is $F(c\vec{v})$ = $cF(\vec{v})$ condition of linear transformation the same thing as $F(\vec{a})+F(\vec{b})=F(\vec{a}+\vec{b})$?
I am a physics undergraduate student and studying linear algebra. 
The 2 conditions of linear transformation seem quite same to me. But, I know if these 2 were the same then it wouldn't be mentioned in explicitly in the literature.
Why do I think the 2 conditions are the same thing:
For any real number (rational or irrational) $c$, the vector $c\vec{v}$ can be expressed as a sum of $n$ equal vectors, $\frac{c}{n}\vec{v}$.
So, $$c\vec{v}=\sum_{i=1}^{n}\frac{c}{n}\vec{v}$$ 
Then, applying the first condition on 2 constituents we get $F(\frac{c}{n}\vec{v})+F(\frac{c}{n}\vec{v})=F(\frac{2c}{n}\vec{v})$ $\implies2F(\frac{c}{n})=F(\frac{2c}{n})$. Then applying induction, we can get the second condition $F(c\vec{v})$ = $cF(\vec{v})$
So, why is the second condition necessary?
One reasoning: If $c$ is an irrational number, then this repeated addition never ends; because $\frac{c}{n}$ is also an irrational number. While it may not be a problem to a physicist, it can be a real problem to a mathematician. But, I don't know that deep about number theory. So, I can't figure out the problem.
So, mathematicians, help me figure out the reason of explicitly stating the 2nd condition. You are welcome to use advanced stuff to make it clear, but first give a simple explanation, then rigorously explain it.
 A: You have proved that if $F(a) + F(b) = F(a+b)$ then given any $n \in \mathbb{N}$ and any real $c$ and any vector $v$ $F(cv) = nF(\frac{cv}{n})$. Following the same argument you can prove that $F(nv) = nF(v)$ with $n \in \mathbb{N}$. Using this two proofs you managed to do it with rational numbers. As you say, you miss the irrational numbers, that is important, indeed they are almost all $\mathbb{R}$. And if you consider a $\mathbb{C}$-Vector Space you are missing much more. Now if you consider an arbitrary space vector, say $(V,\mathbb{K},+,.)$ with $\mathbb{K}$ an arbitrary field?
A: The two conditions are not equivalent since we need both to guarantee that a trasformation is linear. 
The reason can be found in the properties which define a subsbace and notably
$0) \vec{0} \in W\\
1) \vec{cv}\to c \cdot \vec{v}\in W\\
2) \vec{v}+\vec{w} \in W$
Indeed let consider for example the following subset
$$S=\{\vec v,\vec w:\vec v=t(1,0), \vec w=s(0,1)\}\subseteq\mathbb{R}$$
then "1" is satified but "2" is not.
We can also give of course examples of transformation for which


*

*$F(c\vec{v})$ = $cF(\vec{v})$ is satisfied, but

*$F(\vec{a})+F(\vec{b})=F(\vec{a}+\vec{b})$ is not


Let consider for example the transformation
$$\vec v(a,b) \to \begin{cases}\vec v \quad |a|\ge |b|\\-\vec v \quad |a|< |b| \end{cases} $$
