Linear transformation: Map $T:V\to W$ is said to be Linear transformation when it satisfy property :
1) $L(u+v)=L(u)+L(v)\forall u,v \in V$
2)$L(cv)=cL(v) \quad \forall c\in \mathbb{F}$
I know when $c \in \mathbb N$ then property 2 can be obtained from property 1, since $$L(nv)= \underbrace{L(v)+L(v)......+L(v)}_{\text{$n$ times where $n \in \mathbb{N}$}}$$
Now I have read in somewhere that if our scalar $c$ is rational then property 2 also follows from property 1, but I am unable to prove this. How can I show that if $c \in \mathbb{Q}$, when (2) follows from (1)?
I know that if our scalar is irrational, then we cannot use property 1 to prove property 2.
Suppose our scalar $c$ is rational; that is $c = \frac{p}{q}$ for some integers $p$ and $q$. Then, if $L$ satisfies property (1), $$L(cv) = L \left(\frac{p}{q}v\right) = p L \left(\frac{1}{q} v \right)$$ by your previous result for integer scalars. On the other hand, $$L(v) = L\left( q\cdot \frac{1}{q} v \right) = q L\left(\frac{1}{q}v\right)$$ so, multiplying both sides by $1/q$, $$L\left(\frac{1}{q}v\right) = \frac{1}{q} L(v)$$ returning to our first equation, this gives us that $$L\left(\frac{p}{q} v\right) = \frac{p}{q}L(v)$$ so property (2) follows from property (1).