$\alpha (a + b) = \alpha (a) + \alpha (b)$ for all $a,b \in \mathbb{R}.$ show that $\alpha$ is a linear transformation. The question is:
Let $\alpha: \mathbb{R} \rightarrow \mathbb{R}$ be a cts. function which satisfies $\alpha (a + b) = \alpha (a) + \alpha (b)$ for all $a,b \in \mathbb{R}.$ show that $\alpha$ is a linear transformation.
I had seen this other question here on this site: 
Show that additive functions are linear if the vector spaces are defined over $\mathbb{Q}$.
I want someone to explain to me how continuity here makes the function $\alpha$ a linear transformation? it seems like continuity substitutes for the field being $\mathbb{Q}$ in the mentioned link.
If anyone can show me why $\alpha $  is linear transformation and  compare between the 2 exercises, it will be greatly appreciated. 
 A: You need to prove that $\alpha(ra) = r\alpha(a)$ for all $r,a\in \mathbb R$. (You are considering $\mathbb R$ as vector space over itself.)
$\alpha(0) = \alpha(0+0) = \alpha(0) + \alpha (0)$ so $\alpha (0)=0$.
And $\alpha(a) + \alpha(-a) = \alpha(a+(-a)) = \alpha(0) = 0$ so $\alpha(-a)=-\alpha(a)$ for all $a\in \mathbb R$.
So $\alpha(0*a) = \alpha (0) = 0 =0\alpha(a)$ so the proposition is true if $r=0$
And $\alpha(1*a) = \alpha(a) =1*\alpha (a)$ so the proposition is true if $r = 1$.
If the proposition is true for $r=n \in N$ then $\alpha ((n+1)a)=\alpha (na + a) = \alpha(na) +\alpha (a) = n\alpha(a) + \alpha(a)=(n+1)\alpha(a)$ so it is true for $r=n+1$.
So by induction it is true for all $r\in \mathbb N\cup \{0\}$ and as $\alpha(-na)=-\alpha(na) = -n\alpha(a)$ for $n\in \mathbb N$ the proposition is true for $r\in \mathbb Z$.
And for $r = \frac mn \in \mathbb Q; m\in \mathbb  Z; n\in \mathbb N$ we have $\underbrace{\alpha(\frac mna) + ... + \alpha(\frac mna)}_{n\text{ times}}= \alpha (n*\frac mna) =\alpha(ma)=m\alpha(a)$ so $\alpha(\frac mn a) = \frac mn\alpha(a)$.  So the proposition is true if $r\in \mathbb Q$.
Now here is where you need continuity:
Let $r\in \mathbb R$ and let $q_i\in \mathbb Q$ so that $q_i\to r$.
Then $r\alpha(a) = (\lim_{i\to\infty} q_i)\alpha (a) = \lim_{i\to\infty} q_i\alpha (a))=\lim_{i\to\infty}\alpha(q_ia)$.  But $\alpha$ is continuous so $\lim_{i\to\infty}\alpha(q_ia)=\alpha(ra)$
So the proposition is true for all $r \in \mathbb R$.
