I need to find all functions $f:\mathbb R \rightarrow \mathbb R$ which are continuous and satisfy $f(x+y)=f(x)+f(y)$ I need to find all functions $f:\mathbb R \rightarrow \mathbb R$ such that $f(x+y)=f(x)+f(y)$. I know that there are other questions that are asking the same thing, but I'm trying to figure this out by myself as best as possible. Here is how I started out:
Try out some cases:
$x=0:$
$$f(0+y)=f(0)+f(y) \iff f(y)=f(0)+f(y) \iff 0=f(0) $$
The same result is for when $y=0$
$x=-y:$
$$f(-y+y)=f(-y)+f(y) \iff f(0)=f(-y)+f(y) \iff 0=f(-y)+f(y)\iff \quad f(-y)=-f(y)$$
I want to extend the result of setting $x=-y$ to numbers other that $-1$, perhaps all real numbers or all rational numbers. I got a little help from reading other solutions on the next part:
Let $q=1+1+1+...+1$. Then 
$$f(qx)=f((1+1+...+1)x)=f(x+x+...+x)=f(x)+f(x)+...+f(x)=qf(x)$$
I understood this part, but I don't understand why this helps me find all the functions that satisfy the requirement that $f(x+y)=f(x)+f(y)$, but here is how I went on:
Thus 
$$f(qx)=qf(x)$$ and it should follow that
$$f \bigg (\frac {1}{q} x\bigg)= \frac{1}{q}f(x)$$ where $q\not =0$, then it further follows that 
$$f \bigg (\frac {p}{q} x\bigg)= \frac{p}{q}f(x)$$ where $\frac{p}{q}$ is rational, and lastly it further follows that 
$$f (ax)= af(x)$$ where $a$ is real. Thus functions of the form $f(ax)$ where $a$ is real satisfies the requirement of  $f(x+y)=f(x)+f(y)$.
I don't know how much of what I did is correct\incorrect, and any help would be greatly appreciated. Also is there any way that I can say that functions of the form $f(ax)$ where $a$ is real are the only functions that satisfy the requirement of  $f(x+y)=f(x)+f(y)$? Or do other solutions exist?
Again, thanks a lot for any help! (Hints would be appreciated, I'll really try to understand the hints!)
 A: For $f(\frac1q x)$:
$$f(x) = f(q\cdot\frac1q x) = f(\frac1q x+\ldots+\frac1q x) = f(\frac1q x)+\ldots+f(\frac1q x) = qf(\frac1q x)$$
For $f(\frac pqx)$: Set $y=\frac xq$ to get
$$f(\frac pqx) = f(p\frac xq) = f(py) = p\,f(y) = p\,f(\frac1q x) = p\cdot\frac1qf(x)$$
So now you know $f(\alpha x)=\alpha f(x)$ for all $x\in \mathbb Q$. Let $(\alpha_n)$ be a sequence of rational numbers converging to the real number $r$. Since $f$ is continuous, we have $\lim_{n\to\infty}f(\alpha_n x)=f(r x)$. On the other hand, since all $\alpha_n$ are rational, we have $\lim_{n\to\infty}f(\alpha_n x) = \lim_{n\to\infty} \alpha_n f(x) = r\,f(x)$. Since for every $r\in R$ there exists a sequence of rational numbers converging to it, we therefore have $f(rx)=r\,f(x)$ for all $r\in \mathbb R$.
Finally, we can get an explicit form by observing that $f(x) = f(x\cdot 1) = x\,f(1)$. Therefore with $f(1)=c$ arbitrary, we get
$$f(x) = cx$$
A: $$f(x+y)=f(x)+f(x)$$
$$f(0)=0$$
$$f(-x)=-f(x)$$
All these are enough to state a linear function with no constant.
Also,$$f'(x)=\dfrac{f(x+h)-f(x)}h$$
if $$f'(x)=\dfrac{f(h)}{h}=c$$ 
So, The only solution set is $f(x)=cx$
A confirmation too from wolfram.
.
A: If you do not assume continuity of $f$, then it depends on the Axiom of Choice, stated as Zorn's Lemma.  Consider the family $\cal L$ of linearly independent over $\mathbf Z$ subsets of $\mathbf{R}$. Given a chain of such sets,
$$L_1 \subset L_2 \subset \dotsm \subset L_n \subset \dotsm$$
we can prove $L = \bigcup_n L_n \in \cal L$.  Suppose $L$ were not linearly independent over $\mathbf Z$. Then, there would be distinct $x_i \in L$ and nonzero integers $a_i$ such that
$$
a_1 x_1 + \dotsm + a_m x_m = 0
$$
But each $x_i$ must be in $L_n$ for some $n$. Let $N$ be the maximum of all such $n$.  Then, each $x_i \in L_N$, and so $L_N$ is a $\mathbf Z$-dependent set, contrary to the assumption.
Since every chain has an upper bound, Zorn's Lemma states that there is a maximal member, $M$, of $\cal L$.  Every real number $r$ must be a $\mathbf Z$=linear combination of a finite set of elements of $M$; otherwise, $M \cup \{r\}$ would be $\mathbf Z$-independant, contrary to $M$ being maximal. $M$ is a Hamel basis of $\mathbf R$.
And now, any permutation of $M$ extends by linearity to a discontinuous $f$ such that $f(x+y) = f(x) + f(y)$. 
