Finding all $f:\Bbb{R}\to\Bbb{R}$ such that $f(x+y)=f(x)+f(y)$ for all real $x$ and $y$, and $f(1)=0$ 
If $f:\Bbb{R}\to\Bbb{R}$ satisfies $f(x+y)=f(x)+f(y)$ for all real $x$ and $y$, and $f(1)=0$, find $f(x)$.

By definition,
$$f'(x)= \frac{f(x+h)-f(x)}{h}$$
when $h$ tends to zero.
Using the given statement,
$$f'(x)= \frac{f(h)}{h}$$
Since $f(0)=0$,
$$f'(x)= f(h)+\frac{f(0)}{h}$$
or,
$$f'(x)=f'(0)$$
which means $f(x)=mx+C$. However, I am not able to determine these constants.
Moreover, is there any way to do it without calculus?
Please help.
Thanks.
 A: If $f$ is assumed to be continuous then it is a well knwon fact that $f(x)=cx$ for some constant $c$. From the condition $f(1)=0$ it follows that $f(x)=0$ for all $x$.
If $f$ is not assumed to be continuous the there are all kinds of 'wierd'solutions. Take any addive dis-continuous function $g$ and take $f(x)=g(x)-g(1)x$.
Ref: https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation
A: We are given $f(1)=0$.
Then $f(2)=2f(1)\implies f(2)=0$, $f(4)=2f(2)\implies f(4)=0,...$
Moreover, $f(3)=f(2)+f(1)=0,...$
We also have $f(1)=2f(\frac12)\implies f(\frac 12)=0$
Continuing the pattern, we see $f(x)=0\ \forall x\in\Bbb Q$, which is enough to say $f\equiv 0$ if we assume $f$ is "nice".
A: One cannot find $f$ using this information alone.
Proof: consider $\mathbb{R}$ as a $\mathbb{Q}$-vector space. Then $\mathbb{R}$ must have a basis $B$.
Note that if $A$, $B$ are $\mathbb{Q}$-vector spaces, a function $f : A \to B$ satisfies $\forall x, y \in A, f(x + y) = f(x) + f(y)$ if and only if $f$ is $\mathbb{Q}$-linear.
Thus, we are asked to describe the $\mathbb{Q}$-linear functions $\mathbb{R} \to \mathbb{R}$. Without loss of generality, we assume that $1 \in B$.
Then $\mathbb{Q}$-linear functions $f : \mathbb{R} \to \mathbb{R}$ are in bijective correspondence with functions $g : B \to \mathbb{R}$, where we have $g(b) = f(b)$ for all $b \in B$.
Now clearly $B$ cannot be equal to $1$, so the constraint that $g(1) = 0$ does not give us enough information to determine $g : B \to \mathbb{R}$, and hence does not give us enough information to determine the linear function $f$.
