function relationship proving I have a function $f:\mathbb{R}\rightarrow\mathbb{R}$ for which is true: $$f(x+y)=f(x)+f(y)$$
I have already prove that $f(0)=0$ and $f$ is odd.
Now I want to prove the $f$ is One-to-One .
 If $f(x)=0$ and $x=0$ is the only root , but I want some help
 A: The one-to-one statment is not necessary true Ex: $$f\equiv 0\qquad \text{satisfies }\quad f(x+y)=f(x)+f(y)\quad \text{but $f$ is one-to-one }.$$

Result $f(1)\neq 0$ then and $f$ measurable then $f$ is One-to-one Indeed Continuity implies $f(x) =xf(1)$.

Proof:
If $f$ is measurable then form this result $f$ is continuous therefore 


*

*For $n\in\mathbb N$ $$f(n)  =f(\underbrace{1+1+\dots+1}_{n-times}) =\underbrace{f(1)+f(1)+\dots+f(1)}_{n-times}) =nf(1) $$

*For $-n\in\mathbb Z_{-}$ $$0=f(0)=f(n-n) = f(n)+f(-n) = nf(1)+f(-n)$$
that is $$f(-n) = -nf(1).$$

*For $\frac{p}{q}\in\mathbb Q$ with  $p\in\mathbb N$ and $q\in\mathbb Z\setminus\{0\}$ 
$$f(1) = f(\frac{q}{q})=f(\underbrace{\frac{1}{q}+\dots+\frac{1}{q}}_{q-times}) =\underbrace{f(\frac{1}{q})+\dots+f(\frac{1}{q})}_{q-times}= qf(\frac{1}{q}) \\\implies f(\frac{1}{q}) =\frac{1}{q}f(1)$$
Hence, 
$$f(\frac{p}{q}) =f(\underbrace{\frac{1}{q}+\dots+\frac{1}{q}}_{p-times}) =\underbrace{f(\frac{1}{q})+\dots+f(\frac{1}{q})}_{p-times}) =pf(\frac{1}{q}) = \frac{p}{q}f(1). $$
For $x\in\mathbb R$.
By density of  $\mathbb Q$ in  $\mathbb R$ there is  $ r_n \in  \mathbb Q$ such that $r_n\to x$
by continuity, $$f(x) = f(\lim_{n\to\infty} r_n)=\lim_{n\to\infty} f(r_n)=\lim_{n\to\infty} r_nf(1) = xf(1).$$
 from this we see that $f$ is one-to-one 
since $f(1)\neq 0$
$$f(x) = f(y) \implies f(1)(x-y) = 0\implies x=y $$
A: If $f\not\equiv 0$ and continues then $f$ is one to one. To see this prove that $f(x)=ax$. so you should prove the following: 


*

*$$f(n)=nf(1)\quad \forall n\in \Bbb Z$$ 

*$$f(\frac{m}{n})=\frac{m}{n}f(1)\quad \forall m,n\in \Bbb Z$$

*Let $q_n\in\Bbb Q^c\to q\in\Bbb Q$ $$f(q)=\lim_{n\to \infty}f(q_n)$$

*Let $p_n\in\Bbb Q\to p\in\Bbb Q^c$ $$f(p)=\lim_{n\to \infty}f(p_n)$$

A: The question as stated is wrong, assuming the axiom of choice, even if we assume $f$ is not the zero function.
Consider a $\mathbb Q$-basis $B$ of $\mathbb R$. Let $f:\mathbb R \to \mathbb R$ be the function that maps a real number to the sum of its coefficients when expressed as a $\mathbb Q$-linear combination of elements from $B$.
This is a $\mathbb Q$-linear map $\mathbb R \to \mathbb R$, its image is actually just $\mathbb Q$. Since $\mathbb R$ is an infinite-dimensional $\mathbb Q$ vector space and $\mathbb Q$ is one-dimensional, $f$ cannot possibly be one-to-one.
