If $f\colon\mathbb R\to\mathbb R$ is continuous and satisfies $f(x+y) = f(x) + f(y)$ and $f(1)=1$, then $f(x)=x$ for all $x$. Let 
    $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that:
$$f(x+y) = f(x) + f(y)$$
$$f(1) = 1$$ Show that $$f(x) = x$$ I have been having trouble approaching this problem. I have shown, through a system of equations, that $f(x+y) = x + y$, but that's about as far as I can get. Appreciate any help anyone has to offer!
 A: See that
$$f(0)=f(0)+f(0)=2f(0)\implies f(0)=0$$
$$\begin{align}f(a+b+c+\dots+z) & =f(a)+f(b+c+\dots+z) \\&=f(a)+f(b)+
f(c+\dots+z)\\&=\dots\\&=f(a)+f(b)+f(c)+\dots+f(z)\end{align}$$

For natural numbers $x$, we have
$$f(x)=f(\underbrace{1+1+1+\dots+1}_x)=\underbrace{f(1)+f(1)+\dots+f(1)}_x=1+1+1+\dots+1=x$$
$$f(x)=x\ \forall\ x\in\mathbb N$$

For positive rational numbers,
$$1=f(1)=f(\underbrace{\frac1x+\frac1x+\frac1x+\dots+\frac1x}_x)=f(\frac1x)+f(\frac1x)+\dots+f(\frac1x)=xf(\frac1x)$$
$$1=xf(\frac1x)\implies f(\frac1x)=\frac1x\ \forall\ x\in\mathbb N$$
$$f(\frac yx)=f(\frac1x+\frac1x+\dots+\frac1x)=f(\frac1x)+f(\frac1x)+\dots+f(\frac1x)=yf(\frac1x)=\frac yx\\f(\frac yx)=\frac yx\ \forall\ (x,y)\in\mathbb N\times\mathbb N$$
$$f(x)=x\ \forall\ x\in\mathbb Q^+$$

For positive real numbers, every real number is the cauchy sequence of rational numbers, which, since $f(x)$ is continuous, gives us
$$f(x)=x\ \forall\ x\in\mathbb R^+$$

Finally, for all negative numbers, we have
$$0=f(0)=f(x-x)=f(x)+f(-x)$$
$$0=f(x)+f(-x)\implies f(-x)=-f(x)=-x$$
A: If we let $y = 1$:
$$f(x + 1) = f(x) + 1$$
This is a recurrence relation and can be solved with simple techniques. However, we already know what the solution is supposed to be. Proving that $f(x) = x$ is a solution:
LS:
$$f(x + 1) = x + 1$$
RS:
$$f(x) + 1 = x + 1$$
Because LS = RS, $f(x) = x$ is a solution to our problem. Note that this does not prove that it's the only solution.
