A problem on analysis specifically on functions Let $f(x)$ be a function from reals to reals obeying the following: $f(x)$ is continuous, $f(0)=1$, and $f(m+n+1)=f(m)+f(n)$. Show that $f(x) =1 +x$ for all real numbers $x$. 
I am a bit confused on how to start with the sum as I am new to analysis. Assuming $f(x)=1+x$. I can indeed show that it satisfies properties $f(m+n+1) =f(m) + f(n)$, (where m and n are reals) by checking it for RHS and LHS alternatively but can't prove it using the condition itself
 A: At first, we have the following relation
$$f(x+y+1)=f(x)+f(y);$$
or
$$f(x+y)=f(x+(y-1)+1)=f(x)+f(y-1);\ \ \ \ \ (*)$$
for all $x, y\in \mathbb{R}$. Now define the funtion $g:\mathbb{R}\rightarrow \mathbb{R}$ with $g(x):=f(x)-1$ for all $x\in \mathbb{R}$. Now, by replacing $x:=0$ in relation $(*)$ , we get $f(y)=f(0)+f(y-1)=f(y-1)+1$ we have 
$$g(x+y)=f(x+y)-1=f(x)+f(y-1)-1=(f(x)-1)+((f(y-1)+1)-1)=(f(x)-1)+(f(y)-1)=g(x)+g(y)$$
for all $x, y\in \mathbb{R}$. Therefore the function $g$ satisfies the Cauchy functional equation (see https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation) and since the function $g$  is continuous and $g(0)=0$, implies that $g(x)=x$ and so $f(x)=x+1$ for all $x\in \mathbb{R}$. Note that in question, if suppose that the relation $$f(x+y+1)=f(x)+f(y);$$ is true only for rationals, then the solution is same as above, but if the functional equation is true only for integers, then the functions $f(x):=x+1+\sin(2\pi x)$ and  $f(x):=x+1+\cos((\pi/2)+2x\pi)$ satisfy in the given functional equation.
