I've been having a hard time with this proof because I do not know where to go from where I am. The proof we were assigned to in class is as follows:
Let $f : \mathbb Z \rightarrow \mathbb Z$ be a function with $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb Z$.
Prove that if $x_1, x_2, \ldots , x_n \in \mathbb Z$ then $f(x_1+x_2+\ldots+x_n)=f(x_1)+f(x_2)+\ldots+f(x_n)$
I was thinking maybe if we get $f(x_1+x_2+\ldots+x_n)$ into two such that the definition will work, but my teacher said that won't be possible because there's a shorter way and also because it would take forever to get to the original $x_1...x_n$. So I have been staring at this proof and don't know where to go! HELP!
A proof by induction works here : If $n=1$, then $$ f(x_1) = f(x_1) $$ by definition. If $n=2$, $$ f(x_1+x_2) = f(x_1) + f(x_2) $$ again, by definition.
If $n=3$, then let $y_1 = x_1, y_2 = x_2+x_3$, then $$ f(x_1+x_2+x_3) = f(y_1+y_2) = f(y_1) + f(y_2) \qquad\text{by the case for } n=2 $$ $$ = f(x_1) + f(x_2) + f(x_3) \qquad\qquad\qquad\qquad\qquad\text{by the previous case} $$
Can you extend this to $n>3$ as well?