So I am given a function f:R→R such that $f(x+y)=f(x)+f(y)$ for all real $x$ and $y$, and I have to prove $f(-x) = -f(x)$.
I was considering doing $3$ different cases: one where $x<0$, one where $x= 0$, and one where $x>0$ to prove this, with $y = 0$ the entire time.
This is what I've done so far:
Step $1$. Suppose $x = 1$, $y=0$
Step $2$. $f(-x+y) = -f(x)$
Step $3$. $f(-1+0) = -f(1)$
Step $4$. $f(-1) = -f(1)$
Step $5$. $-f(-1) = f(1)$ therefore by double negatives(??)
Step $6$. $f(1) = f(1)$
Not sure if what I did between Step $5$ and $6$ is legal. Thanks, in advance, for your help!
It is not legal (you are using the fact that you are trying to prove), nor is it going to get you anywhere anyway.
Instead, take $y=-x$, so that $$ f(x)+f(y) = f(x)+f(-x), $$ and $$ f(x+y) = f(x-x) = f(0) = 0, $$ where the last equality follows since $$ f(0+0) = f(0)+f(0) \quad\Leftrightarrow\quad f(0) = 2f(0) \quad\Leftrightarrow\quad f(0)=0. $$
Finally, we see that $$f(x+y) = f(x) + f(y) \quad \Leftrightarrow \quad 0 = f(x) + f(-x) \quad \Leftrightarrow \quad f(-x) = -f(x).$$