Determine the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x, y \in \mathbb{R}$

$$
f(x+y)=f(x)-f(y)
$$
1. We can see that $f(t)=0$ for all $t \in \mathbb{R}$ works. Indeed, $f(x+y)=0$ and $f(x)-f(y)=0-0=0$. But for now, we can't find any others.
2. I am looking for $f(0)$. With $x=0$ and $y=0$ we have

$$
f(0)=f(0+0)=f(0)-f(0)=0
$$

So $f(0)=0$

3. If I fix $y=0$ I get

$$
f(x)=f(x+0)=f(x)-f(0)=f(x)
$$

which is $f(x)=f(x)$, meaning nothing interesting. However, if I fix $x=0$ I get

$$
f(y)=f(0+y)=f(0)-f(y)=-f(y)
$$

and thus $f(y)=-f(y)$ for $y$. But then for all $y, f(y)=0$. We deduce that $f(y)=0$ is the only possible solution. For the synthesis, we have already seen that $f(y)=0$ works.

Another proof method: (being clever). We notice that if we swap $x$ and $y$, the left term does not change. We then have these two equations for all $x$ and $y$

$$
\left\{\begin{array}{l}
f(y+x)=f(x)-f(y) \\
f(y+x)=f(y)-f(x) \quad(\text { by swapping }) x, y
\end{array}\right.
$$

If we sum the two lines, we get

$$
f(y+x)+f(y+x)=f(x)-f(y)+f(y)-f(x)=0
$$

and thus $2 f(x+y)=0$ for all $x$ and for all $y$. The function $f(t)=0$ for all $t$ is therefore the only solution.