Question 103
Let $\ominus$ be an operation on the set of real numbers such that
$$
(x \ominus y)+(y \ominus z)+(z \ominus x)=0
$$
for all real $x$, $y$, and $z$. Prove that there is a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
x \ominus y=f(x)-f(y)
$$
for all real $x$ and $y$.
$$
(x \ominus y)+(y \ominus z)+(z \ominus x)=0
$$
for all real $x$, $y$, and $z$. Prove that there is a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
x \ominus y=f(x)-f(y)
$$
for all real $x$ and $y$.