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An additive function on an additive semigroup satisfies $f(x+y)=f(x)+f(y)$ whenever $x$, $y$, and $x+y$ belong to its domain.

For real numbers, an additive function is a group homomorphism from $(\mathbb{R},+)$ to itself. It preserves scalar multiplication by every rational number, so it is a linear map when domain and codomain are viewed as vector spaces over the rational numbers. Being a linear map over the real numbers additionally requires $f(tx)=tf(x)$ for every real number $t$. On the real line this stronger condition is equivalent to $f(x)=xf(1)$. Additivity alone does not imply it.

First, $1\in G(p)$. Otherwise there would be a nonconstant rational function with one simple pole and no others, defining a degree-one finite morphism $X\to {\mathbb{P}}^1$. Such a map is an isomorphism, which would make $X$ have genus zero.

Part (iii) says that there are exactly two gaps. If $2$ is a gap, they must therefore be $1$ and $2$. If $2$ is not a gap, choose a function with its only pole at $p$, of order two. Pole orders form an additive semigroup under multiplication of functions. If $3$ were also a nongap, the semigroup generated by $2$ and $3$ would contain every integer at least two, leaving only the single gap $1$, a contradiction. Hence $3$ is the second gap. This proves the Genus-two Weierstrass gap sequence