Consecutive Integers are Coprime/Proof 2

Theorem
$\forall h \in \Z$, $h$ and $h + 1$ have only two common factors, $1$ and $-1$.

That is, consecutive integers are always coprime.

Proof
Let $k \in \Z: k \mathop \backslash h$.

Also assume $k \mathop \backslash \left({h + 1}\right)$.

Thus:
 * $\exists a, b \in \N: a k = h, b k = \left({h + 1}\right)$

Then:
 * $\left({h + 1}\right) - h = b k - a k$

and so:
 * $1 = \left({b - a}\right) k$

Since the integers form an integral domain, $\left({b - a}\right) \in \Z$.

Thus either $k = 1$ and $b - a = 1$, or $k = -1$ and $b - a = -1$.

Therefore, only $1$ and $-1$ can be factors of both $h$ and $\left({h + 1}\right)$.