Let $a \in \mathbb{Z}$ and let $b \in \mathbb{N}$. To prove the division algorithm, we defined the set $S$ by $$S = \{a-kb \, | \, k \in \mathbb{Z} \text{ and } a-kb \geq 0\}$$ In order to use the well-ordering property, we first must demonstrate that $S \neq \emptyset$.
The texts I've read usually adopt a general approach in showing that $S$ is nonempty, like one book says that because $b\geq 1$we have $$a - (-|a|)b = a+ |a|b \geq a + |a| \geq 0$$ Choosing $k = -|a|$, it follows that $a - kb \in S$.
However, my question is if this more abstract approach is necessary. Would it be sufficient to show that just one particular number belongs to the set, like $0$? Choosing $a = k = 0$, it follows that $a - kb \geq 0$, and thus $0 = a-kb \in S$.