Can it extend to 0?

Note that any triple $(a, b, c) = (1, 2, c)$ satisfy the given inequality. So, $\dfrac{a}{b}$ takes the value $\dfrac{1}{2}$ .

Now we prove that there is no other solution for $b \ge 2a$ .

Let $t = |a^c - b!|$ .

If $t > 0$ , $1 =\left|\dfrac{a^c}{t}-\dfrac{b!}{t}\right|$ .

Since $t \le b$ , $\dfrac{b!}{t}$ is an integer, so $\dfrac{a^c}{t}$ is also an integer.

Furthermore, $2a \le b \Rightarrow a, 2a in \{1, 2, . . . , b\}$ . At least one of $a$ and $2a$ is different from $t$ , so it is not canceled out from the product in $\dfrac{b!}{t}$ . So, $a \mid \dfrac{b!}{t}$ .

Therefore, since the difference between $\dfrac{b!}{t}$ and $\dfrac{a^c}{t}$ is 1, $\gcd \left(a,\dfrac{a^c}{t}\right)=1$ implying $t = a^c$ .

Thus, we are now left with two cases only: $|a^c - b!| = 0$ or $|a^c - b!| = a^c$ .

These cases reduce to $b! = a^c$ and $b! = 2a^c$ respectively.

Either way, $2a - 1 \in\{1, 2, . . . , b\}$ , so $(2a - 1) \mid b!$ or $(2a-1) \mid 2a^c$ .

Now $\gcd(2a - 1, 2a^c) = 1$ , implying $2a - 1 = 1$ and $a = 1$ .

If $a = 1$ , $b! - b \le 1 \Rightarrow b le 2$ . So, $(a, b) = (1, 2)$ is the only solution at $b \le 2a$ .

So minimum value of $\dfrac{a}{b}$ is $\boxed{\dfrac{1}{2}}$ .