If a different factorial function would've been defined, what would the graph of this function look like?

Question

Let's define a new, different factorial function, such that $\frac{a}{b}!$ returns $\frac{a!}{b!}$. How would the graph of the function look like? For example, $\frac{1}{2}!$ does not return $\frac{\sqrt{\pi}}{2}$, instead $\frac{1!}{2!}$ (or just $\frac{1}{2}$).

Answer 1

There would be a number of dissatisfying properties that such a function would have:

• First, note that the fraction $\frac{a}{b}$ would have to be in lowest terms. Otherwise, the function wouldn't be well-defined: would $\left(\frac{1}{2}\right)!$ equal $\left(\frac{2}{4}\right)!?$
• It would also be highly erratic. Note that $1!=1$, but $\left(\frac{400,001}{400,000}\right)!=400,001$. In any neighbourhood of $1$, we could find arbitrarily large and arbitrarily small outputs. This makes it very hard to graph the function.
• It would only be defined on the positive rationals.

Answer 2

Let $f:x\mapsto f(x)$ be such a (real-valued) function, on a real domain to be determined which includes the positive integers. Since $f(x/y)=f(x)/f(y)$ for all $y\in\operatorname{dom}f$, we must have $0\notin\operatorname{dom}f$ and $0\notin\operatorname{range}f$. Setting $y=1$ gives $f(1)=1$. Also, for $y\neq0$, we have $f(x)=f(xy/y)=f(xy)/f(y)$. Hence $$f(xy)=f(x)f(y).$$ Power functions such as $f:\Bbb R_{>0}\to\Bbb R:x\mapsto x^a$, with constant $a\in\Bbb R$, fulfil this and the original condition. There are doubtless lots of pathological noncontinuous functions that do so as well