# Negative Factorials and their Quotients

Dante

A factorial of a number is, for most cases, an equation that multiplies an integer by every other integer less than itself. Or, mathematically speaking,

$$1!=1$$
$$2!=2\cdot 1=2$$
$$3!=3\cdot2\cdot1=6$$
$$4!=4\cdot3\cdot2\cdot1=24$$
$$5!=5\cdot4\cdot3\cdot2\cdot1=120$$

Notice how 5! is just $$5\cdot4!$$, or $$5\cdot4\cdot3\cdot2\cdot1$$. It's not a particularly interesting function, outside of the caveat that 0! is also defined to be 1. However, looking through some of my old math journals, I came across an interesting side note for this subject. As you might have guessed, negative numbers are pretty much a no-no for the factorial function. (-1)! is undefined. That stated, I apparently found that certain ratios of factorials might not be so undefined, in particular, consider the following equation,

$$n^2 = \frac{(n+1)!}{(n-1)!} - n \Rightarrow{\frac{(n+1)!}{(n-1)!}=n^2+n}$$

$$\Rightarrow{\frac{(n-1)!}{(n+1)!}=\frac{1}{n^2+n}}\Rightarrow{(n-1)!=\frac{(n+1)!}{n^2+n}}$$

$$\Rightarrow{n!=\frac{(n+2)!}{(n+1)^2+n + 1}}\Rightarrow{n!=\frac{(n+2)!}{n^2+3\cdot n+2}}$$

That is, this last equation is another way of writing n!, and plugging in a few values we see that our old faithful values from before show up. Of course, what use is it? We're defining n! based on (n+2)!? That seems almost to be a waste, and as with n!, this equation is also undefined for all negative numbers. -1, -2 result in a 0 for the denominator quadratic equation, while all values less than -2 result in a negative number for the factorial itself. However, what is interesting about this equation, is that factorial quotients are defined. In particular, for any negative numbers,

$$\frac{(n+2)!}{n!}=n^2+3\cdot n+2$$

So that, using negative numbers in our factorials quotients, we get...

$$\frac{(-1)!}{(-3)!}=2$$
$$\frac{(-2)!}{(-4)!}=6$$
$$\frac{(-3)!}{(-5)!}=12$$

Of course, maybe this is just a bit of slight of mathematical hand, or maybe not. In particular, if we take the original definition of a factorial function and attempt to extend it to negative numbers, what do we get?

﻿$$\require{cancel} \frac{(-1)!}{(-3)!}=\frac{-1\cdot{-2}\cdot{\cancel{-3}}\cdot{\cancel{-4}}\cdot{...}}{\cancel{-3}\cdot{\cancel{-4}}\cdot{\cancel{-5}}\cdot{...}}=-1\cdot{-2}=2$$
﻿$$\require{cancel} \frac{(-2)!}{(-4)!}=\frac{-2\cdot{-3}\cdot{\cancel{-4}}\cdot{\cancel{-5}}\cdot{...}}{\cancel{-4}\cdot{\cancel{-5}}\cdot{\cancel{-6}}\cdot{...}}=-2\cdot{-3}=6$$
﻿$$\require{cancel} \frac{(-3)!}{(-5)!}=\frac{-3\cdot{-4}\cdot{\cancel{-5}}\cdot{\cancel{-6}}\cdot{...}}{\cancel{-5}\cdot{\cancel{-6}}\cdot{\cancel{-7}}\cdot{...}}=-3\cdot{-4}=12$$

Which matches up nice with the values we acquired before, and tells us what we should have considered all along. That the quotient of two factorials is equal to the product (either in the numerator or the denominator, depending upon the situation) of those elements that are not in  either factorial.

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