Does the definition of factorial fail for $0!$

Question

How does one understand $0!=1$. Is it purely a theoretical value for convenience? as the definition of factorial i.e., $n!=n\times(n-1)\times(n-2)\times...1 $ actually fails here. How can it be logically understood by the definition of factorial.

score 3 · Answer 1 · answered Jul 10 '17 at 18:19

3

If you want to look for a logical way why $0! = 1$. Don't look at the definition, but at what it means in other fields. A factorial is also the amount of ways you can arrange a certain amount of objects. So $0!$ can be seen as the amount of ways you can arrange $0$ objects, which is in only $1$ way.

answered Jul 10 '17 at 18:19

TSpoon

339

Seems bit dubious. How can one arrange 0 objects by 1 way? one can easily argue that 0 objects can only be arranged in 0 ways. That is why it seems derived from inference rather than truly from a mathematical rigor or even logic – user6775 Jul 10 '17 at 20:13

score 2 · Accepted Answer · answered Jul 10 '17 at 18:16

2

It is somewhat for convenience. One reason is that we can think of the factorial as satisfying the recurrence relation $(n+1)! = (n+1)n!$, which if we extend the definition to $n=0$ then it is only compatible if we have $$1 = 1! = 1\cdot 0!$$ which means that we have to take $0! = 1$.

answered Jul 10 '17 at 18:16

DMcMor

9,407

this seems closer to reality. this also means that $0!$ is taken from inference as it satisfies the formula rather than from a purely mathematical logic in a real sense – user6775 Jul 10 '17 at 20:17

score 2 · Answer 3 · answered Jul 10 '17 at 18:54

Another approach is $n! = \prod_{i=1}^{n} i$. Now similarly to: the sum of all elements of the empty set is the additive neutral element ($\sum_{i\in\emptyset} i = 0$) we can say that the product of all elements of the empty set is the multiplicative neutral element ($\prod_{i\in\emptyset} i = 1$).

Taking this to account we receive $$0! = \prod_{i=1}^{0} i = \prod_{i\in\emptyset} i = 1$$

score 1 · Answer 4 · answered Jul 10 '17 at 18:35

1

Count backwards: $$(n-1)! = \frac{n!}{n}$$ $$0! = \frac{1!}{1} = \frac{1}{1} = 1$$

answered Jul 10 '17 at 18:35

md2perpe

26,770

1

yes but this is derived by inference not through the rigor of mathematics – user6775 Jul 10 '17 at 20:12
We are talking about extending a definition. That can not even be derived rigorously, only motivated. I gave one motivation, which is essentially the same as the one DMcMor gave. Nathanael Skrepek gave another based on the interpretation of an empty product. – md2perpe Jul 10 '17 at 20:35

Does the definition of factorial fail for $0!$

4 Answers4