Well, I can think about 2 more ways why we should define 0!=1
I. n!=1*2*3*4*...*(n-1)*n
— this is definition of n!.
n! is multiplication of n factors starting from 1,2,… till n.
For example, 5!
is multiplication of 5 factors 1*2*3*4*5 or 5!=1*2*3*4*5
.
Similarly, 2!
multiplication of 2 factors 1*2
or 2!=1*2
.
How much is 1!
? 1! is multiplication of 1 factor 1, 1!=1
(you can ask what does multiplication of 1 factor mean? Well, there is “natural” definition. Because n*1=n
you can think of multiplication of 1 factor n as simple multiplication n by 1, that is multiplication of 1 factor 1 is m*1=m
)
How much is 0!
? 0!
should be multiplication of 0 factors, so called “empty product”. The value of “empty product” is 1. (Again, it is because n*1=n
, so for any product we can just add 1 to the product. Alternatively, it can be derived from “empty sum”) as following:
Since logarithms turn products into sums:
they should map an empty product to an empty sum. If we define the empty sum to be 0, then the empty product should be e⁰= 1
.
(Conversely, if we define the empty product to be 1, then the empty sum should be ln 1 = 0.
).
II. Another way.
- From the formula
n!=1*2*3*4..*(n-1)*n
we can define recursive formula as follows:
For n>2n!=n*(n-1)!
(note, that n-1>1>0)
For n=22!=2
(and1!=1
by definion)
Let try it:
5!=5*4!
4!=4*3!
3!=3*2!
2!=2 (base case)
so
3!=3*2=6
4!=4*6*24
5!=5*24=120
But why we chose as base case n=2? Let put it n=1. - For n>1
n!=n*(n-1)!
(note, that n-1>0)
For n=11!=1
So, if from previous derivation we have2!=2
by definition (of the base case), now we can calculate it:2!={2>1}=2!=2*(2–1)!=2*1!={1!=1 by definition of the base case}=2*1*2
that is consistent with the previous definition.
Can we go one step further? Can we define base case forn=0
? Let’s try. - For n>0
n!=n*(n-1)!
(note, that n-1>-1, that is for n=0,n-1
will be negative, and for n=1,n-1
will be 1)
For n=00!=?
Well, let’s try to figure out what is 2!
2!={2>0}=2*(2–1)!=2*1!
1!={1>0}=1*(1–1)!=1*0!=0!
In order that this definition will be consistent with previous ones above, we should define 0!=1. That is we should define:
For n>0n!=n*(n-1)!
(note, that n-1>-1, we can’t put n=0 to this formula, only n>0)
For n=00!=1
In this case
1!=0!=1
as expected.
Originally published in my blog https://www.toalexsmail.com/2012/04/why-zero-factorial-equals-one-english.html