How can I prove from first principles that $0!$ is equal to $1$?
Prove $0! = 1$ from first principles
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0Simple. Instead of defining $0!$, define $1!$ instead, then prove that $0! = 1$. – 2016-06-24
18 Answers
I'm not sure that there is anything to prove. I think it follows directly from the definition of factorial:
$ n! := \prod_{k = 1}^n k$
So if $n=0$ the right hand side is the empty product which is $1$ by convention.
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0@Rawling You said that " "n! = number of ways of arranging n objects" convention: there is only one way of arranging no objects." I just wanted to ask that doesn't there exist infinite number of ways of arranging n objects ?? – 2016-08-14
We need $0!$ to be defined as $1$ so that many mathematical formulae work. For example we would like $n! = n \times (n-1)!$ to work when $n=1,$ ie $1! = 1 \times 0!.$ Also we require that the formula for the number of ways of choosing $k$ objects from $n$ is valid for $k=n.$ ie ${n \choose k} = \frac{n!}{k!(n-k)!}$ is valid when $k=n.$
Things need to work when we extend our definition of the factorial via the gamma function.
$\Gamma(z) = \int\limits_0^\infty t^{z-1} e^{-t} \,\mathrm{d}t,\qquad \Re(z)>0.$
The above gives $\Gamma(n)=(n-1)!$ and so we require $0!=1,$ since $\Gamma(1)=1.$
One of the simplest ways of doing this is to observe that if you have $ 6!= 720 $ then divide both sides by $6$ to get $ 5!=120 $ then divide both sides by $5$ to get $ 4!=24 $ then divide both sides by $4$ to get $ 3!=6 $ then divide both sides by $3$ to get $ 2!=2 $ then divide both sides by $2$ to get $ 1!=1 $ then divide both sides by $1$ to get $ \text{[fill in the blank here]} $
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0@MorganRodgers : ha ha...... Probably I can do that, but it's not worth doing in this context. – 2017-05-17
$0! = 1$ is consistent with, and for reasons related to, how we define the empty product.
See, for example, this entry on empty product. This is simply the name of the phenomenon Michael Hardy alludes to:
Empty product:
The empty product of numbers is the borderline case of product, where the number of factors is zero, that is, the set of the factors, is empty. In such a "borderline" case, the empty product of numbers is equal to the multiplicative identity, which is $1$.
Some of the most common examples are the following:
- The zero$^{\text{th}}$ power of a number $a$: $a^0 = 1$,
- The factorial of $0$: $0! = 1$,
- The prime factor presentation of unity, which has no prime factors.
Just as ${n^0 = 1}$ for any $n$, and the "prime factorization" of $1$ = $1$, we define, as a matter of convention, $0! = 1$.
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0`empty product of numbers is equal to the multiplicative identity` - means defined to be multiplicative identity ? – 2018-10-09
Because there is only one way to do nothing.
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1@columbus8myhw, haha. – 2015-03-26
The empty product is taken to be equal to 1. Take logs and you get an empty sum equal to zero, which is somehow more intuitive, but this trick of taking logs to convert a product into a sum never seems to get a mention in the literature. [Assumes products have positive terms]
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3@robjohn:L Indeed - I still have my slide rule, and ten years ago I used it as a "mystery object" for a youth group - they didn't know what it was. I remember using it in A-level chemistry practicals to set up the instants, and do the calculations quicker than schoolmates with their Sinclair calculators! But the point I was making was more that taking logs can take the mystery out of infinite products (and the notion of convergence applicable) by turning them into (less elementary) infinite sums. – 2011-09-26
One can prove this by convention (that is based on what 'works', based on base cases for product of a list).
But for a proof from 'nothing' as it were...define what factorial is supposed to mean. I think of it as the number of one-to-one, onto (=bijective) functions from a set of size $n$ to itself. if$n$ is 0 then the set is the empty set. How many bijective functions are there from $\emptyset$ to $\emptyset$? (a function is a set of ordered pairs (with some restrictions)? There are no legal pairs for these restrictions, and so there are no legal subsets of these pairs...other than the empty set. So $\emptyset$ -is- a function (it -is- a set of pairs (empty of course), all of whose pairs satisfy the function criteria). No other functions will work so $\emptyset$ is the only function that works. So there is only 1 bijective function on a set of size 0. So $0! = 1$.
Yes, this is weird, but it works. Negative numbers, complex numbers, they're all weird even when you just manipulate their properties. But you'll get over it.
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2I remember in my college days getting drunk and using this argument in an attempt to settle this $0! = 1$ matter. See also https://math.stackexchange.com/a/789411/432081 – 2017-09-09
Let's try a different approach from my other answer to this:
To multiply a number $N$ by $6!$ is to multiply it by six factors and get $ N\cdot1\cdot2\cdot3\cdot4\cdot5\cdot6. $ Similarly to multiply $N$ by $0!$ is to multiply it by $0$ numbers: $ N. $ But that is the same as multiplying it by $1$ $ N\cdot1. $ Multiplying $N$ by no numbers at all is multiplying $N$ by $1$.
(This answer doesn't apply only to factorials; it may be taken as a general explanation of why, when one multiplies no numbers, one gets $1$.)
Explanation 1: We define $n!$ as the product of all integers $k$ with $1\le k \le n.$ When $n = 0$ this product is empty so it should be 1.
Explanation 2: If $n$ is a nonnegative integer, we define $n!$ to be the number of orderings on a set with $n$ distinct objects. If $n = 0$, this set is empty. Vacuously, it has 1 order.
You can define $\exp(x)$ as
$ 1 + \frac {x} {1!} + \frac {x^2} {2!} + \frac {x^3} {3!} + ... $ but the following seems more uniform: $ \frac {x^0} {0!} + \frac {x^1} {1!} + \frac {x^2} {2!} + \frac {x^3} {3!} + ... $ These are only equal if $0! = 1$
Not a proof, but makes sense to me!
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3In think that, in the context of power series, $0^0$ is always taken to be $1$. This is even used in other power series like $\left(1+x\right)^{-1}=\sum \left(-1\right)^n x^n = 1-x+x^2-x^3+\cdots$, so that it has nothing to do with the factorial at all – 2015-01-06
$0! = 1$ usually is one of the "first principles".
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0That would make sense. I thought it was always proven by definition. – 2016-06-03
Similar to @Mitch's answer, but with a slight twist at the end.
For any $n$, we define $n!$ to be the the number of invertible functions from a set of size $n$ to itself. The question is thus restated as "Prove that there is exactly one invertible function from the empty set to itself."
Now, it should be clear that there can't be more than one function from $\emptyset$ to itself, because in order for two functions to be different, they have to give different values for at least one input (clearly impossible if the domain is empty). The question therefore reduces to "Prove that there is at least one invertible function from the empty set to itself."
Well, assuming we can all agree that the empty set is a set, meaning an object in the category of sets, the empty set must have an identity function (morphism) defined on it; every object in every category is required to have an identity morphism. Further, identity morphisms are always invertible because $Id\circ Id = Id$, so we are done. $\Box$
http://en.wikipedia.org/wiki/Category_of_sets
This answers the question if and only if the "first principles" include axioms of category theory (and a few facts about sets).
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0Why isn't this the accepted answer? It is by far the best. – 2015-03-26
${n \choose R} = \frac{n!}{R!(n-R)!}$
You know: ${n \choose 0}=1$, so: $\frac{n!}{0!n!}=1\implies\frac{1}{0!}=1\implies 1=0!$ We know every set has $2^n$ subsets we know empty subset is unique so
First we state that $n!=\frac {(n+1)!}{(n+1)}$. Now let $n=0$ then $0!=\frac{(0+1)!}{(0+1)}=\frac{1!}{1}=1$ Hence $0!=1$
The way, I think about it is via permutations. Suppose you have $n$ objects then to permute them you have $n!$ factorial ways. So, for example, if you had $3$ objects then to permute them you would do $3!$ which is $6$. Suppose, you had $0$ objects then how many times could you permute $0$ things? Once. Therefore, $0!$ should be $1$. This is just an intuitive explanation and a proof by no means.
$n!$ is defined as the number of way n distinct object can be arranged.
For example,
$1!$ is described to arrange $1$ object,i.e $1$ way.
$A$
$2!$ is described to arrange $2$ object,i.e $2$ way.
$AB,BA$
$3!$ is described to arrange $3$ object,i.e $6$ way.
$ABC,ACB,BAC,BCA,CAB,CBA$
Similary, $0!$ describe the way to arrange $0$ object.$0$ object means nothing.There comes a hypothetical thinking,there is $1$ way to arrange $0$ object or nothing,that is nothing.
So, $0!=1$
You cannot prove that $0!=1$ from first principles, because it is a convention. First, we define "$!$" as follows:
$n!=\begin{array}{c}n\\ \prod\\ k=1 \end{array}k=1\cdot2\cdot3\cdot...\cdot(n-1)\cdot n$
This has no meaning for $n=0$ . But it would be useful if it had one. So, we make the convention that $0!=1$ . This is the best choice, because:
For $n>1$ , $n!=n\cdot(n-1)!$ If we want to extent this for $n=1$ , we must have: $1!=1\cdot0!\Rightarrow1!=0!\Rightarrow0!=1$.
There are $n!$ ways to arrange $n$ objects into a sequence, for $n\geq1$. For zero objects, there is only one arrangement. So, if we want to extent the formula $n!$ to be applicable to zero objects, we must have $0!=1$.
A taylor series is an infinite series of the form: $f(a)+\dfrac{f'(a)}{1!}\left(x-a\right)+\dfrac{f''(a)}{2!}\left(x-a\right)^{2}+\dfrac{f'''(a)}{3!}\left(x-a\right)^{3}+...$
If we aggre that $0!=1$ , we can write the same thing using the sigma notation: $\begin{array}{c} \infty\\ \sum\\ k=0 \end{array}\dfrac{f^{(k)}(\alpha)}{k!} \left(x-a\right)^{k}$
From these and other cases (see previous answers), we conclude that the best choice is to give $0!$ the value 1.
Also remember that $n! $ is the number of ways to arrange $n$ objects. We have only one way to arrange $0$ objects, i.e; $0!=1$.