Why is 0! equal to 1?

Answer 1

0!= 1! = 1

The definition incorporates the convention that

as an instance of the convention that the product of no numbers at all is 1. This fact for factorials is useful, because

---the recursive relation (n + 1)! = n! × (n + 1) works for n = 0;
---this definition makes many identities in combinatorics valid for zero sizes.
In particular, the number of combinations or permutations of an empty set is, clearly, 1.

Answer 2

It is not possible to equate the value of 0! using the normal formula [n! = n x (n-1) x (n-2) x ... x 2 x 1] because 0 is a lower number than 1. But it is possible to equate 0! because 0 isn't a negative number. To equate the value of 0!, you have to use a different method, here's how it works:

5! equals 120
4! equals 24, this means that 4! also equals 5!/5 (120/5 = 24)
3! equals 6, this means that 3! also equals 4!/4 (24/4 = 6)
2! equals 2, this means that 2! also equals 3!/3 (6/3 = 2)
1! equals 1, this means that 1! also equals 2!/2 (2/2 = 1)
Generally speaking, you can state that n! = (n+1)!/(n+1)

Therefore, 0! has to equal 1!/1 and 1!/1 equals 1

Therefore, 0! = 1

Answer 3

when thinking about combinations we can derive a
formula for "the number of ways of choosing k things from a collection
of n things." The formula to count out such problems is n!/k!(n-k)!.
For example, the number of handshakes that occur when everybody in a
group of 5 people shakes hands can be computed using n = 5 (five
people) and k = 2 (2 people per handshake) in this formula. (So the
answer is 5!/(2! 3!) = 10).

Now suppose that there are 2 people and "everybody shakes hands with
everybody else." Obviously there is only one handshake. But what
happens if we put n = 2 (2 people) and k = 2 (2 people per handshake)
in the formula? We get 2! / (2! 0!). This is 2/(2 x), where x is the
value of 0!. The fraction reduces to 1/x, which must equal 1 since
there is only 1 handshake. The only value of 0! that makes sense here
is 0! = 1.

And so we define 0! = 1.

Answer 4

Your simple and reasonable question has produced some really bad answers; either ignorant (0! IS = 1) and offensive (why?) or far too mathematical.

I like to think of it like this. You are a photographer. You wish to take a photo of 5 people lined up against a wall. How many different photos can you take? 5x4x3x2x1 = 5! (=120). If you have 4 people, you can take 4! photos. For 3 people, it's 3!. For 2 people it's 2! and for 1 person it's 1!. It therefore makes sense that for 0 people, it is 0! and clearly you can take one photo (showing the empty wall).

Of course, ultimately, it's just a definition and (believe me) LOADS of formulae work really very nicely if 0! = 1 and would have lots of exceptions and problems if 0! = 0.

Hope you aren't too dis-spirited by the poor responses and hope this one helps.

Answer 5

From http://mathforum.org/library/drmath/view/57128.html

Date: 03/18/98 at 16:39:40
From: Doctor Sam
Subject: Re: 0 factorial = 1

Denise,

You are correct that 0! = 1 for reasons that are similar to why
x^0 = 1. Both are defined that way. But there are reasons for these
definitions; they are not arbitrary.

You cannot reason that x^0 = 1 by thinking of the meaning of powers as
"repeated multiplications" because you cannot multiply x zero times.
Similarly, you cannot reason out 0! just in terms of the meaning of
factorial because you cannot multiply all the numbers from zero down
to 1 to get 1.

Mathematicians *define* x^0 = 1 in order to make the laws of exponents
work even when the exponents can no longer be thought of as repeated
multiplication. For example, (x^3)(x^5) = x^8 because you can add
exponents. In the same way (x^0)(x^2) should be equal to x^2 by
adding exponents. But that means that x^0 must be 1 because when you
multiply x^2 by it, the result is still x^2. Only x^0 = 1 makes sense
here.

In the same way, when thinking about combinations we can derive a
formula for "the number of ways of choosing k things from a collection
of n things." The formula to count out such problems is n!/k!(n-k)!.
For example, the number of handshakes that occur when everybody in a
group of 5 people shakes hands can be computed using n = 5 (five
people) and k = 2 (2 people per handshake) in this formula. (So the
answer is 5!/(2! 3!) = 10).

Now suppose that there are 2 people and "everybody shakes hands with
everybody else." Obviously there is only one handshake. But what
happens if we put n = 2 (2 people) and k = 2 (2 people per handshake)
in the formula? We get 2! / (2! 0!). This is 2/(2 x), where x is the
value of 0!. The fraction reduces to 1/x, which must equal 1 since
there is only 1 handshake. The only value of 0! that makes sense here
is 0! = 1.

And so we define 0! = 1.

Answer 6

This is an instance of the convention that the product of no numbers at all is 1. For a discussion of this convention see http://en.wikipedia.org/wiki/Empty_product

This fact for factorials is useful, because
the recursive relation (n + 1)! = n! × (n + 1) works for n = 0;
this definition makes many identities in combinatorics valid for zero sizes. In particular, the number of combinations or permutations of an empty set is, clearly, 1

An alternate route to 0!= 1 is as follows

n! = 1.2.3.4. ... .(n-1).n

So, by definition,

(n!)(n+1) = (n+1)!

5! = 5x4x3x2x1 = 120
4! = 4x3x2x1 = 24
3! = 3x2x1 = 6
2! = 2x1 = 2
1! = 1 = 1
0! = 1

As we go down the lines, we find the next one by dividing by 5,4,3,2 and 1

Answer 7

It needs to be defined this way so that the formulas in combinatorics work properly. Otherwise the forumlas would break down and we wouldn't be able to figure out, for example, how many subsets of size 0 or size 1 we can take from a set. In the same way, zero exponent is defined as 1 because it works well this way in terms of the rules for exponents.

So why don't we just define division by zero? Well, 0! can be easily defined in a way that is consistent with the other basic facts of combinatorics. But division by zero cannot be defined in any nice consistent way. About the best we can do is define it in specific circumstances such as limits of trig ratios, using big guns like the squeeze theorem and L'hospital's rule. Even then, it's iffy, applying only to specific circumstances. There is no easy general definition for division by zero like there is for 0!, that's why we leave division by 0 undefined.

Answer 8

Practically all solutions showed arrive at the point where they intend to do:
0! = 1
But I am not sure if this is correct, because if you think the factorial like a sequence of multiplications, then the final value for 0! must be 0 and not 1 because all numbers *0=0.

Answer 9

0! = 1 is a fact because
We have (n+1)! = n!*(n+1), put n = 0
We now have 1! = 0! * 1 => 0! = 1!/1 = 1. QED

Like wise: X^0 = 1 is a fact because
X^0 = X^(1-1) = X^(1)* X^(-1) = X/X = 1 (where X is not 0). QED.

Answer 10

This (0!=1) can be seen as a sensible convention, but also it is simply a fact when n! is defined as the number of permutations of n objects. A permutation of n is a one-to-one function from a set with n elements into the same set. When n=0, there is a single function defined on the empty set: its is the empty function (its graph is the empty set). It may seem strange, but it does satisfy the usual definition of a function in set theory.

Answer 11

Where do you start to define?

With integers only, factorization can be defined as:
0!=1 , n!=(n-1)!*n where n>=1
From this it automatically follows, that 0!=1!=1

You can also start with higher mathematics, with complex numbers (z=x+y*i, where i is defined by i*i = -1), with Euler's Gamma-function: Gamma(z+1)=Pi(z) for complex numbers z (z may not be any negative integer: z<> -1, -2,...)

(Note that Gamma and Pi should be written as greek upper case letters.)

And Pi(n)=n! for integer numbers n>=0

From the definition of the Gamma function it automatically follows, that Gamma(1)=Pi(0)=0!=1 and Gamma(2)=Pi(1)=1!=1

See also http://numbers.computation.free.fr/Constants/Miscellaneous/gammaFunction.html