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# Why is 0!=1?

Bikerman
One of my pupils asked me this a couple of weeks ago (he was busy revising for a maths exam in my IT lesson). I had to say that I didn't have a definitive answer - or even one in which I was confident.
I played with it for a while and ended up giving him one possible way of getting it....pattern continuation.
So 5!=6!/6
4!=5!/5
3!=4!/4
2!=3!/3
1!=2!/2
therefore
0!=1!/1
This works but I'm aware it isn't very rigorous and certainly wouldn't be considered a proof. Any mathematicians care to comment?
SonLight
I'm not into formal definitions either, but I think it's a matter of how you define the product of a small set of numbers.

By definition, n! is prod [i=1,n] i

prod(a,b) = a times b, for any a, b within the appropriate set of numbers.
prod(a,b,c) = prod(a,b) times c
in general, prod( set a, set b) is prod(set a) times prod(set b) if set a is at least two numbers and set b is at least one number.
now, 1! = prod(1), which already requires an extension to the product function to be defined.
so define prod(a) for single number a = 1 times a or a
and also define prod( null set ) = 1, which is the only consistent extension.

Then 0! = prod( null set ) = 1
1! = prod(1) which does not involve any multiplication, but is equal to 1 by our definition.

I see this as 1 being the multiplicative identity, just as zero is the additive identity.

Maybe someone else can use clearer terminology and formalize this.
kelseymh
 Bikerman wrote: One of my pupils asked me this a couple of weeks ago (he was busy revising for a maths exam in my IT lesson). I had to say that I didn't have a definitive answer - or even one in which I was confident. I played with it for a while and ended up giving him one possible way of getting it....pattern continuation. So 5!=6!/6 4!=5!/5 3!=4!/4 2!=3!/3 1!=2!/2 therefore 0!=1!/1 This works but I'm aware it isn't very rigorous and certainly wouldn't be considered a proof. Any mathematicians care to comment?

One way to do it is to consider the extension of factorials to the real (complex) plane, namely the Gamma function (http://en.wikipedia.org/wiki/Gamma_function).

The function, Gamma(t) = Int[0 to +infinity] x^(t-1) exp(-x) dx, is continuous for positive real numbers, with the property that Gamma(n) = (n-1)! for all integers n (which you can confirm since the integral is analytic in that case).

Plugging in n=1, we have 0! = Gamma(1) = 1.

EDIT: The integrand is x^(t-1) exp(-x), not "exp(-1)" as I originally wrote.
Bikerman
Ahhh thanks to both of you. I'll go away and play with those ideas until I understand them properly
manfer
 SonLight wrote: Maybe someone else can use clearer terminology and formalize this.

http://en.wikipedia.org/wiki/Empty_product
SonLight
Thanks, manfer. They use the proper "pi" notation for a product, which clarifies it a lot. Also I loved their comment,

 en.wikipedia.org/wiki/Empty_product wrote: The notion of an empty product is useful for the same reason that the number zero and the empty set are useful: while they seem to represent quite uninteresting notions, their existence allows for a much shorter mathematical presentation of many subjects.

And the way logarithms map empty products to empty sums, and exponentiation does the opposite.
Bikerman
Well done chaps - that is just what I needed. I don't mind admitting to a pupil that I don't know something, but I try to make sure that when we next meet I DO know
thejoyofcominghome
n! is defined as 1x2x3x...xn, but this only works for all integers > 0. So one has to extend this definition to define 0!. And there is really a choice to make, but there are several reasons to make 0!=1:

Reason 1.
Consider the row:
5!=120
4!=24
3!=6
2!=2
1!=1
When asking yourself to find the next item in the row of numbers, you'll try to see a pattern. And the pattern is /5, then /4, /3, /2, so the next one will be /1.

The same holds for 2^0=1, because it completes the row 2^3=8, 2^2=4, 2^1=2, 2^0=1 beautifully.

Reason 2.
As said before: 1 is the identity in multiplication, like 0 is in addition:
1x10=10 and 10x1=10
0+10=10 and 10+0=10
! has to do with multiplication, so "don't multiply" should mean, multiply with 1

Reason 3.
Consider binomial coefficients, where (n \above k) means the number of combinations to be formed when i pick a group of k elements from a group of n elements.
When n=k, for example i pick 4 out of 4 elements, of course the number of combinations is 1. The general formula for binomial coefficients is n!/(k!(k-n)!) . When applied to n=k=4, you get:
4!/(4!(0)!) and this is equal to 1/0!. Since this should be equal to 1, 0! should be 1.
Bikerman
Thanks for that.
I need to declare a personal interest, since I am a moderator. I am liable to treat posters who make postings which are on topic, relevant, original and interesting, with more respect than other posters. To those who think that this is unfair all I can say is .. I don't really care.
Afaceinthematrix

n! = n(n-1)!

Let n = 1

1! = 1(0)! ===> 1 = 0!

Of course... This is exactly what you did (I just abstracted it a little). I actually can't think of a more rigorous proof right now...
Afaceinthematrix
You can also think of combinations... How many combinations are there of 5 students from a pool of 5 students? We know that the answer is 1.

nCr = n!/(r!(n-r)!): 5C5 = 5!/(5!0!) = 1

The permutation formula would do the same thing.

The number of permutations of 10 objects from a pool of 10 is obviously 10!

nPr = n!/(n-r)!: 10P10 = 10!/0! = 10!

While this isn't a proof - and neither is the one that I put before - it may simply help your students believe you when you say that 0! = 1 because they can simply see that it works....

Of course the obvious issue with this demonstration (I won't call it a proof) is that we're dividing by 0! - which is something that we have not yet proved is non-zero. Given that we have not yet proved it's 1, we cannot rule out it being 0 in both cases.
johans
i don't have any idea.
Bikerman
And you thought that we needed to know that for what reason exactly? Do you normally go around writing 'I don't know' in threads like this? Seems an odd way to behave. Aside from being irritating to those contributing to or reading the thread (picking past meaningless postings is a pain), why would you waste your own time in this way?

There are tens of thousands of threads in the various groups - the science groups alone probably have several thousand threads. Please feel free to not post anything in any of them, unless you actually DO know. I think that would be best all round.

PS - In these forums there is me (GCMG)*, then there are other staff (KCMG)**, then there are the science professionals I have invited to help out, then the regular posters, and finally the lurkers and occasional contributors - IN THAT ORDER. God can join the queue at the end - in these parts he is certainly not 'first'. In fact, unless he has something useful or interesting to contribute, on matters scientific (which seems doubtful given his publication record) he had better stick to forums better suited to his 'special needs' and leave the difficult sums to us....

* GCMG - God calls me God
** KCMG - Kindly Call Me God
krackers
Essentially, because it is defined that way. Doing such makes things work out nicely in combinatorics and other branches of math. One way to gain intuition is to consider the factorial in terms of its combinatorial application, for finding the number of permutations of all the objects (nPn). In this case you can see that there is only 1 way to arrange 0 objects, and hence 0! = 1.
afsoeiro
It is certainly easier to discuss God's plans to humanity than certain Math and Physics questions...