0!=1

For the discussion of math. Duh.

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NOT Steve
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0!=1

Postby NOT Steve » Wed Jun 06, 2007 1:13 am UTC

Can someone explain to me why this is the case? I know it makes things work nicely in probability problems, but is their any actual mathematical explanation for it?
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Postby crazyjimbo » Wed Jun 06, 2007 1:18 am UTC

If you consider factorials to be special integer cases of the gamma function, then 0! = 1 is no longer by definition, but is instead the same case as Gamma(1) = 1.

I don't know how or why it was initially defined, but this is a good enough reason for me.

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Postby dp » Wed Jun 06, 2007 1:23 am UTC

An intuitive way to think about it is that n! = n*(n-1)!. So 1! = 1 * 0!, which suggests 0! = 1.

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Postby Woxor » Wed Jun 06, 2007 1:35 am UTC

dp wrote:An intuitive way to think about it is that n! = n*(n-1)!. So 1! = 1 * 0!, which suggests 0! = 1.

Heh, that sort of works for negative integers, too.

0! = 0*(-1)!, so (-1)! = 1/0, which is undefined, just like the gamma function at zero. It's increasing to infinity from one side, and decreasing to negative infinity on the other side.

(-1)!=-1*(-2)!, so (-2)!=-1/0. Undefined again, but with opposite limits from either side, and so on for the rest of the negative integers.

It makes intuitive sense, even if the mathematics is nonsense. Actually for 0!, the given method is totally valid for computing Gamma(1) from Gamma(2)=1!. But the rest is Pretty Picture Mathematical Fun Time more than a proof.
Last edited by Woxor on Wed Jun 06, 2007 1:39 am UTC, edited 1 time in total.

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Postby EradicateIV » Wed Jun 06, 2007 1:38 am UTC

Hahaha, flashback from my first course in discrete mathematics (tons of easy subjects as a simple CS credit), I asked why 0! was 1.

The professor just said because if it wasn't, nothing would work properly.
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Postby SpitValve » Wed Jun 06, 2007 1:48 am UTC

Ha! I have been doing too much programming, because I thought somebody was asking us to prove that zero is not equal to one.

<ahem> never mind :P

Also 0! = 1 fits the form for the 0th order term in the Taylor series...

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Postby crazyjimbo » Wed Jun 06, 2007 1:52 am UTC

SpitValve wrote:Ha! I have been doing too much programming, because I thought somebody was asking us to prove that zero is not equal to one.


Well clearly it is equal... 0! = 1! => 0 = 1. Q.E.D.

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Postby Buttons » Wed Jun 06, 2007 2:12 am UTC

If we define n! (for nonnegative integers n) to be the number of bijections from a set of cardinality n to itself, then 0! should be 1, since there's exactly one function (the "empty function") from the empty set to itself. This is also the justification for why we have 0^0 = 1, rather than 0.

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Postby NOT Steve » Wed Jun 06, 2007 2:13 am UTC

All right. I'm just fourteen so I don't really know gamma yet, but I can accept the other explanation of x!=x*(x-1)! since that seems to work out well. Thanks.
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Postby themandotcom » Wed Jun 06, 2007 2:33 am UTC

But what I don't get is why 1^infinity is undefined. I can clearly express it as a limit 1^x as x->infinity should equal one. Does anyone know why?
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Postby skeptical scientist » Wed Jun 06, 2007 2:50 am UTC

It's an "indefinite form" with respect to limits, since if x_n tends towards 1 and y_n tends towards infinity, x_n^y_n could take on infinitely many values. Sure if x_n stays 1 you get 1, but if x_n is the sequence (1+1/n) and y_n is the sequence n, then the limit is e, for example. For other applications it may make sense to treat 1^oo differently.
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Postby Blatm » Wed Jun 06, 2007 3:07 am UTC

crazyjimbo wrote:
SpitValve wrote:Ha! I have been doing too much programming, because I thought somebody was asking us to prove that zero is not equal to one.


Well clearly it is equal... 0! = 1! => 0 = 1. Q.E.D.


I loled.

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Postby ptveite » Wed Jun 06, 2007 3:38 am UTC

n! is the number of permutations on n elements. There is exactly one permutation on 0 elements, the empty map.

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Postby Cosmologicon » Wed Jun 06, 2007 7:00 am UTC

As far as I know, every possible line of reasoning you could use to give 0! a value leads to the value 1. It's not like 0^0 or anything. Am I right? Or is there any reason at all to give it another value?

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Postby Peshmerga » Wed Jun 06, 2007 7:31 am UTC

Shit, joke stoled. Should read the thread...
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Postby deadly penguin » Wed Jun 06, 2007 10:16 am UTC

Buttons wrote:This is also the justification for why we have 0^0 = 1, rather than 0.

I always thought that was undefined (it gives me errors in my calculator >.<)

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Postby PaulT » Wed Jun 06, 2007 11:07 am UTC

If you think of it as n! being the product of x over all 1<=x<=n (x in the naturals), then 0! is the product over the empty set, and makes sense to define this as one, because if you're taking products, you have to 'start' with a 1 seeing as its the multiplicative identity. Plus it makes everything else work properly.

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Postby kira » Wed Jun 06, 2007 11:45 am UTC

deadly penguin wrote:
Buttons wrote:This is also the justification for why we have 0^0 = 1, rather than 0.

I always thought that was undefined (it gives me errors in my calculator >.<)


Unless they have changed math in the past few years, 0^0 IS undefinied. Indeterminate form, L'Hopital's Rule, all that.

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Postby Andrew » Wed Jun 06, 2007 1:58 pm UTC

SpitValve wrote:Ha! I have been doing too much programming, because I thought somebody was asking us to prove that zero is not equal to one.

Yeah, and there was nothing in the post that suggested otherwise -- if 0==1 then probability problems would indeed be troublesome.

kira wrote:
deadly penguin wrote:
Buttons wrote:This is also the justification for why we have 0^0 = 1, rather than 0.

I always thought that was undefined (it gives me errors in my calculator >.<)


Unless they have changed math in the past few years, 0^0 IS undefinied.


Nullity!

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Postby Buttons » Wed Jun 06, 2007 4:24 pm UTC

kira wrote:Unless they have changed math in the past few years, 0^0 IS undefinied. Indeterminate form, L'Hopital's Rule, all that.

I guess it depends on the context, then. In set theory, 0^0 is most definitely defined, but then again set theorists don't have to worry about continuity as much.

Google seems to think that it's 1, whereas my calculator doesn't. Interesting.

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Postby Yakk » Wed Jun 06, 2007 4:52 pm UTC

kira wrote:
deadly penguin wrote:
Buttons wrote:This is also the justification for why we have 0^0 = 1, rather than 0.

I always thought that was undefined (it gives me errors in my calculator >.<)


Unless they have changed math in the past few years, 0^0 IS undefinied. Indeterminate form, L'Hopital's Rule, all that.


Set theoretically, a^b is the set of functions from b to a.

Numbers are sets whose cardinality is their value.

So 1 is a set with 1 element, 2 is a set with 2 elements, etc.

Then 2^2 is the number of functions from a set with 2 elements to a set with 2 elements -- 4.

1^1 is the number of functions from a set with 1 element to a set with 1 element -- 1.

0^0 is the number of functions from a set with 0 elements to a set with 0 elements -- 1.

A "function" being a set of pairs {(a,b)} such that the 'a's are unique. If (x,y) is in a function f, then the function f is said to map x to y (ie, f(x)=y).

The domain of a function is the set of 'a's, and the range of a function is the set of 'b's. A function is said to belong to A->B (the set of functions from A to B) if the domain of the function is A, and the range is contained in B.

This means there is one function in {}->{}: namely {}. ({} is a set of pairs -- an empty set of pairs, but still a set of pairs. The domain of {} is {}. The range of {}, which is {}, is a subset of {}. Thus, {} is a function from {}->{}.)

In {1}->{1} there is one function: namely {(1,1)}. Note that {} does not qualify -- the domain of {} is {}, which does not equal {1}.

Note that 1 and {1} are both "sets containing 1 element", so for this purpose they are relatively interchangeable. Often in set theory, the natural number n is denoted as the set containing it's predicessors: so 0 = {}, 1 = {0} = {{}}, 2 = {1,0} = {{0},0} = {{{}},{}}, etc.

In 2->2 (aka 2^2), there are 4 functions. {Identity, {(1,1),(2,1)}, {(1,2), (2,2)}, {(1,2),(2,1)}}


etc. :)

So by that definition, 0^0 is clearly 1.

Edit: changed at 0 to a 1, d'oh! Also added green-text more explicit descriptions of what is going on.
Last edited by Yakk on Wed Jun 06, 2007 5:45 pm UTC, edited 2 times in total.

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Postby EradicateIV » Wed Jun 06, 2007 5:07 pm UTC

*anger*

The empty set {} is not the same as { {} } or { *null char* }...

Becoming a math major is using your intuition but not letting it get in the way of new ideas, at least for me. 3 more years to go... *sigh*

0! = 1, why? I'm sure the proof is in "The Book".
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Postby skeptical scientist » Wed Jun 06, 2007 5:07 pm UTC

Yakk, you're contradicting yourself.
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Postby Yakk » Wed Jun 06, 2007 5:43 pm UTC

skeptical scientist wrote:Yakk, you're contradicting yourself.


Thanks. Fixed my typo (sigh), and added explanations as to why {} is a function. :)

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Postby SpitValve » Wed Jun 06, 2007 10:11 pm UTC

dammit! Everytime I read this I read "zero not equal to one"

I still can't read != as "factorial equals"... it just feels wrong to me...

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Postby Sygnon » Wed Jun 06, 2007 10:41 pm UTC

This is really just a rewording of some posts but sometimes phraseology can make all the difference,

if the factorial function describes the number of arrangements of n elements, there is only one way you can arrange zero elements.

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Postby hencethus » Wed Jun 06, 2007 10:45 pm UTC

SpitValve wrote:dammit! Everytime I read this I read "zero not equal to one"

I still can't read != as "factorial equals"... it just feels wrong to me...


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

I'd say that the thread title is ambiguous, but this is the mathematics forum.

Also, for anyone that's interested, the Wikipedia article on the empty product concept is pretty relevant and informative.

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Postby deadly penguin » Wed Jun 06, 2007 11:24 pm UTC

The reason I thought 0^0 was undefined was because it seems like 0^0=0/0

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Postby PaulT » Wed Jun 06, 2007 11:40 pm UTC

deadly penguin wrote:The reason I thought 0^0 was undefined was because it seems like 0^0=0/0

This is what I thought, admittedly mostly based on the nullity guy who 'proved' that 0^0 was nullity by defining 0/0 as nullity and reasoning 0^0 = 0^(1-1) = 0^1 * 0^-1 = 0/0 = nullity! Essentially he'd come up with a new way to pronounce "0/0" and hence was featured in television news broadcasts. I feel I could contribute to mathematics similarly.

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Postby SpitValve » Wed Jun 06, 2007 11:47 pm UTC

hencethus wrote:(0! = 1) != (0 != 1)


heh, no way that would confuse anyone :D

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Postby shill » Thu Jun 07, 2007 1:23 am UTC

Sygnon wrote:This is really just a rewording of some posts but sometimes phraseology can make all the difference,

if the factorial function describes the number of arrangements of n elements, there is only one way you can arrange zero elements.

Nah, that phraseology sucks, because one might be inclined to say that you can't arrange zero elements at all, and thus the answer should be zero.

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Postby aguacate » Thu Jun 07, 2007 2:41 am UTC

shill wrote:
Sygnon wrote:This is really just a rewording of some posts but sometimes phraseology can make all the difference,

if the factorial function describes the number of arrangements of n elements, there is only one way you can arrange zero elements.

Nah, that phraseology sucks, because one might be inclined to say that you can't arrange zero elements at all, and thus the answer should be zero.


I'm inclined to say that there are an infinite number of ways to arrange zero elements.

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Postby PainGod » Thu Jun 07, 2007 3:11 am UTC

0/0 is undefined, and yet somehow I still remember doing summation problems where the prof just says,"and let's assume 0/0=0 for simplicity".

Being a math major, I don't care what the definition is. I just care that it is defined.

Question: Name one way that you can arrange 0 elements. Now, tell me why doing nothing is considered something.

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Postby EradicateIV » Thu Jun 07, 2007 3:16 am UTC

PainGod wrote:Question: Name one way that you can arrange 0 elements. Now, tell me why doing nothing is considered something.


Before someone who is more education jumps on this question... I'll try to answer.

Consider the empty set, it is a set containing no elements. It is still something.
A function is just a subset of a Cartesian product.
The Cartesian product of an empty set onto an empty set is *alas* an empty set.
Therefore, the only function that can be a subset of that, would be, the empty set function.
And that, in itself, is one thing.

I hope I didn't butcher any definitions with this.
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Postby Yakk » Thu Jun 07, 2007 5:06 am UTC

An arrangement is a bijective function from your set into the set of numbers {1 ... n}, where n is the size of your set.

It assigns each element of your set the order it is in the arrangement.

A function is, if you recall, a set of pairs (source, dest), such that source ranges over everything in your source set exactly 1, and dest ranges over some subset of your destination set.

A bijection is simply a function that if you flip each element (source, dest) to (dest, source) you end up with another function (ie, dest also exactly covers the destination set exactly once).

So an {a, b} arrangement would include: {(a, 1), (b, 2)} or {(a, 2), (b, 1)}.

Now, what are the arrangements of {}? As it happens, {} is a valid arrangement under the above definition. It is also the only arrangement. So there is exactly 1 arrangements of the empty set.

Thus, 0! = 1.

Note that this means you can calculate things like aleph_0! as a bonus. (with a slight redefinition of the destination set {1...n}).

Which leads to the question: what is the value of aleph_0!?

Isn't math fun!

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Postby skeptical scientist » Thu Jun 07, 2007 5:42 am UTC

Aleph_0! is the number of permutations of the natural numbers. Every such permutations is a function from the natural numbers to themselves, which can be thought of as a subset of Aleph_0xAleph_0 (i.e. the graph). Therefore Aleph_0! is at most 2^(Aleph_0xAleph_0)=2^Aleph_0, since the cardinality of Aleph_0xAleph_0 is Aleph_0. On the other hand, you can code an infinite and coinfinite subset of the natural numbers into a permutation as follows: let f send the nth smallest element of the set to the nth odd number, and the nth smallest element of the complement to the nth even number. Then f is a permutation of the natural numbers so that the set in question is f^(-1) of the set of even numbers. So Aleph_0! is at least 2^Aleph_0, and therefore there are the same number of permutations of the natural numbers as there are subsets of the natural numbers (as there are functions from the natural numbers to themselves).

Aleph_0^Aleph_0=Aleph_0!=2^Aleph_0
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Postby Ended » Thu Jun 07, 2007 2:02 pm UTC

From an analytical perspective, it's possible to show that the gamma function is the (unique) analytic continuation of the factorial.

I.e., if we want an analytic function which is equal to the factorial on positive integer points, we have to take the gamma function (actually, gamma(z+1)). So, we define 0! = gamma(0+1) = 1.
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*tries to explain his point

Postby PainGod » Thu Jun 07, 2007 2:33 pm UTC

Question: Name one way that you can arrange 0 elements. Now, tell me why doing nothing is considered something.


Consider the empty set, it is a set containing no elements. It is still something.
A function is just a subset of a Cartesian product.
The Cartesian product of an empty set onto an empty set is *alas* an empty set.
Therefore, the only function that can be a subset of that, would be, the empty set function.
And that, in itself, is one thing.


My point is that considering the empty set as a valid function is essentially an arbitrary choice. You can think of as many reasons as you like to say why it is clear that one exists, but I will still say that the existence of a empty set is just our own choice. This applies to probability, too.

In general, when you apply a function to a set that does not intersect the function's domain, the set does not change. Thus, applying any function to the set {} will do nothing. For good reason, we only include functions that have exactly the given set as a domain (who knows why). Thus, in order to have a function "of interest", as above, on {}, it must have a domain of {}. What can this function possibly do to the empty set? Well, do nothing. BUT who's to say that a function from {} to {} is otherwise a function? Granted it does make the math easier and does eliminate the need for a special case, but I still say it's a choice that we have made.

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Re: *tries to explain his point

Postby Woxor » Thu Jun 07, 2007 2:51 pm UTC

PainGod wrote:My point is that considering the empty set as a valid function is essentially an arbitrary choice. You can think of as many reasons as you like to say why it is clear that one exists, but I will still say that the existence of a empty set is just our own choice. This applies to probability, too.

But really, the definition of a function implies the existence of the empty set function. A function is simply a relation such that for any element a of the domain, there is exactly one pair (a, b) in the relation. Since the domain has no elements, the condition holds for the empty relation. It's no more arbitrary than set theory itself.

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Postby Yakk » Thu Jun 07, 2007 3:34 pm UTC

PainGod wrote:Granted it does make the math easier and does eliminate the need for a special case, but I still say it's a choice that we have made.


*nod*, it is a statement true by Occam, not by Observation or Consequence.


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