Exploring the Math of "0!=1": Debunking Misconceptions

[Pete Mcg]

TL;DR Summary

Problems with defining 0! using n!=n(n-1)! formula

I am not a Mathematician so the terms I use are probably not correct in Mathematical language. My concern is with the equation n!=n(n-1)! should only to be used for numbers > 2. I contend that using the ! 'function' on the right hand side applies ONLY when one wants to make a shortcut: ie 6!=6(6-1)(6-2)(6-3)(6-4)(6-5) = 6x5x4x3x2x1 = 720. To work out 7! one could go through the whole process again but why? Just simplify the equation to 7!=7(7-1)! or 7 x6! = 5040.

Now, what seems to happen in almost all the 'proofs' I see is that the fact that the process can be simplified is then applied retrospectively for want of a better word. I have heard it said in some of these explanations that 'We can now apply n!=n(n-1)! for ALL other numbers or at least to 1. And this 'logic' is always claimed after some numbers bigger than 4 are looked at- to me it resembles a magic trick. Why should we apply it smaller numbers?

My contention is that this is terribly flawed and contend that if this time saving tool of using the ! 'function' is not used at all (as I said it is only a device for convenience) we get not a different result but an honest and consistent one.

Using the 'tool' of 1! on the right hand side of the equation allows one to make a seemingly rationale inference that 0!=1 by what appears to me to be circular and faulty reasoning - faulty because, as I have said, the usage of 1! on the right for 1! is wrong. The so called reasoning goes: 1!=1(1-1)! = 1!=1 (0)! Since 1! cannot be 1x0=1, it must follow that 0! is 1. To me, this is almost Pythonesque.

So: why use an 'equation' at all? Obviously, for convenience when calculating increasingly large numbers. For smaller numbers it is simply not needed.

1!=1, 2!=2x1 =2, 3!=3x2x1=6, 4!=4x3x2x1=24, 5!=5x4x3x2x1= 120 and so on. As far as 0! goes, there is a way of showing its possible truth - as I said, I am a layman so don't know if this constitutes a 'proof', That is the pattern 120 - 24 - 6 - 2-1 - ie the divisors from 5! are 5 (120 divided by 24, the next number in the descending sequence) ,4,3,2 and finally 1 to get (logically according to the sequence) to 0.

As I said, I'm not sure if this constitutes a proof and if it does not, the 0! question may only be resolved in an essentially philosophical basis: I won't bore with you all the 'arguments' about 0 and its properties or lack of...Or: (and this is my thinking but I'm sure someone else has said this in the same or different ways): sometimes when there are a number of choices we have to choose the one that makes the most sense... It seems to me that 0!=1 makes the most sense in a Mathematical sense at least...

Finally: it drives me to distraction every time I see the above n!=n(n-1)! equation used to justify the 0! =1 proposition. If it is, for example, being taught in schools, it is maybe tantamount to ... I hesitate in using too strong language ... at least misleading. I make comments on You Tube channels that insist on repeating what I see as a fallacy but I have had no responses. I hope this forum can offer some insights because my 'proof' that using the equation is wriong may indeed be flawed... Comments PLEASE!

 

[PeroK]

$0! = 1$ (by definition)

$n! = n(n-1)!$ (for $n \ge 1$)

 

[Frabjous]

Also check out the Γ function which is a “generalization” of factorial.

 

[fresh_42]

$n!=n(n-1)!$ is a recursion. It does not tell us what happens at the start. The initial and additional equation $0!=1$ solves this problem.

$0!=5$ could also be used, but it turned out to be of very limited use that way.

Formally, we have an empty product, a product without factors. Setting it zero is not an option since zero has nothing to do with multiplication. We therefore set it to be equal to the neutral element of multiplication which is $1.$

It is the same principle with empty sums. A sum without summands is set to be the neutral element of addition, $0.$

 

[PeroK]

$0!=5$ could also be used, but it turned out to be of very limited use that way.

In that case we could have $$\binom{n}{k} = \frac{5(n!)}{(n-k)!k!}$$A bit clumsy, perhaps!

 

[PeroK]

Alternatively, if we take $n!$ to be the number of distinct ways of choosing $n$ things, then $0! = 1$.

For the benefit of anyone who thinks that $0! = 0$: If a shop has $n$ items for sale, then you can still go to the shop, buy nothing and come home with $0$ items. That is an outcome, so $0! = 1$

 

[Pete Mcg]

$0! = 1$ (by definition)

$n! = n(n-1)!$ (for $n \ge 1$)

Also check out the Γ function which is a “generalization” of factorial.

OK, thanks, I will. A question: does the pattern I mentioned in my first post constitute a 'proof'? And, I guess, what then is a 'proof'? Is there any kind of arbitrer - like some kind of ... I don't know...an International Court of Arbitration on Deciding Whether Mathematical Proofs Are Legitimate... Or what? - is it just kind of open slather? (I saw in a book I guess is reputable a list of factorials - it included 0!=1 so I guess there has to be some kind of AUTHORITY behind this... like a definitively recognised proof... )

 

[Frabjous]

OK, thanks, I will. A question: does the pattern I mentioned in my first post constitute a 'proof'? And, I guess, what then is a 'proof'? Is there any kind of arbitrer - like some kind of ... I don't know...an International Court of Arbitration on Deciding Whether Mathematical Proofs Are Legitimate... Or what? - is it just kind of open slather? (I saw in a book I guess is reputable a list of factorials - it included 0!=1 so I guess there has to be some kind of AUTHORITY behind this... like a definitively recognised proof... )

No, it is essentially a definition like the n(n-1)! “proof” is.

 

[Pete Mcg]

$n!=n(n-1)!$ is a recursion. It does not tell us what happens at the start. The initial and additional equation $0!=1$ solves this problem.

$0!=5$ could also be used, but it turned out to be of very limited use that way.

Formally, we have an empty product, a product without factors. Setting it zero is not an option since zero has nothing to do with multiplication. We therefore set it to be equal to the neutral element of multiplication which is $1.$

It is the same principle with empty sums. A sum without summands is set to be the neutral element of addition, $0.$

Sorry mate, you've lost me - I am not a mathematician, I'm just an old but still reasonably intelligent Australian fella who likes playing around with numbers. Do you know the film Philadelphia? - the Denzel Washington character says (not verbatim) in one scene: explain this to me as if I'm a 5 year old... I'm not a 5 year old but I don't get it, could you ecxplain in language i might get? ... What do you mean 0!=5 could also be used...? There are lots of other things you said I have no idea about but I would like to learn. Can we start with the thing you said about 5?

 

[fresh_42]

Sorry mate, you've lost me - I am not a mathematician, I'm just an old but still reasonably intelligent Australian fella who likes playing around with numbers. Do you know the film Philadelphia? - the Denzel Washington character says (not verbatim) in one scene: explain this to me as if I'm a 5 year old... I'm not a 5 year old but I don't get it, could you ecxplain in language i might get? ... What do you mean 0!=5 could also be used...? There are lots of other things you said I have no idea about but I would like to learn. Can we start with the thing you said about 5?

If you explain something by what it was before, then you need a starting point. Otherwise, you haven't explained anything. Before before before before ... has no content. At some point in time, something has to happen without referrencing another 'before'. This point in time is here $0!.$ We have to give it a value since there isn't another 'before'. And setting $0!=1$ is the only reasonable possibility. Any other choice would be: illogical, inconvenient, unnatural, contradictory, or whatever.

 

[PeroK]

Sorry mate, you've lost me - I am not a mathematician, I'm just an old but still reasonably intelligent Australian fella who likes playing around with numbers.

I'm going to have a function $F(n)$ defined on the non-negative integers such that $F(n) = nF(n-1)$ for $n \ge 1$.

You can't prove $F(0) =1$ because I can choose any value I like for $F(0)$. There is a whole family of such functions depending on the choice for $F(0)$.

What we can do is take the case where $F(0) = 1$ and call that function the factorial.

 

[Hornbein]

I figure that 0! = 1 in order to get binomial coefficients to work neatly.

 

[Frabjous]

Trying to simplify further …

0 is the identity for the addition operation; a+0=a
1 is the identity for the multiplication operation; a×1=a

In the same spirit of “if you do not add anything you have 0,”
you have ”if you do not multiply anything you have 1”

 

[Pete Mcg]

If you explain something by what it was before, then you need a starting point. Otherwise, you haven't explained anything. Before before before before ... has no content. At some point in time, something has to happen without referrencing another 'before'. This point in time is here $0!.$ We have to give it a value since there isn't another 'before'. And setting $0!=1$ is the only reasonable possibility. Any other choice would be: illogical, inconvenient, unnatural, contradictory, or whatever.

OK, got that

 

[Pete Mcg]

$0! = 1$ (by definition)

$n! = n(n-1)!$ (for $n \ge 1$)

Ok but superfluous for 2 !=2 (2-1) = 2 X 1= 2= Adding the ! to the right is not needed.

 

[Pete Mcg]

I'm going to have a function $F(n)$ defined on the non-negative integers such that $F(n) = nF(n-1)$ for $n \ge 1$.

You can't prove $F(0) =1$ because I can choose any value I like for $F(0)$. There is a whole family of such functions depending on the choice for $F(0)$.

What we can do is take the case where $F(0) = 1$ and call that function the factorial.

OK then, you are not listening and I have no idea what you are talking about so see you later!

 

[vela]

Ok but superfluous for 2 !=2 (2-1) = 2 X 1= 2= Adding the ! to the right is not needed.

If you say 2! = 2(2-1) instead of 2! = 2(2-1)!, you're not using the definition.

The answer to your original question is 0! = 1 by definition. This raises the question of why we define it this way. The answer to that question is ultimately "because it's convenient and useful to do so, both conceptually and practically."

For example, if there are n objects, we currently say there are n! ways to arrange them. However, if we defined 0! = 5, we'd have to say there are n! ways to arrange them if n≥1 or 1 way if n=0. If we didn't define 0!=1, we'd generally have to carve out an exception for the case n=0 in numerous places.

 

[Vanadium 50]

Summary:: Problems with defining 0! using n!=n(n-1)! formula

I am not a Mathematician

But nonetheless are sure that all the mathematicians are doing their mathematics wrong. That's confidence!

As pointed out, this is a definition. A definition can't be wrong - by definition.

 

[Pete Mcg]

But nonetheless are sure that all the mathematicians are doing their mathematics wrong. That's confidence!

As pointed out, this is a definition. A definition can't be wrong - by definition.

No, I am not saying that - I am saying that Mathematicians who use the formula n!=n (n-1)! to prove 0!=1 are using that formula wrongly, and I gave my reasons in a cogent manner for thinking so in a cogent manner. Do you agree with me and if not can you show me where I am wrong?

 

[MidgetDwarf]

No, I am not saying that - I am saying that Mathematicians who use the formula n!=n (n-1)! to prove 0!=1 are using that formula wrongly, and I gave my reasons in a cogent manner for thinking so in a cogent manner. Do you agree with me and if not can you show me where I am wrong?

It is a definition. You do not prove a definition in mathematics. You take it on face value. I suggest looking up what a definition, axiom, lemma, theorem mean in mathematics. At this point you are being silly.

 

[Vanadium 50]

Do you agree with me

No, because it is a definition. Nothing you write is remotely relevant,

 

[Frabjous]

You are correct in recognizing that there is something weird with 0!
That weirdness is because 0! is a defined value. When doing a recursive definition, one has to define the initial value. It is simpler in some situations if 0! exists and has a value of one, so mathematicians chose to extend the simple definition (n!=n*(n-1)*…*2*1)to include it. For example, the binomial coefficient which is used for combinations and permutations.

 

[Pete Mcg]

It is a definition. You do not prove a definition in mathematics. You take it on face value. I suggest looking up what a definition, axiom, lemma, theorem mean in mathematics. At this point you are being silly.

OK then, I'll be even sillier. I get what you're saying - I hope - please correct me if I'm wrong. 0!=1 is a 'definition' in the same way that 1=1 or any other self evident statement is a definition and 'You do no prove a definition in Mathematics'. Does that therefore mean that all the Mathematicians who 'prove' 0!=1 either don't understand this rule (assuming it is a universal 'rule') or are, like me, just plain silly?
I will look up those terms, thank you for that. Maybe 'definition' has a different definition in Maths but yes I am being silly - substitute 'meaning' for definition. Maybe I don't understand all the meanings of the word - I have a Dictionary of philosophy and that does appear to be the case...The dictionary is open as I write.
And even sillier: (I can almost see you slapping your forehead in frustration after you read what follows): Can we not say that a self evident statement is also its own proof ? But we then get into a silly world of antinomy, are we not? Or I am just not understanding Mathematical language again (eg what does 'proof' mean in Maths language?) OR just being playful.

On a hopefully more serious note, it occurred to me that perhaps what you might be saying is that long ago in the distant past 1=1 was indeed 'proved' to be self evident and is now a totally immutable axiom or truth and 'proof' is redundant, a waste of time, silly. If that is the case, some time in the long distant past a proof morphed into a definition... Antimony?

Cheers.

 

[Pete Mcg]

You are correct in recognizing that there is something weird with 0!
That weirdness is because 0! is a defined value. When doing a recursive definition, one has to define the initial value. It is simpler in some situations if 0! exists and has a value of one, so mathematicians chose to extend the simple definition (n!=n*(n-1)*…*2*1)to include it. For example, the binomial coefficient which is used for combinations and permutations.

Thanks for that. I am on a little journey into Maths - I will look up those terms, sound intriguing.

 

[Pete Mcg]

If you explain something by what it was before, then you need a starting point. Otherwise, you haven't explained anything. Before before before before ... has no content. At some point in time, something has to happen without referrencing another 'before'. This point in time is here $0!.$ We have to give it a value since there isn't another 'before'. And setting $0!=1$ is the only reasonable possibility. Any other choice would be: illogical, inconvenient, unnatural, contradictory, or whatever.

Yes, I do get that - I have no problem with 0!=1 for the very reasons you have set out and thank you for that. At the risk of repeating myself, this is not what I am interested in - I am interested in why some Mathematicians use the formula n!=n(n-1)! - please see my original post.
Apparently 'proving' 0!=1 is not needed anyway because 0!=1 is a definition and apparently we don't prove definitions in Maths. If that is true, it - this whole 0!=1 proving thing - is all a bit silly.

 

[MidgetDwarf]

Yes, I do get that - I have no problem with 0!=1 for the very reasons you have set out and thank you for that. At the risk of repeating myself, this is not what I am interested in - I am interested in why some Mathematicians use the formula n!=n(n-1)! - please see my original post.
Apparently 'proving' 0!=1 is not needed anyway because 0!=1 is a definition and apparently we don't prove definitions in Maths. If that is true, it - this whole 0!=1 proving thing - is all a bit silly.

Yup, that is the gist of it.

If you want to understand what mathematicians call a proof https://www.people.vcu.edu/~rhammack/BookOfProof/

It is pdf made free by the author. Happy reading.

 

[MidgetDwarf]

Caz has perfectly answered the question you asked me in a previous post.There are some definitions that one can argue, are not really definitions, but deep powerful theorems.
Most commonly seen in the area of "foundational" mathematics. )

A perfect example is the set of all real numbers. You can google the history behind the set of all real numbers.

There is nothing more to 0!. Take it at face value and move on.

 

[PeroK]

Does that therefore mean that all the Mathematicians who 'prove' 0!=1 either don't understand this rule (assuming it is a universal 'rule') or are, like me, just plain silly?

These "proofs" will generally not be a proof but a justification. There's a difference between a proof and justification. One justification that $0! = 1$ is to satisfy the general formula $n! = n(n-1)!$. Many mathematical definitions have justifications, to explain why something is so defined.

I found this, which is probably the sort of mathematics you object to:

https://www.chilimath.com/lessons/intermediate-algebra/zero-factorial/

Note that whoever wrote that page put "proof" in quotes, and in the final paragraph confirmed that it's effectively a definition. He says "we must force the value of $0!$ to equal one". That's a justification of the definition.

 

[Pete Mcg]

These "proofs" will generally not be a proof but a justification. There's a difference between a proof and justification. One justification that $0! = 1$ is to satisfy the general formula $n! = n(n-1)!$. Many mathematical definitions have justifications, to explain why something is so defined.

I found this, which is probably the sort of mathematics you object to:

https://www.chilimath.com/lessons/intermediate-algebra/zero-factorial/

Note that whoever wrote that page put "proof" in quotes, and in the final paragraph confirmed that it's effectively a definition. He says "we must force the value of $0!$ to equal one". That's a justification of the definition.

Wow! Thank you! I did read the Chilmath page and it is now making sense, kind of... I note that the writer says 'The general formula can be written fully expanded form ... or in partially expanded from as n!=nx(n-1)!.' The use of the word 'or' is misleading. It would seem to imply that you can use either where in fact if you use the 'expanded form' you will either get a nonsensical result for 1! - ie 1!=1(1-0) = 1x0 = 0 OR 1!=1 (because the whole object of the exercise is to get to 1 so you don't need to complete the formula - you are already there or, as you may say, 1!=1 is defined)... , neither of which helps for 0!. So, it needs to be specified: the 'partially expanded form must be used...'

BUT WAIT! I think there is light bulb moment happening right now as I write! The 'general formula DOES work! as I've just shown above... Okay, maybe I can let this go now...

 

[PeroK]

PS Wikipedia has a list of motivations for $0! = 1$:

https://en.wikipedia.org/wiki/Factorial#Factorial_of_zero

 

[sysprog]

 

[fresh_42]

I am interested in why some Mathematicians use the formula n!=n(n-1)! - please see my original post.

We want to abbreviate $1\cdot 2\cdot 3 \cdots n$ and choose the notation $n!$ for this product. There are 3 possible ways to do so:

• by words
• by dots, like I just did
• by recursion

The latter means basically by its precedessor, i.e. $(n-1)!$ This is commonly seen as the most elegant way since it avoids dots or phrases like 'and so on'. As explained before, this needs an anchor, a starting point, which is $0!=1$ for various reasons.

 

[sysprog]

@fresh_42, so do you suppose it would be more explanatory or just more inelegant (or something else) to write $0!=0.9^* =1$ instead of just writing $0!=1$?

 

[PeroK]

@fresh_42, so do you suppose it would be more explanatory or just more inelegant (or something else) to write $0!=0.9^* =1$ instead of just writing $0!=1$?

In general, we have a universal constant $0! = f_0$, which appears as a factor in the relevant equations. E.g. $$\binom n k = \frac{(n!)f_0}{(n-k)!k!}$$For simplicity, mathematicians generally work with the choice of $f_0 = 1$.

 

[Pete Mcg]

We want to abbreviate $1\cdot 2\cdot 3 \cdots n$ and choose the notation $n!$ for this product. There are 3 possible ways to do so:

• by words
• by dots, like I just did
• by recursion

The latter means basically by its precedessor, i.e. $(n-1)!$ This is commonly seen as the most elegant way since it avoids dots or phrases like 'and so on'. As explained before, this needs an anchor, a starting point, which is $0!=1$ for various reasons.

Thank you.