Can 0=1? The Math Behind the Assumption

[mcfetridges]

Can you prove that 0!=1 or is it just an assumption?

 

[chroot]

It is a convention, chosen simply because it makes the most sense.

- Warren

 

[AKG]

It depends on how "factorial" is defined. If n! is defined as the product of all natural numbers less than or equal to n, for natural n, then 0! = 1 would just be conventional (0 isn't even natural). On the other hand, if you want to define n! to be the number of ways you can permute/arrange n distinct objects, then there's of course 1 way you can permute zero objects.

 

[600burger]

Too much programming...i read

0!=1 as "Zero not equal to One"

Which in itself is an interesting question. Why can't it be equal? I must only convince myself that they are! But if i do, it will make everything much more difficult...

...then there's of course 1 way you can permute zero objects.

This confuses me, how can we say that there is one way to arrange nothing. It would make more sense that there would be infinte.

 

[Sirus]

It would make more sense that there would be infinte.

...or none. Highly peculiar.

 

[ZeAsYn51]

?/

If 0!=1 then what the heck does 1! equal?

 

[shmoe]

1!=1 also.

One reason to define 0!=1 would be so the formula for binomial coefficients works out, making the binomial theorem easier to state.

Another reason is the Gamma function, which very nicely extends the definition of factorial from all natural numbers to all complex numbers. This extension would tells us that 0!=1, and it would be a shame to disagree with such a nifty complex function.

Another reason based on convention and possibly not very satisfying- just as an 'empty sum' is often defined to be 0, an 'empty product' is often defined to be 1. So if n! is the product of all natural numbers less than or equal to n, 0!=1.

 

[Gokul43201]

Here's another : C(n,0) = number of ways of choosing 0 objects out of n = $\frac{n!}{n!~0!} = \frac{1}{0!}$

So, it's useful to define 0! = 1. It allows you to do more math.

 

[AKG]

If you have zero objects, then every arrangment of them will look the same, in fact, they will be the same. What's the difference between nothing and nothing? None, they're all the same. If you have nothing, there's only one way to arrange it. It is peculiar to talk about arranging nothing, but I don't think it's that peculiar. You can add 5 + 0 and get 5, and most people don't find it too peculiar to think that if you have 5 apples, and add nothing, you still have 5 apples. If you can add nothing mathematically, and can do it physically, then arranging nothing mathematical and physically shouldn't seem too peculiar.

 

[ZeAsYn51]

Here's another : C(n,0) = number of ways of choosing 0 objects out of n = $\frac{n!}{n!~0!} = \frac{1}{0!}$

So, it's useful to define 0! = 1. It allows you to do more math.

I still don't follow. So does this mean that when C(n,2) there is only a half of a way to choose 2 objects out of n?

 

[AKG]

No ZeAsYn51:

$$C(n, k) = \frac{n!}{(n - k)!k!}$$

If k = 0:

$$C(n, k) = \frac{n!}{(n - 0)!0!} = \frac{n!}{n!0!} = \frac{1}{0!}$$

If k = 2, then you won't get two "$n!$"s cancelling out.

 

[600burger]

There are few things in math that "just are" but i would think this is one of them. All of this "arranging nothing" makes no sense, and feels more like a (and really is) metaphor to hep people understand, but metaphors are never exact replicas of the things they represent, or else they'd just be that thing!

If anyone ever asks me this question i will just say "because it is". I find it to be pure convention. A made up defintion of something that will never happen so that we can go onto bigger and better equations.

 

[Tide]

If you relate the factorial to the gamma function then $0! = \Gamma (1) = 1$ follows by direct integration.

 

[AKG]

There are few things in math that "just are" but i would think this is one of them. All of this "arranging nothing" makes no sense, and feels more like a (and really is) metaphor to hep people understand, but metaphors are never exact replicas of the things they represent, or else they'd just be that thing!

Factorial is defined a number of ways. I don't think it's a wrong definition to define it as the number of ways to permute things. As such, it's not metaphorical to say 0! is the number of ways to arrange nothing. It's literal, it's just based on a more simplistic, probably unconventional definition (used mostly to help people understand). The idea of arranging nothing, however, is not really nonsense, although I can see you might have trouble understanding at first. However, definitions like n! = n x (n - 1) x ... x 1 and definition by the Gamma function are better definitions. That's a different matter, however.

If anyone ever asks me this question i will just say "because it is". I find it to be pure convention. A made up defintion of something that will never happen so that we can go onto bigger and better equations.

I'm not sure you know what you're saying here. What exactly is it that will never happen? If factorial is defined in a more abstract sense, then there is no hypothetical event to speak of which we can say will never happen. If we define 0! by relating it to the Gamma function, what exactly isn't happening? On the other hand, if we say 0! is the number of ways of arranging zero things, what exactly is "not happening" there? The Greeks, I have read, had great trouble with the concept of zero, and thought that it made no sense to have zero as a number, so it would make no sense to say that you were adding 0 apples to the pile, nor did it make sense to have 5 + 0 = 5, since 0 wasn't an acceptable number, i.e. a number plus a number would always have to give a bigger number, that's how they thought "plus" should work. However, if that legend is true, then things that smart Greek mathematicians couldn't accept are things that children today can easily accept. I think there might be some conceptual difficulty in saying that 0! is the number of ways to permute zero things, but it's not a necessarily flawed notion.

Also, you speak as though this is some peculiar things which is made up, defined, and accepted by convention. What mathematical fact is true which does not follow from definition, convention, or being made up? Are there god-given definitions? All definitions are made-up conventions.

 

[matt grime]

There are few things in math that "just are" but i would think this is one of them. All of this "arranging nothing" makes no sense, and feels more like a (and really is) metaphor to hep people understand, but metaphors are never exact replicas of the things they represent, or else they'd just be that thing!

If anyone ever asks me this question i will just say "because it is". I find it to be pure convention. A made up defintion of something that will never happen so that we can go onto bigger and better equations.

Actually most, if not all, things in maths "just are": that's why it's maths and not an empirical science. Why is 5! 120? Why should seeing the ! sign make me multiply all those numbers together. All things in maths are definitions essentially. After all, what's sqrt(2) if not by definition a/the number whose square is 2

If we take n! as the number of bijections of a set of cardinality n, then 0!=1 makes perfect sense - there is only the empty function from the empty set to itself.

 

[JasonRox]

If we take n! as the number of bijections of a set of cardinality n, then 0!=1 makes perfect sense - there is only the empty function from the empty set to itself.

That's the way I looked at it.

Sets.

 

[CTS]

Although 0! = 1 by definition only, here's a reason it makes sense using the recursive definition of factorial:

n! * (n + 1) = (n + 1)!
n! = (n + 1)! / (n + 1)

0! = (0 + 1)! / (0 + 1) = 1! / 1 = 1

 

[Swapnil]

Although 0! = 1 by definition only, here's a reason it makes sense using the recursive definition of factorial:

n! * (n + 1) = (n + 1)!
n! = (n + 1)! / (n + 1)

0! = (0 + 1)! / (0 + 1) = 1! / 1 = 1

Why can't we just use this to *prove* that 0! = 1 ?

 

[3trQN]

Maybe you should ask Dr.Anderson ?

Shuzan held out his short staff and said: " If you call this a short staff, you oppose its reality. If you do not call it a short staff, you ignore the fact. Now what do you wish to call this?"

I understand the usefulness of factorials, but how do you justify that there is any numerical way to arrange nothing since you cannot create order from something that has no properties to order it by.

Factorials are like Zen mathematics.

 

[robert Ihnot]

shmoe: One reason to define 0!=1 would be so the formula for binomial coefficients works out, making the binomial theorem easier to state.

Another reason is the Gamma function, which very nicely extends the definition of factorial from all natural numbers to all complex numbers. This extension would tells us that 0!=1, and it would be a shame to disagree with such a nifty complex function.

Another reason based on convention and possibly not very satisfying- just as an 'empty sum' is often defined to be 0, an 'empty product' is often defined to be 1. So if n! is the product of all natural numbers less than or equal to n, 0!=1.

That's what I was going to say, especially about the Gamma function and the "empty product." (Is it really a stretch of the imagination to say there is just one way to make an empty product?)

When I learned about it in school, it was a called a "convention." (That's any easy way to get attention on some other matter.) But it does work VERY NICELY FOR THE MACLAURIN SERIES FOR THE COSINE AND FOR e^x. Thus the idea that it is merely a decided upon "convention" is just too much of a simplification.

3TrQN: I understand the usefulness of factorials, but how do you justify that there is any numerical way to arrange nothing since you cannot create order from something that has no properties to order it by.

Well, the "empty set" is counted as a set.

Swapnil: Why can't we just use this to *prove* that 0! = 1 ?

Well, the idea of using 0! = (0+1)!/(1+0)=1!/1=1 is a little bit suspect, usually you are going to start by saying that n! =n(n-1)! for n an integer greater than 1. We certainly don't want 1! = 1x0 = 0. Though maybe you could fix that difficulity by defining 0! to be 1, but that was what was to be shown.

 

[3trQN]

Well, the "empty set" is counted as a set.

Yes, that's my point. You have basically re-worded the "Its convention" answer to a set theoretic one.

Is it really a stretch of the imagination to say there is just one way to make an empty product?

Ah so its imagination that defines it. Perhaps more like Zen.

 

[robert Ihnot]

The empty set comes out of Zermelo-Fraenkel Set Theory, http://plato.stanford.edu/entries/set-theory/ZF.html

There, the axiom of the null set, I read, asserts the existence of the empty set. I assume it is not mandatory to accept such an axiom.

Wikiipedia tells us: " Various possible properties of sets are trivially true for the empty set."

A good use of the empty set is that it is the intersection of two disjoint sets. Otherwise what would happen at the intersection of such sets? We would have left set theory behind.

 

[Gib Z]

Finally, I was starting to think no one would post the proof CTS did, though this slight variation might be simpler.

$$n!=n\cdot (n-1)!$$
$$1!=1\cdot 0!$$
Since 1! equals 1, 0!=1.

 

[Hurkyl]

I understand the usefulness of factorials, but how do you justify that there is any numerical way to arrange nothing

Who said anything about arranging nothing? I'm not even sure that makes grammatical¹ sense! 0! counts the number of ways to arrange a collection with nothing in it. (i.e. a collection with zero objects). Here's one way to compute 0!:

There is exactly one way to permute zero objects. (do nothing)
Thus, 0! = 1.​

since you cannot create order from something that has no properties to order it by.

It's easy to order the empty set. (In fact, there is exactly one way to do so) And the empty set certainly has properties. For example, $\forall x: x \notin \emptyset$.



1: mathematical grammar, of course

 

[morphism]

Finally, I was starting to think no one would post the proof CTS did, though this slight variation might be simpler.

$$n!=n\cdot (n-1)!$$
$$1!=1\cdot 0!$$
Since 1! equals 1, 0!=1.

Your first equality only holds (by definition) for $(n-1) \geq 1$. Otherwise we can go on and say:

$$1 = 0! = 0 \cdot (-1)! = 0$$