Proving 0!=1: An Analysis of the Fundamental Mathematical Concept

[rock.freak667]

Proof that 0!=1??

Homework Statement

How does one prove that 0!=1, which I've only been told to accept as true.

Homework Equations

n!=n(n-1)(n-2)(n-3)*...*3*2*1

The Attempt at a Solution

well using the def'n for n!

0!=0(-1)(-2)(-3)(-4)(-5)*...*3*2*1

which should be zero as...any number by 0 is 0

 

[arildno]

It is a matter of definition. Period.

For n>=1, we have the recursive relationship $$n!=n*(n-1)!$$

 

[Avodyne]

Yes, it's a matter of definition, but this is a useful defintion. Note that arildno's recursion relation can be written as (n-1)!=n!/n. Plugging in n=1 yields 0!=1.

Also, formulas like $e^x = \sum_{n=0}^\infty x^n/n!$ only work if 0!=1.

 

[rock.freak667]

hm...but then the same goes for 1!...as that is the same as 0! by the definition of n!

 

[ZioX]

How many ways can you order 0 things? 1 way!

I had originally thought that you were asking why 0 was not equal to 1. A very interesting question to pose compared to other more tired equalities.

 

[ZioX]

hm...but then the same goes for 1!...as that is the same as 0! by the definition of n!

Well, no. Formally these things are defined recursively starting at 1.

 

[Avodyne]

Yes, 1!=1 as well.

Do you know about the gamma function?
$$\Gamma(z) = \int_0^\infty dt\,t^{z-1}\,e^{-t}.$$
This has the interesting property that for an integer n, $\Gamma(n)=(n-1)!$
For more, see http://en.wikipedia.org/wiki/Gamma_function

 

[rock.freak667]

never heard of this gamma function...shall read now

 

[dynamicsolo]

We also define 0! = 1 to provide consistency in the equations for nPr and nCr : when r = 0 or r = n, the formula should give values of 1. This will only be possible if (n-n)! equals 1.

There are a number of such situations in mathematics where an operation originally defined only for positive integers ("counting numbers"), as evolved from ordinary human uses, is extended to larger sets of numbers. A widely-used example is x^n .

 

[rock.freak667]

So then I guess there is no way to prove it but it is just taken as 1 just to make things simpler

 

[Avodyne]

Yes, that's basically it.

 

[symbolipoint]

So then I guess there is no way to prove it but it is just taken as 1 just to make things simpler

Actually, there is an algebraic proof. Try a search of physicforums. I know I saw the proof somewhere, but I cannot remember exactly where, and I do not remember this proof myself.

 

[stewartcs]

http://www.jimloy.com/algebra/zero-f.htm

 

[arildno]

So then I guess there is no way to prove it but it is just taken as 1 just to make things simpler

What do you think a "proof" is?

It is a valid deduction from ACCEPTED AXIOMS OR DEFINITIONS.

Thus, that some statement is an axiom or a definition is NOT some flaw with it as if ideally we would have liked to prove it.

Of course, we have the freedom to choose another set of axioms&definitions than the "standard" set; in that case, axioms in another set might be needed to be proved, and vice versa.

Specifically, if we CHOOSE to define the factorial in terms of the gamma function, then we may prove the statement 0!=1.

 

[Avodyne]

Thus, that some statement is an axiom or a definition is NOT some flaw with it as if ideally we would have liked to prove it.

I can't agree with that. We'd like to use the minimal number of axioms. If one can be shown to be derivable from the others, that's an improvement. People spent 2000 years trying to show that Euclid's fifth postulate was derivable before concluding that it couldn't be done.

 

[arildno]

I can agree to that; I should have inserted the word "necesssarily" in "...is NOT necessarily a flaw..", which should cover the instances you mentioned.

 

[rjoshi1906]

To Prove:- 0! =1.

Known:
y! = y x (y-1)!

So,
1! = 1 x (1-1)!
1! = 0!
or
0! = 1! -(a)
Proof:
As We know,

n! = n x n-1 x n-2 x n-3 x n-4 x ... 2 x 1
So,
4! = (3+1)! = 4 x 3! = 4 x 3 x 2 x 1
3! = (2+1)! = 3 x 2! = 3 x 2 x 1
2! = (1+1)! = 2 x 1! = 2 x 1
1! = (0+1)! = 1 x 0! = 1 -(b)

Hence, from (a) and (b)
We get, 0! = 1

 

[Fightfish]

Personally I tend to prefer the combinatorial approach. There is just one way to choose 0 objects from a set of n objects, hence:
$$^{n}C_{0} = \frac{n!}{(n-0)!(0!)} = \frac{1}{0!} = 1$$
which thus implies that 0! = 1

 

[HallsofIvy]

To Prove:- 0! =1.

Known:
y! = y x (y-1)!

This is "known" for what values of y? And how is it known? Exactly what definition of factorial are you using?

So,
1! = 1 x (1-1)!
1! = 0!
or
0! = 1! -(a)
Proof:
As We know,

n! = n x n-1 x n-2 x n-3 x n-4 x ... 2 x 1
So,
4! = (3+1)! = 4 x 3! = 4 x 3 x 2 x 1
3! = (2+1)! = 3 x 2! = 3 x 2 x 1
2! = (1+1)! = 2 x 1! = 2 x 1
1! = (0+1)! = 1 x 0! = 1 -(b)

Hence, from (a) and (b)
We get, 0! = 1

By exactly the same "proof" then, 0!= 0(0-1)!. But 0 times any number is 0 so whatever (-1)! is, 0!= 0.

 

[Mentallic]

By exactly the same "proof" then, 0!= 0(0-1)!. But 0 times any number is 0 so whatever (-1)! is, 0!= 0.

Well then couldn't (-1)! simply be undefined to keep the consistency that 0!=1.

In the same way that $$lim_{x \rightarrow 0}\left(x.\frac{1}{x}\right)=1$$you are similarly saying that the first factor x=0 but whatever the next factor 1/x is equal to, the RHS should equal 0 since 0 times any other number is 0. This is not the case however.

 

[joaquince]

f(x)= x^5 = [5!/5!] x5
f(x)⁽¹⁾= 5x^4 = [5!/4!] x^4
f(x)⁽²⁾= 20x^3 = [5!/3!] x^3
f(x)⁽³⁾= 60x^2 = [5!/2!] x^2
f(x)⁽⁴⁾= 120x^1 = [5!/1!] x^1
f(x)⁽⁵⁾= 120x^0 = [5!/0!] x^0 [Following the observed pattern… ]

f(x)^(k)= n!/((n-k)!) [x^(n-k)] […in general, for coefficient of x^n = 1, and n a positive real integer ]

Since f(x)⁽⁵⁾ = 120, we must conclude that 0! = 1

 

[dragonslayer]

alright, consider the gamma function:
Γ(z)=∫(t^(z-1)e^(-t)dt, 0, infinity)

which has the property Γ(z+1)=zΓ(z) (too long to prove, look it up),
in turn implying that Γ(z+1)=z!

so 0!=Γ(0+1)=Γ(1)

Γ(1)=∫(t^(1-1)e^(-t)dt, 0, infinity)
=∫(e^(-t)dt, 0, infinity)
= -e^(-t), 0, infinity
=0-(-1)
=1

sorry for the poor formatting, but I hope you get the idea here. The gamma function can also show that (1/2)!=sqrt(pi)

 

[stefounet]

What do you think a "proof" is?

It is a valid deduction from ACCEPTED AXIOMS OR DEFINITIONS.

Thus, that some statement is an axiom or a definition is NOT some flaw with it as if ideally we would have liked to prove it.

Of course, we have the freedom to choose another set of axioms&definitions than the "standard" set; in that case, axioms in another set might be needed to be proved, and vice versa.

Specifically, if we CHOOSE to define the factorial in terms of the gamma function, then we may prove the statement 0!=1.

Agreed. There are some sets containing X such that X*a=a (X==1) and X+a=a (X==0) for all a. Depending on the definition of '*' and '+', I suppose there could be more than one such set.
So, I don't know what the OP means, really. I mean, if we're talking Peano's integers, then 1!=0 because 1 is the successor of 0. If we're talking real numbers, then those are an extension of the former...

 

[stefounet]

Agreed. There are some sets containing X such that X*a=a (X==1) and X+a=a (X==0) for all a. Depending on the definition of '*' and '+', I suppose there could be more than one such set.
So, I don't know what the OP means, really. I mean, if we're talking Peano's integers, then 1!=0 because 1 is the successor of 0. If we're talking real numbers, then those are an extension of the former...

lol, I get it now. Not "prove that 0 is different from 1", but "prove that factorial of 0 is 1".
Just refer to the definition of factorial, then, 0! = 1 does not contain ANY information whatsoever.

 

[stefounet]

Homework Statement

How does one prove that 0!=1, which I've only been told to accept as true.

Homework Equations

n!=n(n-1)(n-2)(n-3)*...*3*2*1

The Attempt at a Solution

well using the def'n for n!

0!=0(-1)(-2)(-3)(-4)(-5)*...*3*2*1

which should be zero as...any number by 0 is 0

You've been told to accept it as true because it is a convenient convention. Just like any other word, n! means nothing unless we agree on a meaning for it. Well, n! means n*((n-1)!) if n >0 and 0! means 1. People could have agreed that n! means n*((n-1)!) if n >1, that 1! means 1, and that 0! means nothing, but that would be inconvenient in most applications even though both conventions coincide for any n>0.

 

[rock.freak667]

This thread is some years old!

 

[stefounet]

Indeed. I'm a new joiner. This post was on top of some category.
But, not matter how old the thread is, it is very frustrating that it did not meet its purpose : some high school student asked why 0! = 1 ; most replies were tentative demonstrations, hence gibberish. The answer the OP needed had nothing to do with the Γ function
If some other student googles this question, he may be glad to find my reply, however old it may be.
Regards.

 

[stefounet]

Indeed. I'm a new joiner. This post was on top of some category.
But, not matter how old the thread is, it is very frustrating that it did not meet its purpose : some high school student asked why 0! = 1 ; most replies were tentative demonstrations, hence gibberish. The answer the OP needed had nothing to do with the Γ function
If some other student googles this question, he may be glad to find my reply, however old it may be.
Regards.

Now I realize you are the OP :) Well, you got my point, right ?

 

[dragonslayer]

You've been told to accept it as true because it is a convenient convention. Just like any other word, n! means nothing unless we agree on a meaning for it. Well, n! means n*((n-1)!) if n >0 and 0! means 1. People could have agreed that n! means n*((n-1)!) if n >1, that 1! means 1, and that 0! means nothing, but that would be inconvenient in most applications even though both conventions coincide for any n>0.

The fact that 0!=1 is not a convention, it is a mathematical fact (not something we "agreed" on). And his can be proved, as the OP was asking. The Γ function is unique because it shows step by step that 0!=1, without any subjective reasoning. To further my point, we could not have decided that (1/2)!=sqrt(pi), but we can derive it.

And this post is super old, but it's never too late to revisit it!

 

[stefounet]

If I understand correctly, you define n!=Γ(n+1) for all integer and I define n!=n(n-1)...1. In the scope of my definition, 0! is NOT defined.

Depending on the OP's definition of n!, either the Γ function argument is (obviously) a satisfactory answer to his question, or there is no absolute answer because it is a matter of convention. In this later case, agreed, the Γ function argument would be one among many other justifications of the convenient convention 0!=1.

Considering "The attempt at a solution" of the OP, it seemed obvious to me that he was working in the set of integers, and that he was looking for an (impossible) arithmetic proof. I agree that your analytic definition is on top of the pure arithmeric one, but it does rely on a much heavier set of axioms.

Apologies if I'm starting to sound like a troll :)
Cheers

 

[Dick]

If I understand correctly, you define n!=Γ(n+1) for all integer and I define n!=n(n-1)...1. In the scope of my definition, 0! is NOT defined.

Depending on the OP's definition of n!, either the Γ function argument is (obviously) a satisfactory answer to his question, or there is no absolute answer because it is a matter of convention. In this later case, agreed, the Γ function argument would be one among many other justifications of the convenient convention 0!=1.

Considering "The attempt at a solution" of the OP, it seemed obvious to me that he was working in the set of integers, and that he was looking for an (impossible) arithmetic proof. I agree that your analytic definition is on top of the pure arithmeric one, but it does rely on a much heavier set of axioms.

Apologies if I'm starting to sound like a troll :)
Cheers

Everybody is right here. So this is beating an old dead horse. Defining n!=n(n-1)...1 clearly makes no sense if n=0. Making 0! an arbitrary definition. But then the notation n!=n(n-1)...1 is also vague. Best to define it inductively by (n+1)!=n!*(n+1). But where to start the induction? If you start it by defining 1!=1 that works. But you could also start it by defining 0!=1 and get the same function. Clearly there's a problem with defining (-1)! equal to some number and trying to proceed from there. Starting by defining 0!=1 and going inductively from there agrees with the gamma function definition and gives you the usual definition of n!. So why not 0!=1?