math question: factorials

duke

Golden Member
Nov 22, 1999
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What is zero factorial (0!)? And show proof. Forgive me if this is really a stupid question. I'm not in the mood to think real hard right now.
 

Shmorq

Diamond Member
Aug 10, 2000
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n! = n * (n-1)!

If you let n =1:

1! = 1 * (1-1)!
1! = (1-1)!
1! = 0!
1 = 0!

Hope that makes sense...:)
 

xtreme2k

Diamond Member
Jun 3, 2000
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Shmorq

that is a poor prove.

if you take your formula
n! = n(n-1)!
0! = 0(0-1)!
0! = 0(-1)!
see it doesnt make any sense?
 

duke

Golden Member
Nov 22, 1999
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Thanks xtreme. Yeah, my calculator tells me it is one as well. But if I remember correctly the definition of N!=N*(N-1)*(N-2)*(N-3)*...*(N-(N-1))

But this 0!=1 has got me scratching my head. Anyone else?
 

xtreme2k

Diamond Member
Jun 3, 2000
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I havent come across to proving this 0! = 1. However, looking at your prove, it is like you are explaining the meaning of something, and yet you use the same word in explaining what it is. You get what I mean.


Could n^0 = 1 give anyone any hints?
 

Shmorq

Diamond Member
Aug 10, 2000
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xtreme2k

Let n=1, not n=0.


n! = n * (n-1)!

If you let n =1:

1! = 1 * (1-1)!
1! = (1-1)!
1! = 0!
1 = 0!
 

xtreme2k

Diamond Member
Jun 3, 2000
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Shmorq

you dont understand.

If you let N>0, everything make sense. So proving it doesnt mean anything. He is looking for when N=0.


Anyway your priorities went wrong.

1*(1-1)! = 1(0)! not (1-1)!. We do brackets first remember?
so at the end you have

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

this might be a better prove :)
 

Shmorq

Diamond Member
Aug 10, 2000
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<< 1*(1-1) = 1(0) not (1-1). We do brackets first remember? >>

Don't forget that it's 1*(1-1)! which is 1*(0!) Not 1*(1-1).
 

Shmorq

Diamond Member
Aug 10, 2000
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<< 1*(1-1)! = 1(0)! not (1-1)!. We do brackets first remember?
so at the end you have
1!= 1(0)!
1 = 1(0)!
1/1 = 0!
this might be a better prove :)
>>



That's the exact same proof that I just showed?!? I'm confused now...:confused:
 

Noirish

Diamond Member
May 2, 2000
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being an applied math major, i think shmorq is correct.
after all, all you care about is being able to arrive at 0!=1 (or 1=0!).
 

Relf Lauren

Golden Member
Oct 10, 1999
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Assigning a value to 0!

P(n,r) = n!/(n-r)!
Therefore P(n,n) = n!/(n-n)! <== right side MUST equal n!...therefore the denominator MUST equal 1..therefore 0! MUST equal 1

Booya


Relf Lauren
 

Hanpan

Diamond Member
Aug 17, 2000
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Often it is taken for granted taht 0! is 1. Much like 2+2 = 4. You shouldn't have to prove it anyway.
 

Mister T

Diamond Member
Feb 25, 2000
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you can derive a proof from the Gamma Function:

Gamma Function

Factorial can be defined by an improper integral. Plug in n=1 to the integral,
do the math and you will get Gamma(1)=(1-1)!=0!=1
 

hendon

Senior member
Oct 9, 2000
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It's just a definition to make the formulas work out for n=0, in summation formulae etc...
 

Mister T

Diamond Member
Feb 25, 2000
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By the way, without using the Gamma Function and/or calculus(which extends the domain from just natural numbers), proving that 0!=1 is not actually a rigorous proof. It is more a definition than something you can prove.