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0! - Why does it equal 1?

Out of curiosity the other day I tried to see what 0! was on my calculator. The answer I got was "1" which suprised me rather a lot. Can anyone prove why 0! = 1? Thanks
Reply 1
Avatar for manps
manps
see this link

click here
Reply 2
Avatar for Hoofbeat
Hoofbeat
OP
14
manps
see this link

click here


Thanks :smile:
Reply 3
Avatar for SsEe
SsEe
13
I say this:
n! = n(n-1)!
Let n=1
We know that 1! = 1 but:
1! = 1*0! by that little formula. So 0!=1
Reply 4
Avatar for Newton
Newton
The special case 0! is defined to have value 0!=1, consistent with the combinatorial interpretation of there being exactly one way to arrange zero objects (i.e., there is a single permutation of zero elements, namely the empty set).

Newton.
Reply 5
Avatar for El Stevo
El Stevo
11
i'd have to agree with Newton :smile: but think of it this way, there is only one way to do nothing and that is, do nothing.
Reply 6
Avatar for MC REN
MC REN
14
what is 0^0?
Reply 7
Avatar for El Stevo
El Stevo
11
MC REN
what is 0^0?


1, i think...

edit: calculator says... Error 2. Nuff said, 0^0 = Error 2.
Reply 8
Avatar for k@tie
k@tie
10
MC REN
what is 0^0?

It's undefined.
Think because n^0 is defined as n/n = 0/0
(It's n/n because n^a = n^(a+1)/n and taking a=0, then n^0=n/n) But then you might argue that 0^n is never defined using this logic, but we can see that for n<0, 0^n=(0x0x....0) (n times)=0
Reply 9
Avatar for J.F.N
J.F.N
1
k@tie
It's undefined.
Think because n^0 is defined as n/n = 0/0
(It's n/n because n^a = n^(a+1)/n and taking a=0, then n^0=n/n) But then you might argue that 0^n is never defined using this logic, but we can see that for n<0, 0^n=(0x0x....0) (n times)=0


Actually, 0/0 is called an intermediate. For example, the limit of ax/x as a approaches 0 is a, regardless of what a is. So if you choose x to be 0, then you have 0/0=a. If you want a to be 10, then 0/0=10. You can verify this using L'Hospitals rule. Similarly, you have that lim(sinx)/x as x approaches zero is one.
Reply 10
Avatar for RichE
RichE
15
J.F.N
Actually, 0/0 is called an intermediate. For example, the limit of ax/x as a approaches 0 is a, regardless of what a is. So if you choose x to be 0, then you have 0/0=a. If you want a to be 10, then 0/0=10. You can verify this using L'Hospitals rule. Similarly, you have that lim(sinx)/x as x approaches zero is one.


I think you mean "indeterminate" - which means the same as Katie's "undefined".

You can never justify writing "0/0=10", but in certain examples, in the limit, this can make a sort of sense.
Reply 11
Avatar for J.F.N
J.F.N
1
RichE
I think you mean "indeterminate" - which means the same as Katie's "undefined".


Right--indeterminate. Its still early in America.
Reply 12
Avatar for john !!
john !!
14
J.F.N
the limit of ax/x as a approaches 0 is a, regardless of what a is.


do you mean as x approaches 0?
just checking because I don't understand it as it is.
Reply 13
Avatar for Gauss
Gauss
11
mik1a
do you mean as x approaches 0?
just checking because I don't understand it as it is.


Yes I think he did. It wouldn't make sense otherwise, the limit would just be 0.

Euclid

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