1. Can we prove that 0! = 1?
Also what is 00?  2. Originally Posted by MaxPayne Can we prove that 0! = 1?
I don't think one can "prove" 0! = 1 because that is a matter of definition of the factorial. However, 0! = 1 is necessary for many combinatoric expressions to work. Also, if one generalises the factorial to non-integers using the gamma function, then 0! = 1 is a natural result. Originally Posted by MaxPayne Also what is 00?
An indeterminate form.  3. thank you!!!   4. 1 = 1! = 1x0!.  5. Originally Posted by mathman 1 = 1! = 1x0!.
Good point.  6. Lol , the second equation assumes, that 0! = 1.
So can we really use it to prove the former.  7. Originally Posted by MaxPayne  Lol , the second equation assumes, that 0! = 1.
So can we really use it to prove the former.
No. In the definition of you provided, is axiomatic (part of the definition of ). The second part of the definition is a recursive definition for all integers greater than zero. Being recursive, it requires an initial condition to start from (the first part of the definition). In other words, both parts are necessary to define . Therefore, the second part cannot prove the first part.  8. Originally Posted by MaxPayne  Lol , the second equation assumes, that 0! = 1.
So can we really use it to prove the former.
When I first learned about n!, it was simply defined as n(n-1)...1. So it starts with 1! = 1, not 0! = 1.  9. Originally Posted by MaxPayne Can we prove that 0! = 1?
Also what is 00?
Consider this pattern.     While this isn't a theorem it is proof by logic and it's more about definition.  10. n!=n(n-1)!
=> 1!=1*0!
but 1!=1
so 0!=1  11. Originally Posted by MaxPayne Can we prove that 0! = 1?
Also what is 00?
What do you do to work out the factorial of zero?  Posting Permissions
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