# Thread: Proof by Induction for 0!

1. ## Proof by Induction for 0!

Hi,
Would someone kindly show and explain the proof by induction for 0! = 1

I would be very greatful if you could.
Thanks

2. Your question makes no sense. A proof by induction, by definition, means you're trying to prove a statement that depends on some integer n. There is no such n in what you're trying to prove. To show that 0! = 1, I would use the generalized factorial function, the gamma function, and show that

$0!=\Gamma(1)=1.$

But you'd need to know about the gamma function, which is defined in terms of an integral, and hence would be inappropriate for this forum. I can move this thread to Calculus, if you prefer, where the question would not be out of place.

3. Originally Posted by Abromavich
Hi,
Would someone kindly show and explain the proof by induction for 0! = 1

I would be very greatful if you could.
Thanks
A 'proof by induction' is consequence of the general property in the particular case n=1...

Kind regards

$\chi$ $\sigma$

4. If we want to be 'rigorous' a precise definition of n! is necessary... in my opinion such precise definition is the following...

We call $a_{n}=n!$ the n-th term of a sequence defined by the recursive relation...

$a_{n+1}= (n+1)\ a_{n}$ (1)

... with the 'initial condition' $a_{0}=0!=1$, so that $0!=1$ is true 'by definition'. Alternatively You can have as 'initial condition' $a_{1}=1!=1$ an in this case 0! is defined writing the (1) as...

$a_{n-1}= \frac{a_{n}}{n}$ (2)

Kind regards

$\chi$ $\sigma$

5. Originally Posted by Abromavich
Hi,
Would someone kindly show and explain the proof by induction for 0! = 1

I would be very greatful if you could.
Thanks
In elementary combinatorics i believe 0! is DEFINED to be 1.