Hi,

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

I would be very greatful if you could.

Thanks

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- May 18th 2011, 06:55 AMAbromavichProof 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 - May 18th 2011, 07:07 AMAckbeet
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

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

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. - May 18th 2011, 07:18 AMchisigma
A 'proof by induction' is consequence of the general property http://quicklatex.com/cache3/ql_0c64...1871046_l3.png in the particular case n=1...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - May 18th 2011, 10:06 AMchisigma
If we want to be 'rigorous' a precise definition of n! is necessary... in my opinion such precise definition is the following...

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - May 18th 2011, 10:32 AMabhishekkgp