Hi,
Would someone kindly show and explain the proof by induction for 0! = 1
I would be very greatful if you could.
Thanks
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.
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
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