I was thinking how to show that $0!=1.$ I came up with this arguement.
$n!=n\times(n-1)!,$ so with $n=1$, we have
$1!=1\times0! \implies 0!=1.$ Is this correct ? Or there are some better proofs for this ? Or is it simply an assumption.
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2Previously: Prove $0!=1$ from first principles – Sep 04 '14 at 10:04
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4You used the fact that $1!=1$, how do you know this? – Git Gud Sep 04 '14 at 10:04
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1how do you define factorial? from what i know it is defined like this $n!=\prod_{k=1}^n k$ or, $n!= (n-1)!\times n$ for $n>0$ and $n!=1$ if $n=0$ so from the first one it is easy to see that $0!=1$ and from the second, well this is by definition. – sha Sep 04 '14 at 10:19
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By definition, $$n!=\Gamma(n+1).$$ Then $$0!=\Gamma(1)=\int_0^{\infty }t^0e^{-t}dt=\int_0^\infty e^{-t}dt=\left[-e^{-t}\right]_0^\infty =1-\lim_{t\to\infty }e^{-t}=1.$$
Q.E.D.
idm
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Most approaches I know say $0!= 1$ by definition. However, this definition is consistent with the meaning of factorial. It is also clear from the combinatorical definition of factorial: how many ways are there to order $n$ objects in a row? The answer is $$n! = n \times (n-1) \times (n-2) \times ... \times 2 \times 1 \ . $$ How many ways there are to order $0$ objects in a row? One way, which is "the empty way". Therefore $0!=1$.
LinAlgMan
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