Factorial of 0, Proof and Examples

Factorial of 0 Proof:

since we know that

n!=n(n-1)(n-2)(n-3)\dots\dots.(n-n+2)(n-n+1)(-n-n)

n!=n(n-1)(n-2)(n-3)\dots\dots.3.2.1

where n belongs to positive integer

Examples.

2!=2(2-1)=2(1)=2

3!=3(3-1)(3-2)=3.2.1=6

Now we will prove that 0!=1 is equal to one.

n!=n(n-1)(n-2)(n-3)!

n!=n(n-1)!

Put $n=1$

1!=1(1-1)!

1!=1(0)!

1!=(0)!

\boxed {1=0!}

Hence factorial of 0 is equal to 1

Why it is not possible to have a negative factorial ?

n!=n(n-1)(n-2)(n-3)\dots\dots.3.2.1

for negative integers

put $n=-m\in Z$

(-m)!=-m(-m-1)(-m-2)(-m-3)\dots\dots.(-m-m)(-m-m-1).........

$(-m)!=$ (negative).(negative).(negative).(negative)…………(negative).(negative)……

(-m)!=-\infty

Hence prove that negative factorial does not exist.

x^{n+1}=x^n.x^1.....(a)

put $n=0$ in eq(a)

x^{0+1}=x^0.x^1

x^{1}=x^0.x^1

dividing $x$ on both sides.

\frac xx=x^0.\frac xx

1=x^0.1

\boxed{1=x^0}

hence any number power zero is equal to 1

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