Question

Math Question
1!=how much? How i can prove 0 with factorial? Can you give me your correct answer?

Answer 1

$\Huge\rightarrow0!=1$

$\Large\text{\underline{\underline{Explanation}}}$

The factorial of a natural number is a product of itself and every natural number below it. So, we write the following equation.

$\cdots\longrightarrow n!=n\cdot(n-1)!$

The factorial $0!$ satisfies the previous equation.

$\cdots\longrightarrow 1!=1\cdot(0)!$

$\large\therefore0!=1$

$\Large\text{\underline{\underline{Learn more}}}$

$\large\text{[Gamma function]}$

The previous step was on the extent of the factorial function to the whole number.

If we choose a number from the set of natural numbers, it will produce a graph. A graph that consists of disjoints between each whole number.

Gauss defined a function that satisfies the three conditions.

$\large\text{ \boxed{\begin{aligned}&\text{(1) }x!=x\cdot(x-1)!\\\\&\text{(2) No disjoint points on the graph}\\\\&\text{(3) A graph in reals}\end{aligned}}\\\\ \longrightarrow\boxed{\Gamma(x)=(x-1)!} }$

If we evaulate $\dfrac{1}{2}!$, we get $\dfrac{\sqrt{\pi}}{2}$.

$\large\text{[Factorials in negative integers]}$

However, the graph is defined anywhere except negative integers. As said, using the definition of factorial gives the following equation.

$\cdots\longrightarrow 0!=0\cdot(-1)!$

$\cdots\longrightarrow1=0\cdot(-1)!$

Such a number does not exist.