# Thread: [SOLVED] dividing by 0!

1. ## [SOLVED] dividing by 0!

Can someone show me why the following works out? Thanks!

C(5,5) = (5!) ÷ (5!)(5-5)!
= (5!) ÷ (5!)(0)!
= 5x4x3x2x1 ÷ (5x4x3x2x1)... ?

The calculator says this equals 1.

2. I don't recall why but 0! = 1.

3. $\displaystyle 0!=\prod_{k=1}^0k$
which is an empty product. An empty sum is 0. An empty product is 1. See this

4. ## a proof

Thanks, I will definitely look at this after my final!

5. Originally Posted by Truthbetold
I don't recall why but 0! = 1.
This is one of the reasons:

In calculus, there is something known as the Gamma Function. It is defined as follows:

$\displaystyle \Gamma\left(x\right)=\int_0^{\infty}e^{-t}t^{x-1}\,dt$

When evaluating $\displaystyle \Gamma\left(n\right)$, where $\displaystyle n\in\mathbb{N}$, it has this special property:

$\displaystyle \Gamma\left(n\right)=\left(n-1\right)!$

Now, let's see what happens when $\displaystyle n=1$

This means that $\displaystyle \Gamma\left(1\right)=\color{red}0!$

By definition, $\displaystyle \Gamma\left(1\right)=\int_0^{\infty}e^{-t}\,dt=\left.\left[-e^{-t}\right]\right|_0^{\infty}=\left[-e^{-\infty}-\left(-e^{0}\right)\right]=e^0=\color{red}1$

Thus, $\displaystyle \Gamma\left(1\right)=\color{red}\boxed{0!=1}$

6. ## more research to do

Thanks! Looks like I have some more research to do after Finals.

Best of luck on your finals!

I just have two left to do, Prob & Stats, and Dynamical Systems! Thank goodness this semester is almost over!!!

7. Originally Posted by yvonnehr
Thanks! Looks like I have some more research to do after Finals.

Best of luck on your finals!

I just have two left to do, Prob & Stats, and Dynamical Systems! Thank goodness this semester is almost over!!!
At your level: 0! = 1 by definition so that the number of way of choosing zero objects from n objects is equal to 1 (a highly sensible result).

8. Originally Posted by mr fantastic
At your level: 0! = 1 by definition so that the number of way of choosing zero objects from n objects is equal to 1 (a highly sensible result).
Very sensible! Thank you!

9. You have already seen several excellent reasons why we define 0! = 1, but here is one more.

A fundamental recurrence relation for the factorial function is

$\displaystyle (n+1)! = (n+1) \; n!$

We would like this recurrence to hold even when $\displaystyle n = 0$. This requires

$\displaystyle 1! = 1 \cdot 0!$
$\displaystyle 1 = 1 \cdot 0!$
$\displaystyle 1 = 0!$

10. ## very nice!

Originally Posted by awkward
You have already seen several excellent reasons why we define 0! = 1, but here is one more.

A fundamental recurrence relation for the factorial function is

$\displaystyle (n+1)! = (n+1) \; n!$

We would like this recurrence to hold even when $\displaystyle n = 0$. This requires

$\displaystyle 1! = 1 \cdot 0!$
$\displaystyle 1 = 1 \cdot 0!$
$\displaystyle 1 = 0!$
This worked out very nicely!