# [SOLVED] dividing by 0!

• Dec 9th 2008, 06:18 PM
yvonnehr
[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)... (Thinking)?

The calculator says this equals 1.
• Dec 9th 2008, 06:40 PM
Truthbetold
I don't recall why but 0! = 1.
• Dec 9th 2008, 06:55 PM
meymathis
$\displaystyle 0!=\prod_{k=1}^0k$
which is an empty product. An empty sum is 0. An empty product is 1. See this
• Dec 9th 2008, 07:35 PM
yvonnehr
a proof
Thanks, I will definitely look at this after my final! (Nerd)
• Dec 9th 2008, 08:43 PM
Chris L T521
Quote:

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, $\displaystyle \Gamma\left(1\right)=\color{red}\boxed{0!=1}$
• Dec 9th 2008, 08:48 PM
yvonnehr
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!!!(Rock)
• Dec 9th 2008, 10:10 PM
mr fantastic
Quote:

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!!!(Rock)

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).
• Dec 10th 2008, 05:49 AM
yvonnehr
Quote:

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!(Clapping)
• Dec 10th 2008, 04:43 PM
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

$\displaystyle 1! = 1 \cdot 0!$
$\displaystyle 1 = 1 \cdot 0!$
$\displaystyle 1 = 0!$
• Dec 10th 2008, 07:50 PM
yvonnehr
very nice!
Quote:

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!(Clapping)