Why Does 0! = 1

►WEBSITE

►MY COURSE
Prove It Like A Mathematician! (Intro To Math Proofs)

►BECOME A CHANNEL MEMBER

#math #brithemathguy #shorts

Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information. Viewers should always verify the information provided in this video by consulting other reliable sources.

[Ninja]

now show 0! = 1 using the gamma function.

[lordstevenson]

Some comments are asking what happens to negative factorials.
In fact the factorial is defined for all complex numbers expect purely real negative integers, using what’s called the Gamma function.
Look it up if your interested.

[jirkasirka]

But then for the zero: 0! = 0 × (0 - 1)! = 0 × (- 1)!
And what is (-1)! ?

[tamirerez]

O! = 1 because you can arrange O items in only 1 way.

[lovelynarhe]

Then that means...
0! = 0·(-1)! = 0

Because 0×(any number) = 0

Clearify me!

[TheRecklessGamer]

There is only 1 way to arrange 0 objects

[naivedyam]

A flaw in this method is this way the factorial of all negative numbers will be 1

[GileCalsmalar]

Definite: 1) n!=n.(n-1)! 2) 0!=1
Therefore, there is no proof of 0!=1, it is so by definition

[RhapsodyHC]

This is a bad explanation because if you plug in, it results that 0! = 0 * (-1)! which is undefined

[nopenopenope]

Now tell me when I would ever need this in day to day life

[ArjunJadhav-jk]

1:40:13 ye toh socha hi nhi tha kabhi🧠

[claycreate]

Isn't 0! = 0 * -1 = 0? Why draw the line at negative numbers but not at 0?

[Rudxain]

You know how the Imaginary Unit is defined as "x^2 = -1"? I wonder what kind of numbers we could get by doing *"x! = 0".*

Yes I know that looks impossible. Not even the Gamma function returns 0, regardless of input. But maybe we could find a use case for such an "impossible" number?

[muhtarjohn]

Think about they are going to take a picture how can you take a pic without someone. Its 1 that's why 0! = 1

[VihaanBelani]

Watch Eddie Woo's video on this topic

[BobChess]

So if 1 = 0! Because 1! = 1*(0!)

That mean 0! = 0*(-1)! which is equal to 0.

[PedroHenrique-qhbl]

Just do the opposite:

3! = 6
2! = 6 / 3 = 2
1! = 2 / 2 = 1
0! = 1 / 1
(-1)! = 1 / 0 (sadly)

[Darckseyes]

Don’t really like this explanation because it falls apart when we go one more step and let n=0

[pelasgeuspelasgeus]

Listen up dorks. The definition of factorial is one : n! =1*2*3*(n-1)*n and it makes sense for positive integers and only.
It's obvious that based on the definition 0! should be equal to 0 and not 1. Now, some brainiacs say that 0! must be 1 so that the type n! =n* (n-1)! can be computed for n=1, but it's obvious that the type is valid only for n>1.
I know there is a way to compute non integer factorial but that's another fabrication which is actually an approximation/interpolation based on closest integers and it's not in agreement with the definition.

[leesweets]

I would say this is a bad argument. There are better ways of using the recursive approach to proving this claim. You never once mentioned the 0! at any of your 4! through 2! examples. Its only until you got to 1! that you arbitrarily chose to include the 0!, out of nowhere. Why would you do that? A naive definition of factorial is to multiply all integers up FROM 1 to n. That definition alone precludes any 0! from discussion as undefined. I mean think about it. If youre going to include the 0! as part of the recursion, why end there? Why not also include the (-1)! and the (-2)!, so on, as well? If you want to arbitrarily stop at 0! then you could just as easily arbitrarily stop at 1!, and it really proves nothing to go any furhter.