why 0! = 1 ? | quick and easy proof #maths

Why does 0! (zero factorial) equal 1? It's a question that often stumps students, but the answer is surprisingly simple! This short maths video breaks down the concept of factorials, explaining how they're defined and showing the pattern that leads us to 0! = 1. We'll explore how this definition maintains consistency in various mathematical formulas and applications. Perfect for students and anyone curious about the logic behind this fundamental mathematical concept. #maths #0factorial #factorial #proof
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[adwz]

There are so many reasons as to why we should set 0! = 1. Here are a few:

i) Use the gamma function, proven to be the unique analytic continuation of the factorial to the complex plane (by Wielandt’s Theorem):

Γ(z) = ∫[0 to ∞] t^(z-1) e^-t dt,

Γ(n) =(n-1)! for positive integer n.

0! = Γ(1) = ∫[0 to ∞] e^-t dt

= -e^-t ] ∞, 0 = 0 - (-1) = 1



ii) n! = n(n-1)!

==> 1! = 1(0!) ==> 0! = 1

(video, also can use this to show why factorials of negative integers are (simple) poles)



iii) n! = | number of arrangements of (1, 2, …n) | . For example:

3! = | {1, 2, 3}, {3, 1, 2}, {2, 3, 1}, {1, 3, 2}, {3, 2, 1}, {2, 1, 3} | = 6

1! = | {1} | = 1

0! = | ∅ | = 1, where ∅ = {} is the empty set.



iv) e^x := Σ[n ≥0] x^n / n!

= 1 + x + x^2 / 2! + …

By convention, it only makes sense to set 0! = 1 looking at the constant term.



v) nCr = n! / [ (n-r)! r! ]

nC0 = 1, again the number of ways to arrange 0 items out of n is simply | ∅ | = 1 by definition.

==> 1 = n! / [n! * 0!] ==> 0! = 1

(The same argument works with 0C0 = 0!/(0!)^2 = 1 ==> 0! = 1)



vi) Can use the Beta function; for Re(a), Re(b) > 0:

B(a, b) = ∫[0 to 1] t^(a-1) (1-t)^(b-1)dt

= [Γ(a) Γ(b)] / Γ(a+b) (can be shown)

B(1, 1) = ∫[0 to 1] dt = 1

= [Γ(1)^2] / Γ(2) = Γ(1)^2

==> Γ(1) = 0! = 1 (as Γ(x) positive for x > 0)



vii) This is a stretch, but alludes to point ii) about poles of the gamma function. For n >= 0 integer,

Res(Γ(z), -n) = lim z -> -n [(z+n) Γ(z)]

= … (using different forms of Γ(z))

= (-1)^n / n!

Γ(z) has Laurent expansion about 0:

Γ(z) = 1/z - γ + O(z), where γ is the Euler-Mascheroni constant.

==> Res(Γ(z), 0) = 1. Set n = 0 in above residue formula.

==> Res(Γ(z), 0) = 1 = 1/0!

==> 0! = 1



There are probably a million more reasons I could come up with. None of the above arguments are circular (i.e. they don’t use the fact that 0! = 1 to prove itself), but you may notice a lot of the reasons use an argument similar to “it is sensible to” “it makes sense to” etc… They are plausibility arguments rather than proofs.

[EzioAuditore-ksts]

You can use pi function and gamma function too, but this is more creative, well done!

[AgastyaGD]

because x!/x=(x-1)!
for example 4!=24
and 24/4=6=3!
(3!=6)

[KookyPiranha]

or you could think about it intuitively

0! represents the total permutations of 0 items which is 1

[mathematicsman]

0! Become undefined implies 1! Become undefined and then 2! Become undefined....then n! Become undefined>< contradiction 😊

[Hemangvashishth]

Will you please add more videos on geometrical proofs sir

[RoaAwais-ognk]

Brother you're literally genius thanks😮😮❤❤❤

[MaxSwitzer-ibns]

Γ(z)=(z+1)!, let z=-1, Γ(-1)=Π(0)=1, Euler and Gama functions is a better way to solve it because multiplying infinite seiries' can prove less strait forward like when you get {N}=-1/12, and10^∞=-1

[Logicbhai-Logic.]

Understanding in school ❌😅😅
Understanding in this short ✅😉

[AniketKumarRoyRoboticsEngineer]

what if n is some negative number?? :P, let n = 0, -1, -2...

[alexanderx]

1! = 1 * (1 - 1)! = 1 * 0! since you're trying to prove that 0! = 1 then you can't calculate 1 * 0! without assuming beforehand that 0! = 1 but you're trying to prove that 0! = 1 in the first place so you can't calculate anything using 0! because you don't know beforehand what 0! represents since this is the factorial number you're tring to discover

[YugamGarg-mt]

but it should also be proved that this formula is also applicable for numbers beyond 1 as logically speaking it wont

[shivamshukla]

n!=n(n-1)(n-2)...3×2×1
This implies that n factorial can be opened only upto 1. When we see a series like 100, 99, 98, ...2, 1, this means the series ends at 1 and 0 isn't part of the series.
So making 1!=1×0! doesn't make sense by definition when opening n!, it should end at 1 and cannot go beyond that. So 1!=1 just this it cannot go beyond this. From what I understand.

[AkhiDas-ptyp]

We know that, n!=n‌‌‌‌‌‌‌‌pn Or, n!=n!/n(n-n)! Or, 0!=n!/n!. And the answer is 1.simple logic.

[Omagad]

Well, is 1/n! better to type 1/0×1×2×...×n or 1/n×(n-1) ?
I used the first method that I learned in class but as I don't have a university type calculator so there is no "!", I had to type it by hand but i couldnt type everything

[KavianAM-uvwv]

No arrangement is indeed one arrangement. 😊

[MoxHypKa]

This is s-hit of a proof. By definition multiplying starts from 1. Yet you trick that it may start from 0. You are breaking the definition, no wonder you can "prove" anything with that. 0! is a predefined constant for rare special occasions. Why does one need to prove a constant at all!?

[im_spyder]

WAIT!!!
THIS WORKS!!
Honestly, this is exactly what I did one day randomly to prove when thinking about proving 0!=1, just knowing the definition of a factorial...
But I also found something weird, if we substitute n=0, doesn't that just provide 0!=1?
So, if you're reading this, please consider helping me :)

Edit: It gives 0!=0 not 1 (perhaps)

[Ескендір-бр]

Easiest proof ever:
0!=1
*We know that*

[Benekuri]

You made this? It actually made math easy to understand