Prove that Zero Factorial is Equal to One

#Factorial_ibmath #factorialnotation #Permutation_combination_made_simple #datamanagement_IBSL #Probability_Statistics #permutationandcombination #permutation_combination #combinatorics #permutations #Probability

[ioozeer]

By that logic, if you start with 0! where n=0, then 0! = 0 x (0-1)!
0! = 0 x (-1)!, and since 0 multiplied by any number is 0, then 0! = 0

[andrews]

Thank you, this helped me out in C++ because I did not know how to handle finding the factorial of 0 recursively.

[Steve-trem]

This proof seems incorrect to me. You are defining factorial as:

n! = n * (n-1)!

But the definition of factorial is the multiplication of all integers up to a given integer, starting at 1. When you set 'n = 1', the definition used in your proof is no longer consistent with the definition of factorial.

1! is simply 1.

However, if you define it this way:

n! = (n+1)! / (n+1)
0! = 1! / 1
0! = 1!

This holds true for any integer 'n' that factorial can be applied to.

The fact that you could never use that definition to actually find say 3! bugs me though. Its 4!÷4 duh.

[YodoJakamodo]

I don't think one can prove a convention for the first element of an inductive definition. I think 0! = 1 is simply the convention that allows the definition of the factorial to include the number zero.

[barneylaurance]

There didn't seem to be any justification for the step in the proof where 1! is replaced with 1. What was the proof that 1! is equal to 1?

(Of course we know that that's true, but if you're going to prove something as fundamental as the value of 0! it makes little sense to take the value of 1! on pure assumption)

[ZesteriaX]

Another way of proving that 0!=1 is using permutation
We know P(n, r)=n!/(n-r)!
P(1, 1)=1!/(1-1)!
And the number of permutation(rearrangement) you can make using one object taken one at a time is only one, i.e, P(1, 1)=1
Therefore we get that 1=1/0!
Therefore 0!=1
You can also use the combinations formula to prove this!

[andreyvasyaev]

Что мы знаем о факториалах...
Для начала мы знаем что
факториал следующего числа равен факториалу предыдущего числа умноженному на это самое следующее число...
N!= (N-1)!×N
или по другому... факториал предыдущего числа равен факториалу следующего числа деленному на это самое следующее число...
N!=(N+1)!/(N+1)
есть еще вид (N+1)!= N!×(N+1)...
значит (N-1)!=N!/N и N=N!/(N-1)!
При N=1 получаем 0!=1!/1 и 1=1!/0!
При N=0 получаем (-1)!=0!/0 и 0=0!/(-1)!
При N=(-1) получаем (-2)!=(-1)!/(-1) и (-1)=(-1)!/(-2)!
При N=(-2) получаем (-3)!=(-2)!/(-2) и (-2)=(-2)!/(-3)!
При N=(-3) получаем (-4)!=(-3)!/(-3) и (-3)=(-3)!/(-4)!
При N=(-4) получаем (-5)!=(-4)!/(-4) и (-4)=(-4)!/(-5)!
Видим что вычисление положительных факториалов по действию очень похоже на действие возведения в степень...
только множители различные...
Исходя из полученных формул отрицательный факториал берется не только от отрицательного значения но и имеет смысл обратных значений для положительных факториалов N...
Во всяком случае вполне возможно
N!=(N+1)!/(N+1)
0!=1!/1=1
(-1)!=0!/(0)=1/(0)= 1 неделённая единица
(-2)!=(-1)!/(-1)= 1/(-1)= -1
(-3)!=(-2)!/(-2)=(-1)/(-2)= 1/2
(-4)!=(-3)!/(-3)=(1/2)/(-3)= -1/6
(-5)!=(-4)!/(-4)=(-1/6)/(-4)= 1/24
(-6)!=(-5)!/(-5)=(1/24)/(-5)= -1/120...

Интересно что получаются обратные значения Гамма функциям от положительных значений когда
Г(N+1)=N!
Г(N+1)=N×Г(N)=N×(N-1)!
Немного неожиданно...
Получается что для отрицательных Г(-(N+1))=1/Г(N+1)=1/N!
Но есть "проблема" со знаком...

Видим что постоянно через один изменяется знак при делении "факториалов" от отрицательных значений...
Предположу что нужно брать для отрицательных значений N значение по модулю (а для обобщения и для положительных значений N...)
N!=(N+1)!/|N+1| (N-1)!=N!/|N|
0!=1/1=1
(-1)!=0!/0=1/0= 0 (относительный ноль)
или безотносительно единица неделённая что более верно...
Тогда следует (-2)!= (-1)!/|-1|=1
(-3)!=(-2)!/|-2|=1/2
(-4)!=(-3)!/|-3|=1/6
(-5)!=(-4)!/|-4|=1/24...
Как видим получаем обратные величины факториалов для положительных значений N...
но еще идет сдвиг на один ход относительно факториалов для положительных значений N...
Смею предположить что отрицательные факториалы должны считаться по формуле
N!=(N+1)!/|N|...
Тогда
(-1)!=0!/|-1|=1/1=1
(-2)!=(-1)!/|-2|=1/2
(-3)!=(-2)!/|-3|=1/6
(-4)!=(-3)!/|-4|=1/24
(-5)!=(-4)!/|-5|=1/120...
и получается что эти значения численно равны коэффициентам для нахождения "обратного факториала"...
Кстати по этой же формуле получается
0!=1!/0=1/0=1 единица неделённая
что наверное будет более верно...

Если уж быть совсем дерзким и исходить из того что график этих значений должен бы быть хоть немного математически красив то возможно факториалы от отрицательных значений должны бы быть и сами отрицательными...
Но я пока не нахожу физического смысла отрицательным значениям факториалов...
(самим факториалам от отрицательных чисел смысл проявился очень явно)...
к тому же придется признать что тогда при этом 0!=1/0=0 равен относительному нулю...
Но это пока мои личные фантазии...
и в этом надо сначала разобраться...
а перед этим хорошенько подумать...

Мне все же ближе "вариант с модулями"...

[yabetsedubale]

Sorry but I think this explanation is completely wrong. 0!=1 just because it is a convention. Its not something we can prove.

To start of your explanation about factorial is wrong. Factorial is the the product of an integer and all the integers below it until it reaches 1. So 1! Is just 1 not 1×(1-1)

I hope this was helpful for the viewers.

[dimkit]

ok, but now start to find the 0! from the factorial definition and you will end up with 1=0!=(0-1).0=-1.0=0 !!!!....simply it makes no sense looking for 0! because 0 is excluded according to the definition that states that factorial of a natural number is the product of that number with every whole number less than or equal to 'n' till 1...0 factorial makes no sense, because 0 is not included in the definition (or there is not any permutations for a set of objects that their number is 0, or simply for objects that do not exist)

[JeonJungkook-vxzl]

This came in our 1st unit test of 8 th grade and i couldn't do it, , it came in the half yearly exam too and i still couldn't do it, ,now when i saw the answer i feel so stupid, , its so easy😂

[waltz]

This isn't a valid mathematical proof. You're using the entire argument to prove itself. To make a valid direct proof you need to assume the premise then prove the conclusion. This isn't a proof. The truth is that we just made this up because it fits with the way we invented this math, just like x^0=1, it's not a statement that makes any logical sense, but we decided to fit it into our math anyway.

[yarragogusaiprasannakumar]

If we use n=0 then
0!=0*(0-1)!
0!=0*(-1)!
0!=0 (something multiplied by is zero)
0!=0.
(Comment your opinion regarding this).

[isaac]

This is the part of Mathematics that seem to stop being based on logic and are just assumptions and imaginery like i.

[ilovehonglong]

Using an assumption to prove an assumption... wtf

[waynecook]

I agree that the argument doesn't seem to work since it appears that 0!=0.(-1)! =0 but also how to prove 1!=1 or is it just a definition?

[DracoXul]

n! = n * (n-1)! is a generalized version of the full version n! = n * (n-1) * (n-2) * .. * 2 * 1
If you were calculate 1! by using the full form, you would get 1! =1 and no more.
The generalized version is not accurate for n = 1.
0 is not a Natural Number therefore 0! is undefined.

[ryanperkins]

Well then all factorials are actually zero because 1-0 will emerge in every set.
The proper way to say it is that 0 must be excluded.
Anyways what this proves is that 1factorial is zero not the other way around.

[rashiahuja]

So, I'm in 9th Nd knowing nothing about it but still here to understand it Nd I got it which was unexpected to me 😀

[catvio]

I'm doing an introductory to math course at the moment so I'm definitely no expert.. but I don't see how this is proof. Maybe I'm missing something, but using the formula of n! = n * (n-1)! with the definition of a factorial being that it is the multiplication of all integers up to the given integer, then when finding 0! you would do 0! * (0-1) which if you type this in your calculator is equal to 0, but also, doing 0! * (0-1) doesn't make sense because 1 is an integer that is above 0, not below it. So realistically I guess you would do 0! * 0 ? (But this is also equal to zero).

It's known that zero multiplied by anything = zero. So how come you didn't do the last step of multiplying the 1 by zero?

[waldwassermann]

I think it's fundamentally all about companionship really...