Prove that Zero Factorial is Equal to One

What if n=0 instead of 1?

Im not sure how this proves it to be true. Well, it does but partially. How can we prove that 1! is 1? Since by using this proof, all we know is that 1! = 0! I don't mean this in any negative way, just genuinely curious.

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)

He sounds as oogway

My math teacher told me that this is my homework ez thx

Haha

girl math

?????!!!!!!
but 1! = 1 🙃

0! should really be undefined because n! is the multiplication of n by itself and all POSITIVE integers, and since 0 is neither positive nor negative, it shouldn't really be counted.

I get the proof but it's still a little confusingly mysterious. 😂

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

You forget that n! works only for n>0.

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.

Dai thayoli yevarada nee

If this is only the proof of finding 0!.
Then why 2! Is not equal to :
2! = 2×(2-1)×(2-2)
2!=2×1×0
2!=0.
But 2! Is equal to 2 not 0.
So how we will find 2!

Wouldn't 0! =1! mean 0=1 ?

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.

mathematicians can prove anything they like

There is no proof of 0!=1 ....it's just assumption

Wrong video

Thanks

I don't think this is legit.
Of we remove factorial symbol in the rightside and write down the actuals. It says n-2 .n-3.n-4 ....
In the begining we have 1-1 .and rest are all the products. So obviously we get zero. Not zero factorial.

This explanation is just a snake biting its own tail.

So I think the factorial applies only if n-1 >0

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?

Thank you so much sir!!

well i don't think this explanation is logically correct because , you are just assuming 1 factorial is 1 and then prove that 1fact = 0factorial is equals 1fact , but its ok for a 1 min video

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

Sets out to prove 0! is 1, immediately provwes its 0, since 0 times anything is zero. therefor N * (N-1)! equals zero times doesnt matter.

but factorial sign is where gone

👍👍

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"...

Thank you so much

i still am not sure that 0! is 1 because when you say that a factorial is equal to n. (n-1) its based around whole numbers, and zero is not a whole number... in fact its not even a number. its a simple placeholder of potential. I think a fallacy is afoot. but maybe thats just me.

Why would 0! not be considered 0 * (-1)!

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).

Tq SO much sir, u are the grate mathematician sir🙏🙏🙏🙏

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.

How do u get 1! As 1

but if substitute n=0 we get 0!=0*-1! hence 0 factorial=0

But we can't really accept this fact even though we can prove it mathematically first of all 0 is not a natural no and we define factorial for natural no only so how ??????????

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)

Thanku sir 🙏

Respectfully, your proof is incorrect. 'n' has to be a "non-negative" integer in the definition. This naturally excludes '0'. 0! =1 according to vacuous product convention.

i’m still taking alg 2 but cool vid helped me understand it a lot

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?

Woah
Thank you
Very simple 👏👏👏

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.

if i don't know what is 1! than i can expand it according to formula that 1! = 1*0! than how do i get the both value of 1! and 0!. Please help me out of this

I did not understand the last step of your proof at all: if 1! = 0! then 1 = 0! .... why?! .... the other thing: if you define n! as n*(n-1)! then in your last step 0!=0(0-1)! which is equal to 0 ... isn't it? And this contradicts with the whole thing you try to prove (0!=1). I think your prove is simply incorrect. I think 0!=1 is an convention/axiom. No need to prove it.

But... What if you try sub 0 into n! ?
Then wouldn't it be 0!=0.(0-1)! ?
Which is undefined
Or if you apply the same logic in this video it would be 0
I don't understand man

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.