But Why Does 0 Factorial Equal 1

Показать описание

🙏Support me by becoming a channel member!

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.

0 factorial is 1 proof

0! =1

#math #brithemathguy #0factorial

Рекомендации по теме

Комментарии

🎓Become a Math Master With My Intro To Proofs Course! (FREE ON YOUTUBE)

BriTheMathGuy

If you graph it, you’ll immediately see that you’ll also need a definition for intermediate real numbers, otherwise factorial would not be continuous. Which begs the question, what about negative and complex numbers?

Misteribel

Can it be this ?
3! = 4! / 4 = 6
2! = 3! / 3 = 2
1! = 2! / 2 = 1
Similarly
0! = 1! / 1 = 1

mohdasifkhan

Considering an infinite product consisting of the quotient of two second-degre polynomials. Then for the special case that the roots of the polynomials differ by exactly 1 in either direction (root sum equal) we have the very surprising and simple result: The product (n going through the (strictly) positive integers) (An^2 +Bn + C)/(An^2 + Bn + D) = (A+B+C-D)/(A+B+D-C).

4:20 it would be interesting to see difference in 0! algebras set. like what would change when we move 0! result on ±∞ ± i*(±∞), I c4.

DeadJDona

4:31 careful with "0?" there. ;

God-ldll

I have always find it difficult to understand why people struggle to accept things such as x^0 = 1 and 0! = 1, but I suspect this is a problem with how the curriculum in every education system handles these topics. Schools spend too much time telling you how to compute things, without telling you how the operations are actually defined, and without giving you any conceptual understanding for why we even care about these definitions. If you understand what the definition of the operation n! is, then you will not feel the need to ask what is 0!, since 0! = 1 is an immediate consequence of the definition of n!. The problem is that the definition of n! is generally not taught, you are only taught how to compute n!, not how it is defined. The same goes with x^n. If you know how x^n is defined, then asking why x^0 = 1 is completely redundant, as it immediately follows from the definition of x^n. We should be focusing on explaining the definition of x^n in general, rather than only explaining why x^0 = 1 specifically.

I also think that part of the problem is that schools seem to have something I like to call zero-phobia: an irrational aversion of the number 0. They do this thing where they teach people to treat 0 as being fundamentally different from all other numbers, even though it conceptually makes no sense to do so. They refuse to treat it like a normal quantity you can just do operations with. So they do this very weird and nonsensical thing where they take an operation, they give an intuitive description of what an operation does for every natural number, but then they decide to arbitrarily exclude 0 from the definition, even though the result was perfectly okay, only to then have students ask "why does this operation with 0 work?" only to arbitrarily add 0 again back into the definition, with the justification of "it makes our notation look nice." This is a terrible way to teach maths! Why exclude 0 from the original definition in the first place when the definition already accounts for 0? So, maybe this is the reason why people struggle with something like 0! = 1. Because at face value, there truly is nothing surprising about this. n! is just the product of the n-tuple of positive integers less than or equal to n. So 0! is just the product of the 0-tuple of the positive integers less than or equal to 0. This is, the product of (). This is 1. The only part about this entire argument that perhaps is unconvincing is that the empty product is 1. But the empty product being 1 is really not all that technical or abstract. It makes sense intuitively. I have 0 objects to mutiply. This multiplication of 0 objects is the empty product. If I multiply this empty product, by some object X, then I get a multiplication of 1 object: the object X. But such a multiplication is equal to just the object itself: that is what our intuition tells us. So, EmptyProduct·X = X·EmptyProduct = X. X is arbitrary, right? So there os only thing EmptyProduct could possibly be equal to: 1.

angelmendez-rivera

Thank you so much. I really like your asmr! Always has the best timing

hiroitakezawa

“How do I arrange 0 objects?”
BriTheMathGuy: that’s the neat part, you don’t

threepointone

An example for the special case mentioned below. Numerator: 5n^2 + 37n + 42, (zeros: -6, -7/5), Denominator: 5n^2 + 42n + 60, (zeros: -5, -12/5). (Special case "1 difference") The infinite product converge to Agreeing with the general formula (2.4! • 5!)/(1.4! • 6!)=2.4/6=2/5.

You can also do: x! = (x+1)! / (x+1)
That means: 4! = 5! / 5
This also works with 0 and 1: 0! = 1! / 1 = 1 / 1 = 1
Oh, sorry! You said that.

DoxxTheMathGeek

As soon as I saw this to be a topic, I came up with my own technique (maybe)

Let's start with 3

3!=1×2×3
2!=1×2×(3/3)=3!/3
1!=1=1×(2/2)×(3/3)=3!/6
0!=(1/1)×(2/2)×(3/3)=3!/6=1!

Therefore,
1!=0!
0!=1

anikethdesai

I was explaining to my dad some Taylor Series expansions and I couldn't quite explain to him why 0! =1. Great video!

alexdotdash

1:07 The issue with the 1! = 1 * 0! demonstration is the next obvious question would be “but then what’s the factorial of -1?” If you continue that pattern then 0 * -1! = 0!, but that would mean whatever number -1! equals you would still have 0 multiplied by it is still 0, so 0! would have to be 0 which is a contradiction.

The Gamma function definition of factorial does though allow for factorials of negative numbers, they just end up being complex numbers.

Bodyknock