The replies are wrong. The factorial is still not defined for numbers that aren't nonnegative integers. What Desmos and other calculators are using are probably something like (but not necessarily) the gamma function which extends the factorial to the reals.
@@Deaddev1there is a youtuber who explained this really well called “lines that connect” in his video “how to take the factorial of any number”. You should go watch it! but in a nutshell, if you try to find the factorial of a non integer, you have to make assumptions about the properties of factorials that cannot be proven to be true or false
with a full deck of cards (52), if you properly shuffle it, it is almost certain that that order of cards has never been seen in history the reason why being is that 52! is such a huge number that it can’t fit inside our world
This isn't actually a proof that 0! = 1. If you're defining factorials as n! = n*(n-1)! then how do you know that 1! = 1 without already assuming 0! = 1. You're assuming 0! = 1 when you state the value for 1! and then working back to show your assumption is true, which just doesn't work. 0! does equal 1, but the actual proof is more interesting
@AchevasTV, your math is flawed, if you disagree answer this... In the equation that you derived.. 1! = 1*(0!) We don't know what 0! Is, RHS is unknown, then how can you deduce LHS is 1? It can be zero, matter of fact it can be any number! Until you define it as *"let 1!=1"* that's a whole different story!
@@utubevideos3317Because the definition of a factorial is n! = n*(n-1)*(n-2)…*(n-n+1). Therefore 1! = 1 as the calculation terminates at the first term n, as n-1 < n-n+1.
@@NewChannel-mm2zi devil is in the details ... Definition of a factorial in simple terms is"product of all integers from 1 to n" and so you have it in your equation: n! = n*(n-1)*(n-2)...(n-(n-1)) and when n=2 you have it 2! = 2*1 But for 1! You try to substitute in that equation, you never get 0 because by definition it's product of all integers upto 1 so you aren't allowed to include anything less than 1 Now that's the reason 0! = 1 and that is by *definition* ... If you try to prove it mathematically, it's not possible because you have to invoke it's definition to arrive at the result and that's what am saying!
@@utubevideos3317 There was another explanation that was better for explaining 0!=1, I don't remember by who though. It used dividing previous terms of the series, so ex. 3!=6, 2!=3!/3=2, 1!=2!/2=1, and so 0!=1!/1=1
excellent! I always tell my students by convention (or by definition) 0! =1. Just like 1 can be divided by 1 and itself but by definition, it is not considered to be a prime number
i learned that 2² = 4 2³ = 8 put it in reverse ( divide from 2 starting with 8 ) 2¹ = 0 2½ = 1.414..... so put same logic 1! = 1 2! = 2 so, 1/1 = 1 = 0!
This explanation is wrong. Factorial definition: For any positive integer n, n! is defined by the product of all positive integers not greater than n. This EXCLUDES 0 and all negative numbers from the definition. The DEFINITION 0! = 1 was added mainly to simplify formulas containing 0! to avoid introducing exception for those cases. For example, the binomial coefficient defined as (n k) = n! / (n-k)! k! would need an exception in its definition for (n n) or (n 0) = n! / 0! n! if 0! was not defined. Same problem for e^x = 1 + x + x^2/2! + x^3/3!... which can be written e^x = Sum of x^i/i! for i varying from 0 to infinity (very short formula with mathematical symbols). Hence 0! had to be defined as equal to 1 for the simplicity of many formulas to avoid including exceptions when n=0. Other arguments can be used to choose 0! = 1. In summary 0! = 1 is a DEFINITION added to the general definition of factorial that EXCLUDES (0!). You confuse a definition with the coherence of the definition with a general property of the factorial. Mathematics is a precise field and should, according to me, taught with precision.
@@SamuraiGamer91111 His explanation is totally wrong too. 0! = 1 is a definition and not a consequence of the general definition of n! WHICH IS NOT DEFINED FOR n=0. I have already explained this above.
Well this is also not a correct explanation, not even the correct definition of factorial, the mathematical definition is n!=integral e^(-x)x^(n-1)dx from 0 to inf
I always thought that the last step a factorial can go is 1. Therefore 1! can not have any value and 0! is out of the question. But this video just melted my brain lol
If you're curious about 0! and fun patterns, check out Pascal's triangle using combinations. There's a lot of 0! coming into play to make the 1s at the edges.
Numberphile does a video on this, which I watched recently and I'm sure this is why the algorithm deemed this should show up in my feed, and they explained it pretty much the exact same way, just slightly backwards if that makes sense. But it all makes sense.
Good try! This is the way of learning mathematics, always think out of the box and always challenge yourself and others, even the idea is 'funny' or 'crazy'.
0! = 1 is more of an axiomatic choice. You have to take it as 1; otherwise, so many different parts of mathematics fall apart. Using the integer definition, there is no way to actually calculate 0!.
Thats saying that 1!=0!. If 3! = 3*2*1 then 0!= 0*(0-1)! which no matter how you look at it will be 0. Your formula, not mine. I think 0!=1 because ∅ is a singular object and {∅} number 1 is a set of one object. 0 isn't a positive integer though so cant be factored despite being the basis of natural numbers.
Your penultimate step states: 1! = (1)(1-1)! So, simplify each side by diving by 1. Multiplicative identity property states that any real number n multiplied (or divided) by 1 equals n. Thus, 1! = (1-1)! 1! = (0)! 1! = 0! Therefore, your math does not prove that 0! = 1. On your last step, you just dropped the ! symbol off the 1 without a valid reason.
Factorial only applies to non-negative (positive) integers. It does not apply to all real numbers (including decimals or negatives). And it does not apply to negative integers.
Invalid argument; the generalized case should be n(n-1)! where n is not equal 0. To prove this plug 0 in for n and then 'do the math'. 0! should be undefined.
3! = 3 x 2 x 1, write out 0! In the same way. You can’t. You can’t get to n! = n(n-1)! using 0 as the starting point. The rule of factorial is that you multiple whole numbers descending to or including 1. So 1! is 1x1. 0! violates the rules of the operation. Also how can any multiplicative function including 0 as a factor ever be >0? Finally using the equation of n!= n(n-1)! , if one uses -3 as n then is it true that -3! = -3(-3-1)! … -3! = -3(-4)! …-3 = 12?! No, -3! Is an imaginary number. And so should 0! be seen as well.
Wow! This is the best expansion from a pure mathematical standpoint of why 0! is 1... However, I still think it makes no sense from a practical standpoint.
by definition of n! = n(n-1)(n-2)...(3)(2)(1) we can then find any factorial of any positive integer, except we don't know what happens if n=0, thus we need to find a pattern as shown in the short
This isn’t a fully accurate proof but rather a trick to show how you can manipulate a formula. It’s using the definition you’re trying to prove in the same expression Remember that factorials come from combinatorics, and for nCr, your explanation breaks because it allows division by 0, which is undefined This means that 0! Is conventionally defined then as 1, but why? From set theory, think of 0! as all the ways I can choose something (or nothing) from an empty set. The answer is 1.
I think 0 factorial is not even possible as when we are finding factorials we just find the multiplication of all the natural numbers till the number for positive integers and all negative integers starting from -1 till the number. So 0 is never in the list. No matter which method we choose 0 is not fitting in any one of them. And also he took formula wrong during 1! As actual formula is n!=n[(n-1)!] not n!=n(n-1)!
@@AchevasTVIt seems arbitrary to say that you can’t perform factorial on negative numbers. Why not declare factorial invalid for any non-positive number, and so make 0! undefined? Is there a mathematical reason for this?
If you actually plug n=0 into the formula, you will get 0! = 0(-1)! , Then since 0! = 1, 1 = 0 x (-1)! , So (-1)! = 1/0 which we know to be undefined, therefore (-1)! is also undefined
@@clover-00 That means, 3! = 6(3-3)! 6 = 6(0)! 0! = 6/6 0! = 1 Yeah, seems like it. But how could 0! equal to 1? Is there an explanation for it in real life? Cuz it could be a math error. I'm not a professional so I want to learn why it turned out to be.
@@clover-00 I've been wondering about this for a while now but why do we put 0 or anything that represents it in equations when we haven't figured out yet what anything divided by 0 is equal to? I mean, 0 represents nothing. But you can still take nothing from nothing.
I still don't understand it. 1! = (1) * (1-1)! is then: 1! = (1) * (0)!, and removing the brackets, we have: 1! = 1 * 0!. Anything multiplied by zero is zero. So shouldn't both 1! and 0! be zero? Unless, by convention we accept that 0! = 1, in which case the equation becomes 1! = 1 * 1. But that doesn't explain why 0! = 1. This video seemed to gloss over the idea that (1)(1-1)! = 0!, and that is the part I'm struggling with. I just don't understand how that equation makes anything but zero.
@@mischievous_luffy Sorry, I still don't understand it. You are saying to solve for 0! But that's the problem I'm having. How does dividing by 0! equal anything other than zero?
@@JRCSalter let 0! = some constant (say 'x') Now initially we had the equation, 1! = 1 × 0! Replace 0! by 'x' 1! = 1 × x 1 = 1 × x [since 1! = 1] We get x = 1 Now we initially stated that 0! = x Replace 'x' by 0! You get 0! = 1
@@mischievous_luffy The trouble I'm having is that I don't understand how we got from (1)(1-1)! to 0!. We are essentially saying that 1 * 0! = 0! = 1. 0 doesn't have anything to multiply itself with, and therefore would equal 0.
@@mischievous_luffy I think I understand your reasoning, but it doesn't help explain why 0! = 1. It just seems like we use it as a synonym for 1, rather than as a calculation. With all other factorials, we take the number, and multiply by the next lowest positive number and so on. So we should take 0, and multiply by the next lowest positive number, but there isn't one, so it would be 0*0=0 That logic to me makes much more sense than what was described in the video.
For the last part, you said that 1(1-1)!=0! So you after the parentheses it’s 1x0! 0!=1, so 1x0!= 1. Do you make the equation more like 1(0)! ? So in 1(0)! Do you multiply the 1 and 0 first, or find the factorial of “0” first?
So that's reason why the factorial cannot
It can be less than 0 but can’t be an integer less than 0 because using an expression makes division by 0. But (-0.5)! ≈ 1.7725
The replies are wrong. The factorial is still not defined for numbers that aren't nonnegative integers. What Desmos and other calculators are using are probably something like (but not necessarily) the gamma function which extends the factorial to the reals.
@@lucaswarnke3668desmos uses the gamma function for the graph of a factorial to extend it to real numbers
@@NeptuneMood08I don't understand. How do you calculate (-0.5)! Pls explain
@@Deaddev1there is a youtuber who explained this really well called “lines that connect” in his video “how to take the factorial of any number”.
You should go watch it! but in a nutshell, if you try to find the factorial of a non integer, you have to make assumptions about the properties of factorials that cannot be proven to be true or false
6!=720, if I have a pack of 6 cards, there is 720 ways to arrange them all without duplicates. 0 cards can be arranged one time, which is with 0 cards
Thx man I understood more becuz of u
That makes a lot more sense for me
then why 6! is 720 when you should count arranging with 0 cards
with a full deck of cards (52), if you properly shuffle it, it is almost certain that that order of cards has never been seen in history
the reason why being is that 52! is such a huge number that it can’t fit inside our world
I think 0 cards can't be arranged at all because they don't exist.
I love it when math teachers are enthusiastic about their subjects
This isn't actually a proof that 0! = 1. If you're defining factorials as n! = n*(n-1)! then how do you know that 1! = 1 without already assuming 0! = 1. You're assuming 0! = 1 when you state the value for 1! and then working back to show your assumption is true, which just doesn't work. 0! does equal 1, but the actual proof is more interesting
His computation is wrong. There's no such (n-x)=0, (n-x)>0
You are right
0! has to be 1 in order for 1! to be 1.
Thats what i was thinking!
This just raises more questions.🤔
Unfortunately, you'll need to learn calculus to understand apparently
thank you sir!
You cleared this confusion ...Thnx...Keep it up..n never stop making these useful videos.....
Thank you, I will
@AchevasTV, your math is flawed, if you disagree answer this... In the equation that you derived..
1! = 1*(0!)
We don't know what 0! Is, RHS is unknown, then how can you deduce LHS is 1? It can be zero, matter of fact it can be any number!
Until you define it as *"let 1!=1"* that's a whole different story!
@@utubevideos3317Because the definition of a factorial is n! = n*(n-1)*(n-2)…*(n-n+1). Therefore 1! = 1 as the calculation terminates at the first term n, as n-1 < n-n+1.
@@NewChannel-mm2zi devil is in the details ... Definition of a factorial in simple terms is"product of all integers from 1 to n" and so you have it in your equation:
n! = n*(n-1)*(n-2)...(n-(n-1)) and when n=2 you have it 2! = 2*1
But for 1! You try to substitute in that equation, you never get 0 because by definition it's product of all integers upto 1 so you aren't allowed to include anything less than 1
Now that's the reason 0! = 1 and that is by *definition* ... If you try to prove it mathematically, it's not possible because you have to invoke it's definition to arrive at the result and that's what am saying!
@@utubevideos3317 There was another explanation that was better for explaining 0!=1, I don't remember by who though. It used dividing previous terms of the series, so ex. 3!=6, 2!=3!/3=2, 1!=2!/2=1, and so 0!=1!/1=1
excellent! I always tell my students by convention (or by definition) 0! =1. Just like 1 can be divided by 1 and itself but by definition, it is not considered to be a prime number
His equation
1! = 1*(0!)
We don't know what 0! Is, RHS is unknown, then how can he deduce LHS is 1? It can be zero too!
I wonder whether the reason 1 is not considered prime is because it is a perfect square?
@@utubevideos3317 finally! Someone who reasons beyond.
1!=1 not 0
The teaching video is wrong
i learned that
2² = 4
2³ = 8
put it in reverse ( divide from 2 starting with 8 )
2¹ = 0
2½ = 1.414.....
so put same logic
1! = 1
2! = 2
so,
1/1 = 1 = 0!
Wait how did you do 2½=1.414?
Isn't √2=1.414?
@ace5161 N raised to power of 1/2 is √n
@@nexus6187 No I solved it in calculator and it gave me 1 as the answer
@@nexus6187 Nvm you're wrong it's 1 not 1.414
@ace5161 It's not 2*1/2
Its 2^(1/2)
This explanation is wrong.
Factorial definition: For any positive integer n, n! is defined by the product of all positive integers not greater than n. This EXCLUDES 0 and all negative numbers from the definition. The DEFINITION 0! = 1 was added mainly to simplify formulas containing 0! to avoid introducing exception for those cases. For example, the binomial coefficient defined as (n k) = n! / (n-k)! k! would need an exception in its definition for (n n) or (n 0) = n! / 0! n! if 0! was not defined. Same problem for e^x = 1 + x + x^2/2! + x^3/3!... which can be written e^x = Sum of x^i/i! for i varying from 0 to infinity (very short formula with mathematical symbols). Hence 0! had to be defined as equal to 1 for the simplicity of many formulas to avoid including exceptions when n=0. Other arguments can be used to choose 0! = 1.
In summary 0! = 1 is a DEFINITION added to the general definition of factorial that EXCLUDES (0!). You confuse a definition with the coherence of the definition with a general property of the factorial. Mathematics is a precise field and should, according to me, taught with precision.
Finally
Watch the video by Eddie Woo, he explains it better than this guy and also, yes, 0! = 1.
@@SamuraiGamer91111 His explanation is totally wrong too. 0! = 1 is a definition and not a consequence of the general definition of n! WHICH IS NOT DEFINED FOR n=0. I have already explained this above.
@@michelpitermann5335 well i didnt read the whole thing, thats what a TL;DR is for
Well this is also not a correct explanation, not even the correct definition of factorial, the mathematical definition is n!=integral e^(-x)x^(n-1)dx from 0 to inf
That's amazing sir, thank you
You are welcome!
The way I like to think about it is that 0 items can only be arranged in 1 way. So 0! = 1 permutation
but how can zero items be arranged in any way
@@thebushmaster0544 the “arrangement” is that you have nothing.
@@kingdedede5933Damn bro you are smart.I was always confused about this.Thanks for clarifying it for me.
It is 0 objects... Therefore, there can be AN INFINITE number of ways to arrange "NOTHING" - so this explanation is a cop-out!
I always thought that the last step a factorial can go is 1. Therefore 1! can not have any value and 0! is out of the question. But this video just melted my brain lol
If you're curious about 0! and fun patterns, check out Pascal's triangle using combinations. There's a lot of 0! coming into play to make the 1s at the edges.
Numberphile does a video on this, which I watched recently and I'm sure this is why the algorithm deemed this should show up in my feed, and they explained it pretty much the exact same way, just slightly backwards if that makes sense. But it all makes sense.
Sir put
0!=0(0-1)!
0!=0
✓ or ×
Hi pal, this won't work because you will be introducing (-1)!, which will cause the entire equation to be undefined.
@@AchevasTV
But
0×(any N.o)=0
@@adityarathor745 Haha, (0)x(undefined) is actually undefined. Try press this into your calculator (0)x(1/0).
@@AchevasTV
Thankyou sir
Good try! This is the way of learning mathematics, always think out of the box and always challenge yourself and others, even the idea is 'funny' or 'crazy'.
Thank you for this excellent explanation!
Very good explanation!!
Good explanation ❤️
Best teacher ever👍👍❤❤
Thank you!!
My curiosity makes me come here😅
Thanks sir I understand easyl
Glad you did!
But wouldn't that mean
0!=0((-1)!)
Thank u so much for making this video ❤
Thanks for this sir
You are welcome!
Très clair, thanks
In reality it comes from gamma function integral
Very elegant sir
Thanks for explanation
You're welcome :)
Your a goooooood teacher. thank you so much.
0! = 1 is more of an axiomatic choice. You have to take it as 1; otherwise, so many different parts of mathematics fall apart. Using the integer definition, there is no way to actually calculate 0!.
Thank you sir
Thanks
Welcome
That's good explanation
I think of it as the no of ways of arranging 0 things which is simply 1 hence 0! = 1
Hey u are amazing. Thankx
Thank you!!!
Great explanation👏, could you also explain why anything to the power of zero equal to 1
excellent Sir
Thats saying that 1!=0!. If 3! = 3*2*1 then 0!= 0*(0-1)! which no matter how you look at it will be 0. Your formula, not mine. I think 0!=1 because ∅ is a singular object and {∅} number 1 is a set of one object. 0 isn't a positive integer though so cant be factored despite being the basis of natural numbers.
Thank you 🙏
Amazing😲🔥🔥🔥🔥
love you sir
Your penultimate step states:
1! = (1)(1-1)!
So, simplify each side by diving by 1. Multiplicative identity property states that any real number n multiplied (or divided) by 1 equals n.
Thus,
1! = (1-1)!
1! = (0)!
1! = 0!
Therefore, your math does not prove that 0! = 1. On your last step, you just dropped the ! symbol off the 1 without a valid reason.
Cool 😂
But we didn't specify what n can be. In particular, does the formula hold for negative integers n?
I am in year five but I can kinda understand this
The very definition of factorial precludes negative integers. It applies to non-negative (positive) integers.
So its impossible to use 0 as n? The formula is useless in this case?
Well the factorial sign means there’s 1
Left like algebra
Finally, now I understood why 0! Is 1.
excellent
2>1 or 2
1! = 0 !
1 = 1
thanks from Bangladesh ❣️🥰
will it work in c++?
I have a question. Why do we change 1 factorial to just 1 in the last line?
Because his mathematical reasoning is bogus and wrong.
WRONG PROOF!! You make all mathematicians trigger..😂😂😂
Prove then.
But what if n was 0…..
Factorial only applies to non-negative (positive) integers. It does not apply to all real numbers (including decimals or negatives). And it does not apply to negative integers.
Empty product must be the neutral element.
This is where the universe came from
So thats why we take the 0! as 1
Dayum
Invalid argument; the generalized case should be n(n-1)! where n is not equal 0. To prove this plug 0 in for n and then 'do the math'. 0! should be undefined.
Flaw, you don't assume LHS without getting rhs
Thanks sir
This is my doubt .
Most welcome
3! = 3 x 2 x 1, write out 0! In the same way. You can’t. You can’t get to n! = n(n-1)! using 0 as the starting point.
The rule of factorial is that you multiple whole numbers descending to or including 1. So 1! is 1x1. 0! violates the rules of the operation. Also how can any multiplicative function including 0 as a factor ever be >0?
Finally using the equation of n!= n(n-1)! , if one uses -3 as n then is it true that -3! = -3(-3-1)! … -3! = -3(-4)! …-3 = 12?! No, -3! Is an imaginary number. And so should 0! be seen as well.
Still looking for the practical application in reality tho
Nice
Wow! This is the best expansion from a pure mathematical standpoint of why 0! is 1...
However, I still think it makes no sense from a practical standpoint.
It takes Nothing, to be 1 with itself. Everybody else has to ask who came before…
great🧠
For n greater than 0
This became theoretical proof what about the experimental or practical proof? 😅
1=1 golden rule perfect balance
There is one way to arrange zero objects
dang okay
i see how it is
super
0⁰=1 ?? How??
But then what is 1! ?
If 1!= 1×0!
Then 1! Doesn't have any definite value in this sense as you equate 1! as 1 somehow.. how do you know that 1! Is 1?
by definition of n! = n(n-1)(n-2)...(3)(2)(1)
we can then find any factorial of any positive integer, except we don't know what happens if n=0, thus we need to find a pattern as shown in the short
If 1 = 0!, can 10 = 9! as well? Or 2 = 1! 😅?
So 0!=0(-1)!, therefore -1! can be anything.
This isn’t a fully accurate proof but rather a trick to show how you can manipulate a formula. It’s using the definition you’re trying to prove in the same expression
Remember that factorials come from combinatorics, and for nCr, your explanation breaks because it allows division by 0, which is undefined
This means that 0! Is conventionally defined then as 1, but why?
From set theory, think of 0! as all the ways I can choose something (or nothing) from an empty set. The answer is 1.
Love from india🇮🇳🇮🇳🇮🇳
faktowia
I think 0 factorial is not even possible as when we are finding factorials we just find the multiplication of all the natural numbers till the number for positive integers and all negative integers starting from -1 till the number. So 0 is never in the list.
No matter which method we choose 0 is not fitting in any one of them.
And also he took formula wrong during 1! As actual formula is
n!=n[(n-1)!] not n!=n(n-1)!
0!=1 returns true cause != checks for "is not" lol XD
This isn’t a proof this is just a list of true statements.
So -1!=0?
No you can't. You cannot perform factorial to a negative number.
@@AchevasTVIt seems arbitrary to say that you can’t perform factorial on negative numbers. Why not declare factorial invalid for any non-positive number, and so make 0! undefined? Is there a mathematical reason for this?
If you actually plug n=0 into the formula, you will get 0! = 0(-1)! , Then since 0! = 1, 1 = 0 x (-1)! , So (-1)! = 1/0 which we know to be undefined, therefore (-1)! is also undefined
@@LargeDivisor Sorry for being 10 months late, but you should look up the pi function. Really mindblowing stuff
0!=0×(0-1)!
0!=0😂
0!=1😂😂😂😂😂😂😂😂😂
Noob in math😂😂😂😂😂😂😂😂😂
don't let bro cook
You wrote,
1!=1(1-1)!
And then,
1=0!
Doesn't that mean 3=3(2!) Which should be false?
Sorry, I'm clueless.
how can it be false??
here 1! = 1
so
1! = 1(1-1)!
by putting 1!=1 equation becomes
1 = 1(1-1)!
1 = 1(0)!
1/1 = 0!
0!= 1
@@clover-00
That means,
3! = 6(3-3)!
6 = 6(0)!
0! = 6/6
0! = 1
Yeah, seems like it. But how could 0! equal to 1? Is there an explanation for it in real life? Cuz it could be a math error. I'm not a professional so I want to learn why it turned out to be.
@@clover-00
I've been wondering about this for a while now but why do we put 0 or anything that represents it in equations when we haven't figured out yet what anything divided by 0 is equal to?
I mean, 0 represents nothing. But you can still take nothing from nothing.
I read this as 0 ≠ 1
Programmers ☕☕
Eddie woo concept was better
So 0! = 0*(-1)!
So 0! = 1 to let 1! work, but 0! is not equal to 0(-1)!.
0!=1!/1😂😂
0! is final step,
Turn up to prove it 👆
👌👌🤭😅
Proof that 0 = 1:
0! = 1 and 1! = 1, therefore 0! = 1! Divide both sides by !, giving 0 = 1, q.e.d.
Just kidding. Don't delete me.
At least 69! > 52!
❤
I still don't understand it.
1! = (1) * (1-1)! is then:
1! = (1) * (0)!, and removing the brackets, we have:
1! = 1 * 0!.
Anything multiplied by zero is zero. So shouldn't both 1! and 0! be zero? Unless, by convention we accept that 0! = 1, in which case the equation becomes 1! = 1 * 1. But that doesn't explain why 0! = 1.
This video seemed to gloss over the idea that (1)(1-1)! = 0!, and that is the part I'm struggling with. I just don't understand how that equation makes anything but zero.
That's 0! and not 0; so you cannot say that 1 × (0)! = (1 × 0)! ❌
Take 0! as a variable and solve for 0! ✅
@@mischievous_luffy Sorry, I still don't understand it. You are saying to solve for 0! But that's the problem I'm having. How does dividing by 0! equal anything other than zero?
@@JRCSalter let 0! = some constant (say 'x')
Now initially we had the equation,
1! = 1 × 0!
Replace 0! by 'x'
1! = 1 × x
1 = 1 × x [since 1! = 1]
We get x = 1
Now we initially stated that 0! = x
Replace 'x' by 0!
You get 0! = 1
@@mischievous_luffy The trouble I'm having is that I don't understand how we got from (1)(1-1)! to 0!. We are essentially saying that 1 * 0! = 0! = 1. 0 doesn't have anything to multiply itself with, and therefore would equal 0.
@@mischievous_luffy I think I understand your reasoning, but it doesn't help explain why 0! = 1. It just seems like we use it as a synonym for 1, rather than as a calculation.
With all other factorials, we take the number, and multiply by the next lowest positive number and so on. So we should take 0, and multiply by the next lowest positive number, but there isn't one, so it would be 0*0=0
That logic to me makes much more sense than what was described in the video.
Cool
For the last part, you said that 1(1-1)!=0! So you after the parentheses it’s 1x0! 0!=1, so 1x0!= 1. Do you make the equation more like 1(0)! ? So in 1(0)! Do you multiply the 1 and 0 first, or find the factorial of “0” first?
💖
🙏👍
great! now is (-1)! equal to (-1)*(-2)! ? I guess it isn’t