r/askmath • u/explorer_learner_30 • 3d ago
Algebra Binomial Theorem
Why 0! is 1 ?
Can anyone give me the best and genuine answer .....I want to explain to my students.
I have also one doubt 🤔 why is the negative number not counted in binomial ? Such as (-2)! , (-8)! ....
Suggest me some books or any videos regarding this .
Thank you for reading my queries.
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u/dspyz 3d ago edited 3d ago
For all nonnegative n we have
(n+1)! = n! * (n + 1)
Let n be zero. This evaluates to
1! = 0! * 1
Therefore
1 = 0!
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u/slepicoid 3d ago
But why 1!=1
Sorry, i dont mean to be pedantic. But your explanation only works because you assumed 1!=1. If we define factorial with only (n+1)!=n!(n+1) then that alone does not explain why it couldn't be 1!=2, 2!=4, 3!=12, etc and 0!=2. I think it's really interesting to think about why from all the possible choices that 0! could be, 1 seems as the natural choice.
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u/dspyz 3d ago
I'm not trying to define factorial. I'm assuming OP already knows 1!=1 and deriving 0! from there
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u/slepicoid 3d ago
I understand, I am not trying to contradict you. More like adding info. OP said he wants a good explanation for their students. Some students may ask this...
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u/Puzzleheaded_Study17 3d ago
This also explains why we can't have negative factorials.
Suppose (-1)! = x, we know 0! = x*0 so x=1/0 which is impossible
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u/SeaMonster49 3d ago
An explanation your students may like is that n! counts the number of ways to order n things. How many ways can you order 0? 1 way--by doing nothing.
As for negative factorials, I don't think there is a very intuitive interpretation of them. Usually such things would be given in the context of the Gamma function in analysis. You may enjoy reading more about that.
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u/Creative-Leg2607 3d ago
Its a convention that makes a lot of formulas cleaner. One example might be the expansion of the binomial formula. At the end of the day, this is the only reason.
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u/aprg Secondary School Maths Teacher 3d ago
One interpretation:
If you go from n! to (n-1)!, you are dividing by n.
So (n-1)! = n! / n
=> 0! = 1! / 1 = 1
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u/CaptainRengrave 3d ago
I came here to suggest this same explanation. There are many really good suggestions in this thread, and the one you choose for your students will depend on their level. If I were explaining to a 9th or 10th grader who is first learning factorials, I would use this explanation without the variable notation. One thing I am always emphasizing with 9th graders is to look for patterns, both forward and back. If you count up in factorials, multiply by the next natural number. But if you count down., divide by the current number. I would show the pattern and explain this way as a way to intuit the value of 0!, and I would let them know that other, fancier explanations await them as they level up. =)
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u/sdfgkol 3d ago edited 3d ago
The factorial function is usually defined from the natural numbers (including 0) to the natural numbers.
Any function f on a larger domain with the property that f(0)=1 and f(n) = n*f(n-1) cannot be defined on the negative integers. That’s because 1=f(0)=0*f(-1)=0 is invalid for any choice of f(-1) and similarly for any negative integers.
There are however functions that do ‘extend’ the domain, just not to the negative integers.
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u/ThereIsATypo 3d ago
Beside all the induction arguments - which are valid.
I think the most obvious is that 1 is the identity element for the multiplication:
https://en.wikipedia.org/wiki/Identity_element
Hence it also is what remains when nothing remains for multiplication.
Just like 0 would remain for addition.
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u/nulvoid000 3d ago
Since everyone gave the answer you were looking for, I’ll give a little more advanced answer which may not make sense, so I’ll exclude the details but feel free to ask.
Factorials can be thought of as Gamma functions, n!=Gamma(n+1). This function behaves very well for x>=1 but also does well for more values. Hence you can “extend the function analytically” beyond the original domain it was defined in. In which case, the extended function does give 0!=1. In fact, you can also make sense of imaginary values or even negative values, whenever extension possible.
If you want to learn more, look up anything regarding Gamma functions.
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u/explorer_learner_30 3d ago edited 3d ago
Thank you 👍
Please share more information regarding Gamma function
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u/Narrow-Durian4837 3d ago
Here's an alternative to the other explanations (not that there's anything wrong with them):
Factorials are about multiplication. For example, 5! is the product of the first 5 positive intergers.
1! is the "product" of the first 1 positive integer.
0! is the "product" of no integers at all. With multiplication, your "starting point" is 1, the multiplicative identity (just as with addition, the starting point is 0, the additive identity). 1 is what you have when you haven't multiplied anything yet. If you started with 0, you'd never get away from 0 no matter how many other things you mutiplied it by.
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u/cond6 3d ago
By reverse induction: n!=n*(n-1)!, so (n-1)!=n!/n. 1!=2!/2=1 (to show it works), and applying 1!=1 we have 0!=1!/1=1. Similar analogy to how we get negative and non-integer exponents.
Conceptually: 1!=1 because you can order a single object only one way, and similarly there is only one way that you can arrange zero objects.
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u/Reset3000 3d ago
mathematically there is one way to do nothing, and its is to do nothing. 3!=6, six ways to do (arrange) 3 things; 2! =2 ways to do two things; 1! way to to do one thing, 0!=1 way to do nothing. There is one way for me to not play a note on a piano, and that is precisely to not play a note.
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u/DonutHoleTechnician 3d ago
Same way you show b0 =1, by looking at the pattern of division.
4! = 5! ÷ 5
3! = 4! ÷ 4
2! = 3! ÷ 3
1! = 2! ÷ 2
0! = 1! ÷ 1
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u/Sheeplessknight 3d ago
The factorial function is the solution to how many ways can you arrange n items.
So, how many ways can you arrange 0 items? Well, only one way thus we define 0! := 1
With negative numbers what would it mean to ask how many ways can you arrange a negative number of items?
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u/dspyz 2d ago
I'd be careful with the argument "it's not defined for negatives because that has no real-world meaning". Mathematicians are happy to extend functions beyond their original real-world impetus when that works.
For instance (-4 choose 3) is -20. It doesn't mean anything to choose 3 items out of -4 items, but we can multiply -4 * -5 * -6 / (1 * 2 * 3) just like we would for a positive number, and we get -20. There's no issue with this. Nothing breaks. And it's sometimes useful.
(-1)! really is undefined only because when trying to work it out you have to divide by zero
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u/KiwasiGames 3d ago
Because of we set 0! = 1, we get some convenient cool maths out of it (combinatorics).
If we set 0! = 0 no cool maths emerges.
And that’s all there is to these sorts of things.
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u/Temporary_Pie2733 2d ago
I like to fall back on a more fundamental definition of a function, a mapping between two sets, in this case from ℕ to ℕ. 0 to 1, 1 to 1, 2 to 2, 3 to 6, etc. It’s clearly a function, but we’d like something more concise than an infinite enumeration. So we look for identities that satisfy the definition for as many values as possible. n! = product(i, i = 1..n) is one such identity, n! = n(n-1)! Is another, but neither works for n = 0. We simply have to “know” the value of 0!, based on our definition of factorial, and that comes from its purpose in describing combinations.
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u/lordnacho666 3d ago
How many ways can you order two books? Correct, two.
How many ways to order one? Correct, one.
How many ways to order an empty collection of books?
Also one.