Prove that 0 exists

Zero is much greater than any number of negative numbers I could name.

Zero is bigger than half the real numbers in the universe!

Basket +5 = If you follow the sequence from my post, the expected answer is zero. As soon as you don't believe in sequences, you cannot believe 0! = 1

https://youtu.be/X32dce7_D48?si=IPmbHwYCVdBrwCez

0! = 1 should be easy to understand once you define factorial of n as the number of permutations of a set of n elements (or the number of bijections from a set of n elements to itself), since the empty set has exactly one permutation. There is exactly one bijection from an empty set to itself. The problem is that many people learn the definition of factorial algebraically, and then they have these doubts. Similarly, at least from a combinatorial perspective, the number of functions from an n element set to a k element set is k^n, and this also works if both sets are empty, since there is exactly one function from an empty set to an empty set. People (including school teachers) don't have enough 'training' working with empty sets, which is why they have these doubts.

And when reading the wise post above, know that "bijection" has nothing to do with gender.

Don't quote me, but I believe (n! / (n - r)!) = the number of ways to choose r objects from among n objects, if the order of choice matters to you.

But if you wish to choose all of the elements, so that r = n, then the formula yields n! /0! = n! / 1 = n!, which is in accord with @kajalmaya's post, unless I misunderstood him or her.

After all, how many ways can you choose no objects? Only one way: choose none of them! :) So 0! = 1.
(Since my explication is circular, I do not mean this to be a proof. I intend it merely as a glimpse into the beauty of the Platonic Universe.

@Noflaps Your argument is not wrong. But the question is what comes first, how you arrive at the formula for the number of combinations. The way (that I prefer) to do this formally is to first define n! as the number of permutations of an n element set, then prove that n! is what it is (including arguments for 0! = 1). Then define nCr (n choose r) as the number of combinations etc, then prove Pascal's identity (purely combinatorially, not using the formula for nCr). Then using that identity, derive a formula for nCr.

Your comment about bijection vs gender reminds me of an episode of Ali G with Chomsky. Very funny. It is there on YouTube.

It is clear you are an excellent mathematician, @kajalmaya .

Indeed, you place egg properly before chicken. (Since egg, from pre-chicken, contained the genetic aberration that led to first "chicken").

And, as I noted (and you apparently noted me note): I was not attempting a proof, but rather an ornament to hang upon the holiday tree of life.

Plus, if zero didn’t exist, 10 would not exist, or 100, or 1000. We need zero

@i-bex said in #2:

dammmmmmmmmmmmmmmmmmmnnnnnnnnnn

No offense, but this is the dumbest question I've seen on this forum, and that's saying something.

I bet you consider infinity a "concept." LOL

Study algebra. Look at the Riemann sphere.

@kajalmaya said in #43:

0! = 1 should be easy to understand once you define factorial of n as the number of permutations of a set of n elements (or the number of bijections from a set of n elements to itself), since the empty set has exactly one permutation. There is exactly one bijection from an empty set to itself. The problem is that many people learn the definition of factorial algebraically, and then they have these doubts. Similarly, at least from a combinatorial perspective, the number of functions from an n element set to a k element set is k^n, and this also works if both sets are empty, since there is exactly one function from an empty set to an empty set. People (including school teachers) don't have enough 'training' working with empty sets, which is why they have these doubts.

I would argue that an easier interpretation is the limit of the analytic continuation of factorial, i.e. gamma.

Combinatorics is way more puzzling than it looks.