$\begingroup$ The theorem that $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ already assumes $0!$ is defined to be $1$. Otherwise this would be restricted to $0 <k < n$. A reason that we do define $0!$ to be $1$ is so that we can cover those edge cases with the same formula, instead of having to treat them separately.

Check to see that $1! = 1 \times (1 - (1 - 1)) = 1 \times (1 - 0) = 1 \times 1 = 1$ If we interpreted $0!$ as conforming to the scheme from (1), though, the result would be $0! = 0$ , which goes against the tradition.

2017年1月22日 · $\begingroup$ @Coltin: Wolframalpha doesn't say that 0^0 is 1. If you enter 0^0 it says "indeterminate". It only says that lim x^x as x->0 is 1, which is perfectly true, but entirely besides the point. $\endgroup$ –

The problem is that the corresponding finite cover, consisting of $(-1,1/2]$ does not extend to a finite subcover of $[0,s+\epsilon]$ for any $\epsilon>0$. $\endgroup$ – Michael Greinecker Commented Aug 23, 2019 at 19:53

Prove $(0,1)$ and $[0,1]$ have the same cardinality. I've seen questions similar to this but I'm still having trouble. I know that for $2$ sets to have the same cardinality there must exist a bijection function from one set to the other. I think I can create a bijection function from $(0,1)$ to $[0,1]$, but I'm not sure how the opposite.

2020年9月23日 · Welcome to MSE! Here $(0 \cup 1)^*$, often written as $(0 | 1)^*$, is actually the set of all bitstrings. Let's break it down. Inside of the parentheses is the expression $0 | 1$, meaning anything that that is a zero or a one is allowed (i.e. everything is allowed). The asterisk for Kleene closure tells you that you can repeat it as many times ...

Slightly modify the above function to prove that $[0,1)$ is equivalent to $[0,1]$ Prove that $[0,1)$ is equivalent to $(0,1]$ Since the "equivalent to" relation is both symmetric and transitive, it should follow that $[0,1]$ is equivalent to $(0,1)$. Hence, there does exist a one-to-one correspondence between $[0,1]$ and $(0,1)$.

2018年9月25日 · The sequence is $f = 0, 1, 0, 1, \\ldots$ I want to find a general formula for the $n$th element. The sequence starts at $n = 0$ (the $0$ here is not the first ...

Possible Duplicate: Bijection between an open and a closed interval How do I define a bijection between $(0,1)$ and $(0,1]$? I wonder if I can cut the interval $(0,1)$ into three pieces: $(0, \\

Then, $$ \phi^{-1}(x,y) = 0.a_1b_1a_2b_2\cdots$$ Some care has to be taken with identification between digital expansions like $0.199999\cdots$ and $0.20000\cdots$, but that is an exercise. Having the bijection between $(0,1)$ and $(0,1)^2$, we can apply one of the other answers to create a bijection with $\mathbb{R}^2$.