There is no general consensus as to whether $0$ is a natural number. So, some authors adopt different conventions to describe the set of naturals with zero or without zero. Without seeing your notes, my guess is that your professor usually does not consider $0$ to be a natural number, and $\mathbb{N}_0$ is shorthand for $\mathbb{N}\cup\{0\}$.

2015年4月29日 · Why is any number to the zeroeth power equal to 1? I would think it would be equal to zero, since nothing multiplied by nothing is, well, I would think 0. But it is 1? Examples: $(-5)^0 = 1$; $0...

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$\begingroup$ 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.

2017年1月12日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

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$\begingroup$ Although n^n > n! does not prove that n!/n^n -> 0 BUT it does prove that n^n/n! -/> 0. That is, it doesn't prove that it's a higher order of growth, but it proves that it's order of growth is at least as high. This is contrary to his professors claims. $\endgroup$ –

2019年5月16日 · "Let P(n) be the statement that (n)! < (n)^n, where is an integer greater than 1. Prove by mathematical induction that P(n) is true for all integers n greater than 1." I've written. Basic step. Show that P(2) is true: 2! < (2)^2 . 1*2 < 2*2. 2 < 4 (which is true) Thus we've proven that the first step is true. Inductive hypothesis

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2016年5月3日 · The set is in $R^1$ and consists of $0$ and the numbers $1/n$. Call it $E$. Take a set of $n$ intervals of radius $r$, centered less than $2r$ apart and such that ...