$\begingroup$ 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.

Mar 15, 2013 · Inclusion of $0$ in the natural numbers is a definition for them that first occurred in the 19th century. The Peano Axioms for natural numbers take $0$ to be one though, so if you are working with these axioms (and a lot of natural number theory …

Oct 28, 2019 · So basically, 1/0 does not exist because if it does, then it wouldn't work with the math rules. Let τ=1/0. 0τ=1. x0τ=x. 0τ=x. τ=x/0. 1/0=x/0 which doesn't work (x represents any number). That means that 1/0, the multiplicative inverse of 0 does not exist. 0 multiplied by the multiplicative inverse of 0 does not make any sense and is undefined.

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$\begingroup$ This definition of the "0-norm" isn't very useful because (1) it doesn't satisfy the properties of a norm and (2) $0^{0}$ is conventionally defined to be 1. $\endgroup$ – Brian Borchers

Jan 22, 2017 · 1) x^a × x^b = x^a+b; for x = 0 and a = 0, you would get 0^0 × 0^b = 0^b = 0, so we can't tell anything -- except confirm that 0^0 = 1 still works here! 2) x^{-a}=1/{x^a} -- so when a = 0 , x^{-0} = 1/x^0 = x^0 , which again does work for 0^0 = 1 ; 3) {x^a}^b = x^{a×b} , thus x^(1/n) is the n-th root -- and 1/n = 0 for no value of n , so ...

Sep 11, 2015 · On the contrary, those limits tell you that the limit of the entire quotient is $0$. This may be easier to see if you rewrite to $$ \lim_{x\to\infty} f(x)\frac1{h(x)} $$ where $\lim_{x\to\infty} f(x) = 0 $ and $\lim_{x\to\infty} \frac1{h(x)}=0 $, and the product of two functions that both have limit $0$ surely also has limit $0$.

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The volume of the parallelepiped determined by the row vectors of the matrix is $0$. The system of homogenous linear equations represented by the matrix has a non-trivial solution. The determinant of the linear transformation determined by the matrix is $0$. The free coefficient in the characteristic polynomial of the matrix is $0$.

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 $\mathbb{N}\cup\{0\}$.