$\begingroup$ 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 …

Jan 12, 2015 · $$\lim_{n\to 0} n^{i} = \lim_{n\to 0} e^{i\log(n)} $$ I know that $0^{0}$ is generally undefined, but can equal one in the context of the empty set mapping to itself only one time. I …

Whereas exponentiation by a real or complex number is a messier concept, inspired by limits and continuity. So $0^0$ with a real 0 in the exponent is indeteriminate, because you get different …

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 …

May 29, 2015 · Is a constant raised to the power of infinity indeterminate? I am just curious. Say, for instance, is $0^\infty$ indeterminate?

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 …

The reason $0/0$ is undefined is that it is impossible to define it to be equal to any real number while obeying the familiar algebraic properties of the reals. It is perfectly reasonable to …

May 14, 2015 · But: I know what I am writing about. I have a PhD mathematics, and have seen all these arguments by people who let $0^0$ undefined, and I have seen even more arguments …

1 x 0 = 0. Applying the above logic, 0 / 0 = 1. However, 2 x 0 = 0, so 0 / 0 must also be 2. In fact, it looks as though 0 / 0 could be any number! This obviously makes no sense - we say that 0 / 0 …

Feb 12, 2018 · Now if we want to solve for $0^0$, well according to $(1)$, we have that $0^0 = 1$. But, $$0^0 = 0^{1-1} = \frac{0^1}{0^1} = \frac{0}{0}.$$ This is where division by $0$ is …