The product of 0 and anything is $0$, and seems like it would be reasonable to assume that $0! = 0$. I'm perplexed as to why I have to account for this condition in my factorial function (Trying …

@Arturo: I heartily disagree with your first sentence. Here's why: There's the binomial theorem (which you find too weak), and there's power series and polynomials (see also Gadi's answer). …

For example, $3^0$ equals 3/3, which equals $1$, but $0^0$ "equals" 0/0, which equals any number, which is why it's indeterminate. Also, 0/0 is undefined because of what I just said.

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 …

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.

Oct 28, 2019 · In the context of limits, $0/0$ is an indeterminate form (limit could be anything) while $1/0$ is not (limit either doesn't exist or is $\pm\infty$). This is a pretty reasonable way to …

Jul 20, 2010 · If you take the more general case of lim x^y as x,y -> 0 then the result depends on exactly how x and y both -> 0. Defining 0^0 as lim x^x is an arbitrary choice. There are …

Jan 12, 2015 · In the context of natural numbers and finite combinatorics it is generally safe to adopt a convention that $0^0=1$. Extending this to a complex arithmetic context is fraught with …

@Swivel But 0 does equal -0. Even under IEEE-754. The only reason IEEE-754 makes a distinction between +0 and -0 at all is because of underflow, and for +/- ∞, overflow. The …

Nov 8, 2013 · That $0$ is a multiple of any number by $0$ is already a flawless, perfectly satisfactory answer to why we do not define $0/0$ to be anything, so this question (which is …