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). …

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 …

@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 …

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 …

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 …

Feb 8, 2014 · In the set of real numbers, there is no negative zero. However, can you please verify if and why this is so? Is zero inherently "neutral"?

Feb 8, 2011 · You can also prove it by moving the space: "0! = 1" $\Leftrightarrow$ "0 != 1", which is computer notation for "0 $\neq$ 1" :-). Then it depends on what you count as "first …

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 …

Nov 17, 2014 · I began by assuming that $\dfrac00$ does equal $1$ and then was eventually able to deduce that, based upon my assumption (which as we know was false) $0=1$. As this is …