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 …

@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 …

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 …

@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). …

Jan 12, 2015 · 0i = 0 0 i = 0 is a good choice, and maybe the only choice that makes concrete sense, since it follows the convention 0x = 0 0 x = 0. On the other hand, 0−1 = 0 0 1 = 0 is …

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 …

11 Several years ago I was bored and so for amusement I wrote out a proof that 0 0 0 0 does not equal 1 1. I began by assuming that 0 0 0 0 does equal 1 1 and then was eventually able to …

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 …

It's slightly more obvious why 0/0 0 / 0 is indeterminate because the solution for x= 0/0 x = 0 / 0 is the solution for 0x= 0 0 x = 0, and every number solves that.

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 …