Why does 0! = 1 0! = 1? All I know of factorial is that x! x! is equal to the product of all the numbers that come before it. The product of 0 and anything is 0 0, and seems like it would be …

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 …

@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 …

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 …

Your title says something else than "infinity times zero". It says "infinity to the zeroth power". It is also an indefinite form because $$\infty^0 = \exp (0\log \infty) $$ but $\log\infty=\infty$, so the …

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 …

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.

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 …