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 to learn …

@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 all this, …

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 clearly false …

Feb 6, 2021 · $$ 0! = \Gamma (1) = \int_0^ {\infty} e^ {-x} dx = 1 $$ If you are starting from the "usual" definition of the factorial, in my opinion it is best to take the statement $0! = 1$ as a part of 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 eternally recurring it …

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 risks, as is …

Is there a consensus in the mathematical community, or some accepted authority, to determine whether zero should be classified as a natural number? It seems as though formerly $0$ was considered i...

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 think about …

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 argument of …

Mar 5, 2025 · Some books say that even numbers start from $2$ but if you consider the number line concept, I think zero($0$) should be even because it is in between $-1$ and $+1$ (i.e in between two …