Reputation
Next tag badge:
88/100 score
15/20 answers
Badges
9 49 142
Newest
 Revival
Impact
~273k people reached

10h
comment Can we use Rolle's Theorem if $f : [a, b] \to \mathbb{R}$ is continuous on $(a, b)$ only?
OK. Note that differentiable on $[a, b]$ implies differentiable on $(a, b)$, so what is your question exactly?
11h
comment Can we use Rolle's Theorem if $f : [a, b] \to \mathbb{R}$ is continuous on $(a, b)$ only?
@WaqarAliShah: You would have to first define what it means for a function to be differentiable on $[a, b]$. The problem here is what it means to be differentiable at the endpoints.
18h
reviewed Reviewed Help to understand material implication
1d
awarded  Revival
1d
revised $f(x) \mid g(x) \iff g(x) \in \langle f(x) \rangle$. Isn't this trivial?
edited title
1d
awarded  Revival
1d
revised When is the quadratic formula $ax^2 + bx +c \equiv 0 \pmod p$ solvable modulo prime $p$?
added 6 characters in body; edited title
1d
answered When is the quadratic formula $ax^2 + bx +c \equiv 0 \pmod p$ solvable modulo prime $p$?
1d
comment $f(x) \mid g(x) \iff g(x) \in \langle f(x) \rangle$. Isn't this trivial?
I've always seen that definition omit the case $a = 0$ (using your letters), but as you say, definitions can vary.
1d
comment $f(x) \mid g(x) \iff g(x) \in \langle f(x) \rangle$. Isn't this trivial?
I don't think so. If $R$ is a Euclidean domain, and $a, b \in R$, $b \neq 0$, then there are unique $q, r \in R$ such that $a = bq + r$; if $r = 0$, we say $b$ divides $a$. You could try to extend the definition to the $b = 0$ and while it is true that $0 = 0q + 0$, $q$ is no longer unique.
1d
comment To find all integers $n > 1$ for which $(n-1)!$ is a zero-divisor in $Z_n$.
A proof of this fact can be found here.
1d
revised Can we use Rolle's Theorem if $f : [a, b] \to \mathbb{R}$ is continuous on $(a, b)$ only?
added 4 characters in body; edited title
1d
answered Can we use Rolle's Theorem if $f : [a, b] \to \mathbb{R}$ is continuous on $(a, b)$ only?
1d
answered $f(x) \mid g(x) \iff g(x) \in \langle f(x) \rangle$. Isn't this trivial?
1d
comment Can one decompose any ring into a (possibly infinite) product of indecomposable rings?
Excellent, thanks. Nice example.
1d
comment Can one decompose any ring into a (possibly infinite) product of indecomposable rings?
The cardinality reason being that the set of eventually constant $\mathbb{Z}_2$-valued sequences is countable, while $\mathbb{Z}_2^I$ is either finite (if $I$ is finite) or uncountable (if $I$ is infinite). Right?
1d
revised If $\{x_n\}$ and $\{y_n\}$ are two bounded sequences then prove that $\liminf x_n -\liminf y_n \leq \liminf(x_n - y_n)$
edited title
1d
comment If $\{x_n\}$ and $\{y_n\}$ are two bounded sequences then prove that $\liminf x_n -\liminf y_n \leq \liminf(x_n - y_n)$
Welcome to math.SE. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level.
1d
revised Silly question about group rings
deleted 3 characters in body
1d
answered Silly question about group rings