Newest Questions - Mathematics Stack Exchange
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(Un-)Countable union of open sets - Mathematics Stack Exchange
A remark: regardless of whether it is true that an infinite union or intersection of open sets is open, when you have a property that holds for every finite collection of sets (in this case, the union or intersection …
If a series converges, then the sequence of terms converges to $0$.
@NeilsonsMilk, ah, it did not even occur to me that this involves a step. See, where I learned mathematics, it is not unusual to first define when a sequence converges to zero (and we have a …
modular arithmetic - Prove that that $U (n)$ is an abelian group ...
Prove that that U(n) U (n), which is the set of all numbers relatively prime to n n that are greater than or equal to one or less than or equal to n − 1 n 1 is an Abelian group. My thought process: for a, b ∈ …
The sequence of integers - Mathematics Stack Exchange
May 9, 2016 · Prove that the sequence $\\{1, 11, 111, 1111, .\\ldots\\}$ will contain two numbers whose difference is a multiple of $2017$. I have been computing some of the immediate multiples of $2017$ …
Double induction example: $ 1 + q + q^2 + q^3 + \\cdots + q^{n-1}
I'm working on a double induction problem with the following prompt: Prove by induction on n n that for any real number q> 1 q> 1 and integer n ≥ 0 n ≥ 0:
Mnemonic for Integration by Parts formula? - Mathematics Stack …
Nov 11, 2018 · The Integration by Parts formula may be stated as: $$\\int uv' = uv - \\int u'v.$$ I wonder if anyone has a clever mnemonic for the above formula. What I often do is to derive it from the …
For what $n$ is $U_n$ cyclic? - Mathematics Stack Exchange
When can we say a multiplicative group of integers modulo $n$, i.e., $U_n$ is cyclic? $$U_n=\\{a \\in\\mathbb Z_n \\mid \\gcd(a,n)=1 \\}$$ I searched the internet but ...
probability - Suppose that $U1, U2, ..., Un$ are iid $U (0,1)$ and $Sn ...
Nov 2, 2022 · I meant it to read: P (S_1 ≤ t) P (S_n ≤t). The product of those probabilities given the assumption is true.
$\\operatorname{Aut}(\\mathbb Z_n)$ is isomorphic to $U_n$.
Aug 3, 2023 · It might be using ring theory in a non-essential way, but it is conceptually simpler because the endomorphisms are easier to describe than the automorphisms, and since the invertible …