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Questions tagged [associativity]

This is the property shared by many binary operations including group operations. For a binary operation $\cdot$, associativity holds if $(x\cdot y)\cdot z = x \cdot(y\cdot z)$.

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2 answers
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Here is the definition of an analytic map. ANALYTIC MAP Let $ E$ and $ F$ be Banach spaces. A map $f \colon E\to F $ is analytic at $ x=0$ if $$ f(x)=\sum_{k=0}^{\infty}a_k(x) \tag 1 $$ where for ...
Laurent Claessens's user avatar
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0 answers
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Here are two definitions. ANALYTIC MAP Let $ E$ and $ F$ be Banach spaces. A map $f \colon E\to F $ is analytic at $ x=0$ if $$ f(x)=\sum_{k=0}^{\infty}a_k(x) $$ where for each $ k$, the map $a_k \...
Laurent Claessens's user avatar
6 votes
0 answers
294 views

By merging together the contributions from: a) this answer, b) the comments under this answer, we come up to the following: Claim. For $n\in\mathbb N$, let $Q=(\{1,\dots,n\},*)$ be a quasigroup. Then, ...
Kan't's user avatar
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2 votes
1 answer
93 views

For $n\in\mathbb N$, the multiplication table of a quasigroup $Q=([n],*)$ induces a transitive subset $\Sigma=\{\sigma_1,\dots,\sigma_n\}\subseteq S_n$ via the position: $$\sigma_i(j):=i*j\tag1$$ and ...
Kan't's user avatar
  • 5,601
2 votes
2 answers
218 views

I am trying to show that the operation $\ast$ gives an associative binary operation on $\mathbb{R}^*$, where $\mathbb{R}^*$ is the set of all real numbers except $0$, and $\ast$ on $\mathbb{R}^*$ is ...
Seungjun Wang's user avatar
3 votes
0 answers
46 views

I'm interested in the following construction: Let $V$ be a vector space and $A(V)$ be the corresponding free associative algebra on $V$, aka the tensor algebra of $V$. Consider a two-sided ideal $I_n$ ...
Astro Log's user avatar
  • 113
1 vote
2 answers
127 views

Is there a two-variable (real) polynomial that is associative, not commutative, and not a projection? That is: Does there exist a polynomial $p$ such that $\forall x, y, z. p(p(x, y), z) = p(x, p(y, ...
user76284's user avatar
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1 vote
2 answers
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I beginning to start math from the start again and I'm trying to understand things more intuitively and I'd like some help to understand associativity. I can understand it as the characteristic of an ...
Linces games's user avatar
0 votes
1 answer
56 views

$\Omega$-algebras are defined as an algebra with $\Omega$ that consists of $n$-ary operations, $n \geq 2$. Is the concept of semidirect product available for them?
Nil's user avatar
  • 1,358
3 votes
1 answer
147 views

I have been investigating commutative, associative magmas in an ad hoc way for the past few days and was curious about idempontent-free magmas. The magma $(\mathbb{Z}_{\ge 1}, +)$ is idempotent-free, ...
Greg Nisbet's user avatar
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1 vote
1 answer
108 views

Here's where the motivation comes from. I was trying to figure out a way to assign a unique natural number to every pure set (that is, I wanted to find a bijective function with a domain of pure sets ...
Mathemagician314's user avatar
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0 answers
21 views

Commonly, a group is defined as follows: Definition. Let $G$ be a set and $\ast:G\times G\to G$ a binary operation on that set. Then the pair $(G,\ast)$ is called a $\textbf{group}$ if the following ...
Vaskara_GRek_O's user avatar
1 vote
1 answer
43 views

Definitions: Given a triple $(S,T,\cdot)$, let us say $(S,T,\cdot)$ has property $R$ if $S$ is a set, $T \subseteq (S \times S)$ is a relation from $S$ to itself, and $\cdot : T \rightarrow S$ is a ...
I.A.S. Tambe's user avatar
  • 3,191
1 vote
0 answers
60 views

I'm trying to understand a property of binary operations where the order of applying the operation to a set of elements always matters. I'm not super advanced in math, but I will try to explain what I ...
n-l-i's user avatar
  • 111
5 votes
1 answer
171 views

Let $S$ be a set with a binary operation $/:S\times S \rightarrow S$ where $(a,b) \mapsto a/b$ such that: There exists an element $1 \in S$, such that $a/b = 1$ if and only if $a=b$. For any ...
Juan Naranjo's user avatar

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