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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Semantics of operator $\times$ with regards to sets

I am trying to understand something about operator $\times$ with regards of sets. In the accepted answer to the following question (The cross product of two sets), the answerer says that $A \times B$ ...
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How smooth can non-nice associative operations on the reals be?

Suppose ${*}:\mathbb R\times\mathbb R\to\mathbb R$ is $\mathcal C^k$ and associative. Does it necessarily satisfy the identity $a * b * c * d = a * c * b * d$? For $k=0$ the answer is "no" -- a ...
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How do I complete this “Cayley table” or binary operation table?

I have an algebraic structure $(S,\cdot)$ and $a,b,c,d \in S$ where $a,b,c,d$ are not necessarily four distinct elements. This is part of a larger problem that I am working on and based on what I ...
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Associativity of arrow composition counter example?

I'm trying to achieve a working understanding of category theory. One of the problems I'm having is that many of the concepts seem too straight-forward or obvious so it's hard to see why they're ...
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1answer
20 views

How to verify an algebraic structure is a ring

I have a problem which ask me to verify that to structures are rings. However, I'm unsure of how exactly to check each property. I believe that the first is closed but not sure how to check the ...
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Just How Strong is Associativity?

A friend of mine is using a lot of Non-associative Algebra for an advanced Chemistry project. We were discussing it recently and I found it rather amusing how often she said things like "brackets ...
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trying to find associativity

Is the binary operation define by: $x*Y = x+y-1$ what my tutor has done: $x*(y*z) = x *(y+z -1) = x+(y+z-1) = x+y+z-2$ My question: how did he get $x+y+z$-2 Where did the '-2' come from? I am ...
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Association, Commutation and Identity Elements on Binary Operations?

Is the following closed, associative or commutative? f(a, b) = (a+b)/2, where a, b ∈ Z. I found that it is not closed but I am not sure how to find whether or not it is associative (I was confused ...
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logical associative expressiveness with no negation operator

Let's suppose we can only use $\wedge$ and $\vee$ operators (we have no negation operator), and by default we have associativity to the left. Is this subset of logic as expressive as the one with the ...
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Prove that the truth value of $x_1 \lor x_2 \lor \ldots \lor x_n$ does not depend on how the formula is parenthesized

So the question is: Generalized Associativity of $\lor$. Prove that, for all positive integers $n$, all ways of parenthesizing the following logical statement have the same truth value: $$x_1 \lor ...
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How to define a taxonomy of non associative operations?

Let $A$ be a set, and let $a,b,c\in A$. Let also $\circ: A\times A\rightarrow A$ be a binary operation on $A$. We agree as usual to write $a\circ b$ to mean $\circ(a,b)$. We say that $\circ$ is ...
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Maximality of the kernel of a quasifield

Setting A (left) quasifield is an algebraic structure $(Q,+,\cdot)$ such that $(Q,+)$ is an abelian group. (As usual, we denote its identity element by $0$.) Each equation $x\cdot a = b$ with ...
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Confusion about the associative property and the mechanics of Parenthesis

This is a follow up question on my earlier post (Updated): Showing that a set $M$ with two elements classifies as a field. I feel this post is necessary because I realize that what confuses me is how ...
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If $x * x = e$ for all $x$ and $(x_i * x_j) * (x_j*x_k) = x_i*x_k$, then $*$ is associative

This should be simple, but for some reason I get stuck on this. Let $G = \{x_1, \ldots x_n\}$ be a set equipped with operation $*$ satisfying the following : 1) $G$ has an identity element $e$ with ...
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1answer
51 views

Which is the property of the functions that correspond to this definition/examples?

I'm looking for a definition for a particular function(-input) property. Considering a function $f$ that takes as input a list of elements and produces in output just one element, which is the ...
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How to prove that: If two binary operations are anti-isomorphic and one of them is associative then the second one also will be associative?

We know what is called an anti-isomorphic operation on a set S. it is just a one two one $ g $ function mapping from $S$ to $S$. $ g: S \rightarrow S$. and it satisfy this condition $ g(xy)= ...
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Are there associative magic squares of any size except $4k+2$?

An associative magic square is a magic square with the additional property that numbers symmetric to the center sum up to $n^2+1$. For example, the square ...
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Examples of reduced associative algebras

An associative $K$-algebra A is called reduced if $A/rad(A)$ has no nilpotent elements. It can be shown that this is equivalent to that $A/rad(A)$ is a isomorphic to a direct sum of division algebras. ...
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Associative Lie algebra without Jacobi identity

1) Is there a name for associative Lie algebra that does not require Jacobi identity to hold? 2) Can such algebra exist, and if it does exist, can this algebra contain infinitely many elements? 3) ...
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Is the number of associative $n$-ary algebraic operations on a finite set with 2 cardinality always 8?

We know that if $n = 2$ then the operation is called a binary operation. $ \circ $ on set $X$ is a function $\circ : X \times X \rightarrow X$. And the number of all associative binary operation on a ...
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Example of operation that is tree associative, but not generally associative

In a lot of algorithms using trees, we need the property that when folding $2^n$ elements with some operator $+$, we can do the first half of $2^{n-1}$ elements and the second half independently. Eg ...
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Associative property for series

Are those equation always valid: $$\sum a_n + b_n = \sum a_n + \sum b_n$$ $$\sum_{k=1}^n(a_{k+1}+a_k)-\sum_{k=1}^na_k=\sum_{k=1}^na_{k+1}$$
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Name of an aglebraic structures $(A,*,\cdot)$ weaker than semirings.

I have a set $A$ with two binary operations on it $(A,*,\cdot)$ STRUCTURE A $(A,*)$ is not commutative, is not associative, it has not an identity $(A,\cdot)$ is a commutative group $(a*b)\cdot ...
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Proving $K$ is a group

Now I have proved certain things are a group before, and I know that it requires: 1)Associativity 2)Inverse 3)Identity But here I have such a strange thing that I wanted to clarify that I am doing ...
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Show that * is associative

Could you show me how to prove the following to be associative? Please take me through the process step by step. $$a*b=a+b+2ab$$ Where $*$ is a binary operation and $a$ and $b$ are real numbers. I ...
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Proof of a corollary about associativity of (differential) convolution operater

I m working on the proof the following corollary for ages... I would appreciate any help!! Cor: Let $h : [0, \infty ) \rightarrow \mathbb{R}$. We define the convolution operator $*$ for the ...
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23 views

Does correlation have to be in the context of (Gaussian) normal distribution?

I am not quite familiar with the concept of correlation. The Pearson's correlation coefficient is defined as: $\rho_{X,Y}=\mathrm{corr}(X,Y)={\mathrm{cov}(X,Y) \over \sigma_X \sigma_Y} ...
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35 views

Is the operation associative

Is it known that the multiplication of matrices is a associative operation ? So,is the relation $(A \cdot B) \cdot C=A \cdot (B \cdot C)$ true?? ($A,B,C$ are matrices)
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Existence of an operation $\cdot$ such that $(a*(b*c))=(a\cdot b)*c$

When we can define a binary operation $\cdot:M\times M\rightarrow M$ on an algebraic structure $(M,*)$ such that $$a*(b*c)=(a\cdot b)*c$$ If $*$ is associative then $\cdot=*$ even if I'm not sure ...
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The condition for associative property

Thank in advance. Is there the condition for associative property in closed number system. It comes from a question. A number system is closed, associative, commutative for some operation, then the ...
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Superassociative operation

Background: Addition and multiplication are associative, but exponentiation is not. Question: Does an operation $\circ_1:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ exist such that ...
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Algebra of pseudo-differential operators

The class of pseudod-ifferential operators form an associative algebra of Fourier integral operators. Moreover, given symbols $a,b,c\in C^\infty$ (each associated to some pseudo differential ...
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tax and discount problem

Problem : In the total purchase amount $z, x\%$ is tax and $y\%$ is discount. Even if the tax is applied first and then discount or if discount is applied and then tax, the final amount is always ...
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Proof of The Associative Law and The Commutative Law.

The associative law of multiplication for three positive integers $a,b$ and $c$ can be proved$^1$ from the Commutative Law and the property of "Number of things" easily. We can prove$^2$ the ...
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Can we have an associative form of “octonions” and hypercomplexes, if we eliminate division?

I'm interested in hypercomplexes, or number systems with many square roots of $-1$. Now, I know that quaternions are non-commutative, but associative. I'm wondering if it's possible to have a number ...
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Why are exponents not associative?

I ran into something that seemed odd to me today: exponents are not associative. The following equation sums that up: $$ 10 * 2^{5x} \not\equiv 20^{5x} $$ Why is this the case? Is there some ...
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Does associativity justify $(f^{-1}gf)(f^{-1}hf) = f^{-1}gff^{-1}hf$?

I'm self-studying abstract algabra (Herstein) and while working on an easy problem became uneasy with a step in my derivation. Given the symmetric group $S_n$ whose elements are bijections $f: S \to ...
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Splitting up bracket terms

I found a statement saying: Let $\circledast $ be an associative binary operation on a set $\mathbb{X}$. A bracket term of length n, consisting of n elements $a_1, ..., a_n$ and arbitrary brackets, ...
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Is convolution associative with regards to the complex unity?

Setup: I need to do a convolution with the function $\cfrac{i}{x}$, and I would like to get rid of the $i$. My functions to be convolved are all real valued. According to the ever-failable wikipedia, ...
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How to correctly do a division using a slash?

What's the correct way to do division using a backslash? If I write $a/bc + d$, will that be equal to $(a/(bc))+d$? Basically, if I place a slash, will I divide by what's directly behind the slash ...
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Real life examples of commutative but non-associative operations

I've been trying to find ways to explain to people why associativity is important. Subtraction is a good example of something that isn't associative, but it is not commutative. So the best I could ...