Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [associativity]

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

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

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Need help with Associative proof

Let$ G = \{x ∈ R | x \neq -1\} $and * a link to ¨G with $x * y: = x + y + xy.$ Show that $(G, *)$ is a group. To determine if $ G$ is a group I have to make the associative proof: Associative? $(x * y) * z = (x + y +xy)+z -(x + y +xy)*z$ $= x + y +…
NoobProgrammer
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Find or describe all associative algebras possessing only inner derivations.

From cohomology theory we know that separables algebras $A$ possess only inner derivations. In fact, they possess only inner Derivation into every $A$-bimodule $M$ which characterize separable algebras. For example, the Algebra of upper triangular…
Sven Wirsing
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Understanding associative operations

If something is associative, then for example $(x + y) + z = x + (y + z)$ is true. I also know that in general subtraction is not an associative operation, but what if subtraction is applied to $\lbrace 0 \rbrace$. Is subtraction an associative…
user6259845
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Associativity Question

Can someone step by step the last line? Confused.
JiHua
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How does one "parenthesize elements"

This is probably something very obvious, but I am a little confused. It's about the associative law. It is known that a binary structure $(S, *)$ is associative if: $(a * b) *c = a * (b * c)$ for all elements in $S$. It's also closed, as well.…
user121615
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Associative Law - Number Sequence

I'm stuck in a problem, need Help. Associative property States that (a + b ) + c = a + (b + c) Which is true but what if i change the position of the numbers (a + c) + b = a + (c + b) Does it hold true for the associative Law because the answer…
Harry47
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Associativity and piecewiseness

I have this confession: Piecewise definitions have always considerably annoyed me (with certain exceptions). For $x,y\in (-1,1),$ let $x\circ y := xy -\sqrt{(1-x^2)(1-y^2)}\in(-1,1).$ (The part that says $\text{“}{\in}\text{”}$ is a simple exercise,…
Michael Hardy
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