Let $Y=\left\{(a,b)\in\mathbb{R}\times\mathbb{R}\ |\ a\ne 0\right\}$. Given $(a,b),(c,d)\in Y$, define $(a,b)∗(c,d)=(ac,ad+b)$. Prove that $Y$ is a group with the operation $*$. I already did the proof that $*$ is an operation on $Y$. To prove that $Y$ is a group with the operation $*$ I do the follow: Associative proof: \begin{align} \left((A,B)*(C,D)\right)*(E,F) &= (A,B)*\left((C,D)*(E,F)\right) \\ (AC,AD+B)*(E,F) […]

Let $f\colon \mathbb{R}^2 \to\mathbb{R} $ be a smooth function. Can there exist an algebraic structure $(\mathbb{R}, \cdot)$ such that for $x,y \in \mathbb{R}$, $x \cdot y = f(x,y)$ that is a non-commutative semigroup that is strictly not a monoid or a group? I can’t think of an example, but it seems so unlikely that you […]

This question already has an answer here: What is the symbol ''$\divideontimes$'' (DIVIDE TIMES) for? 3 answers

In order to show something is associative one must show that $(x*y)*z$ = $x*(y*z)$. I want to show that $x * y = \frac{xy}{x+y+1}$ is associative. This is for self-study (I’m learning algebra over the summer) and need help finishing the proof. Below are my (hopefully not incorrect) steps, 1) $x*(y*z)$ = $x * \left(\frac{yz}{y+z+1}\right)$ […]

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 about the uniqueness (But In right-invertible associative structures this is provable) If $*$ is right-invetible then $a\cdot b=(a*(b*c))\setminus c$ only if $a\cdot b$ doesn’t depends […]

I am trying to understand the relation between exclusive-OR (XOR) and modular addition in the function $f(x,a,b,c,d) = \bigg(\Big(\big((x \oplus a) \boxplus b\big) \oplus c\Big) \boxplus d\bigg)$ over $\mathbb{Z}_{2^n}$, where $\oplus$ denotes XOR and $\boxplus$ denotes addition modulo $2^n$. It seems that when $n=2$, for any value of $(a,b,c,d)$, I can find $31$ different $(a’,b’,c’,d’)$ […]

Addition modulo $1$ defines a group structure on the halfopen unit interval $[0,1)$. Obviously, this construction does not work if one starts with $[0,1]$ instead of $[0,1)$. Is it possible to extend addition on $[0,1]$ to a group operation? More precisely: Does there exist a binary operation $\ast: [0,1]^2 \to [0,1]$ such that $a \ast […]

This question already has an answer here: Does commutativity imply Associativity? 7 answers

This question is inspired by and/or a generalization of this question about the “reciprocal addition” operation. Consider the following: One is tempted to say multiplication is simply “addition under the logarithm”, in the sense that $z = x y$ if and only if $\log z = \log x + \log y$ (for positive reals, anyway). […]

Foreword: I have read R.H. Bruck’s A Survey of binary systems, where the notion of halfoperation is given. A halfoperation $\ast$ differs from a (binary) operation since $a\ast b$ may not be defined for all ordered pairs $(a,b)$. For example, from the Cayley table $$\begin{array}{c|cc}\ast & 0 & 1\\ \hline 0 & 0 & 1\\1 […]

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