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Let's define a binary operation $*$ on $\mathbb{R}$ such as $$ a * b = e^{a+b} $$ and investigate which algebraic structure this is.

Well first of all we notice that the operation is closed under $\mathbb{R}$ since if $a,b \in \mathbb{R}$ and $e^x$ if defined for entire $\mathbb{R}$ then $e^{a+b} \in \mathbb{R}$. From the fact that $\mathbb{R}$ is a field we get commutativity. This is where I start to doubt myself a little. It feels like we do not have associativity since $$ (a * b) * c = e^{a+b} * c = e^{e^{a+b}+c} \neq e^{a+e^{b+c}} = a * e^{b+c} = a * (b * c)$$ I also don't think there exists an universal identity element $x$. Only way I can think of getting $a * x= a$ is when $x = ln(a)-a$, then $a * x = e^{a+ln(a)-a} = e^{ln(a)} = a$, but this is not universal or defined for all $\mathbb{R}$. Since we do not have an identity element, we cannot talk about an inverse either. Is this correct or am I going wrong here? If this is true, what algebraic structure this is? Commutative groupoid?

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  • $\begingroup$ "the operation is closed under $\bf R$...." No; $\bf R$ is closed under the operation. $\endgroup$ – Gerry Myerson Mar 14 at 6:22
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It is easy to check the operation is not associative: $$\begin{aligned} (0 * 0) * 1 &= 1 * 1 = e^2 \\ 0 * (0 * 1) &= 0 * e = e^e \end{aligned}$$ You've already observed there is no unique identity element.

It is commutative, and also satisfies some bizzare properties like $a * (b + c) = b * (a + c) = c * (a + b)$. But perhaps without further observations, it is just a commutative magma.

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You are right; there is no identity element. If $a$ was such element, we would have $a*a=a$, but $a*a=e^{2a}>a$. So, yes, what we have here is a commutative groupoid.

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  • $\begingroup$ Surely anything with “group” in the name must be associative. $\endgroup$ – Joppy Mar 14 at 6:34
  • $\begingroup$ Not at all. $\endgroup$ – José Carlos Santos Mar 14 at 6:42
  • $\begingroup$ Interesting, there is a different definition here: en.wikipedia.org/wiki/Groupoid $\endgroup$ – Joppy Mar 14 at 6:44
  • $\begingroup$ That's about the definition of groupoid in the context of Category Theory. $\endgroup$ – José Carlos Santos Mar 14 at 6:46
  • $\begingroup$ I see that, but it also defines a groupoid as an algebraic structure with a (partial) operation obeying various axioms. I’ve never heard a groupoid mean the same thing as a magma before, learn something new every day! $\endgroup$ – Joppy Mar 14 at 6:48

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