# Need help confirming which algebraic structure this is

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?

• "the operation is closed under $\bf R$...." No; $\bf R$ is closed under the operation. – Gerry Myerson Mar 14 at 6:22

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.
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.