Consider a binary composition $\star:\Bbb R^2\_{>0}\rightarrow \Bbb R\_{>0}:(x,y)\mapsto x\star y$ with the following properties.


(Commutativity)$\quad x\star y=y\star x\;$for all $x,y\in\Bbb R\_{>0};$


(Associativity)$\quad(x\star y)\star z=x\star(y\star z)\;$for all $x,y,z\in\Bbb R\_{>0};$


(Continuity)$\quad x\mapsto a\star x\;(x\in\Bbb R\_{>0})\;$is continuous, for each $a\in \Bbb R\_{>0};$


(Upper monotony)$\quad x\mapsto a\star x\;(x\in\Bbb R\_{>0})\;$is strictly increasing, for each $a\in \Bbb R\_{>0};$


(Surjectivity)$\quad \star\;$is unbounded above and below in $\Bbb R\_{>0}.$


These properties hold for addition and multiplication in $\Bbb R\_{>0}.$ Do they hold for any other composition in $\Bbb R\_{>0}$? If so, is an example of such a composition known? If not, is the above list redundant---that is, can any of the five conditions be weakened or removed?



Which compositions have these sum-like and product-like properties on the positive reals?
As remarked in comments under the question, an example would be $x \star y = \sqrt{x^2 + y^2}$. Indeed, if $f$ is any strictly monotonic bijection on $\mathbb{R}\_{>0}$ (either monotone increasing or monotone decreasing), then $x \star y = f^{-1}(f(x) + f(y))$ or $x \star y = f^{-1}(f(x)f(y))$ furnish examples which address the question. 


This is the usual "transport of structure" trick in mathematics: if we have e.g. a linearly ordered abelian group $(G, \cdot, \leq)$ and a linearly ordered set $(X, \leq)$ and an order-preserving isomorphism $f: X \to G$, then we can transport the group structure on $G$ along $f$ to get an ordered abelian group structure on $X$ which is isomorphic (by construction) to the ordered abelian group structure on $G$. That's all that was done here. 




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However, such examples raise an obvious further question: is every example *isomorphic* (as an ordered semigroup) either to $(\mathbb{R}\_{> 0}, +, \leq)$ or to $(\mathbb{R}\_{> 0}, \cdot, \leq)$? The division between these standard examples is that the additive case has no identity, whereas the multiplicative case does. This observation will to some extent guide the analysis (which I will leave somewhat sketchy at the end). 


Each element $x$ of such a structure $(\mathbb{R}\_{> 0}, \star, \leq)$ satisfies either 


* $x = x \star x$,
* $x < x \star x$,
* $x > x \star x$.


In the first case, we claim $x$ is an identity. Indeed, if $x \star y \neq y$ for some $y$, then $x \star y = (x \star x) \star y = x \star (x \star y)$, which contradicts the assumption that $x \star -$ is injective (since it is strictly monotone = strictly increasing). 


We will say that $x$ is *inflationary* if $x < x \star x$, and *deflationary* if $x \star x < x$. 


**Lemma 1:** If $x$ is inflationary, then $y < x \star y$ for every $y$. If $x$ is deflationary, then $x \star y < y$ for every $y$. 


**Proof:** If $x < x \star x$, then $x \star y < (x \star x) \star y$ since $- \star y$ is strictly increasing. Thus $x \star y < x \star (x \star y)$. Therefore $y < x \star y$ since $x \star -$ is strictly increasing. Similar argument if $x \star x < x$. 


**Lemma 2:** If there exists an inflationary element and a deflationary element, then there exists an identity. 


**Proof:** The collection of inflationary elements is upward closed (if $x$ is inflationary and $x \leq y$, then $y$ is inflationary since by lemma 1 we have $y < x \star y \leq y \star y$). Similarly the collection of deflationary elements is downward closed. These facts imply $y < x$ if $y$ is deflationary and $x$ is inflationary. Put $e = \inf\{x: x < x \star x\} = \sup\{y: y \star y < y\}$. We claim $e$ is an identity. For if $e < e \star e$, then for every $y < e$ we have $y \star e < e < e \star e$ (since $y$ is deflationary). Letting $y$ approach $e$ from the left, this contradicts continuity of $- \star e$. Similarly $e \star e < e$ is impossible. 


If $e$ is an identity, we have $y < e < x$ iff $y$ is deflationary and $x$ is inflationary. 


**Lemma 3:** If $x$ is inflationary, then $x^n \star z = z \star x^n$ is unbounded above for any $z$ (where $x^n := x^{\star n}$). If $y$ is deflationary, then $y^n z = z \star y^n$ is unbounded below. 


**Proof:** Otherwise, if $w$ is the least upper bound of $x \star z, x^2 \star z, x^3 \star z, \ldots$, then $w$ is an upper bound of $x \star w$ by continuity. But this contradicts $w < x \star w$ from lemma 1. Similarly for the second statement. 


**Lemma 4:** If there exists an identity $e$, then $(\mathbb{R}\_{> 0}, \star, \leq)$ is a group. 


**Proof:** Suppose $y < e < x$. Then the set $y^m x^n$ is unbounded above and below by lemma 3. It follows that the images of the functions $y \star -$ and $x \star - = - \star x$ are unbounded both above and below. Since these images are intervals (by strict monotonicity), they must be all of $\mathbb{R}\_{> 0}$. Thus $y \star -$ and $x \star -$ are bijections, so that there exists $z$, $w$ with $y \star z = e$ and $x \star w = e$, as was to be shown. 


At this point, we have that if there exists an identity, then the structure is a linearly ordered abelian group whose underlying order is isomorphic to $\mathbb{R}\_{> 0}$ or to $\mathbb{R}$. Sketching here: it is standard that such a structure is isomorphic to $(\mathbb{R}, +, \leq)$ (or to $(\mathbb{R}\_{> 0}, \cdot, \leq)$). For such a group is divisible, contains a copy of $\mathbb{Q}$ as a dense subgroup, and then one finishes using completeness and continuity. Notice this is true even without assuming the Surjectivity property of the OP (that comes for free if there's an identity). 


Now suppose there is no identity. Then either every element is inflationary, or every element is deflationary, by lemma 2. WLOG, we may assume every element is inflationary. Let me sketch an argument why such a structure should be isomorphic to $(\mathbb{R}\_{> 0}, +, \leq)$. 


The trick is to extend $(\mathbb{R}\_{> 0}, \star, \leq)$ to a linearly ordered commutative topological monoid $(\mathbb{R}\_{\geq 0}, \star, \leq)$ (the topology is the order topology) by treating $0$ as the identity. Here the continuity of the operation $\star: \mathbb{R}\_{\geq 0}^2 \to \mathbb{R}\_{\geq 0}$ at $(0, 0)$ is secured by the Surjectivity property. 


Now we have a *cancellative* linearly ordered commutative monoid, to which we may formally adjoin negatives to get a linearly ordered abelian group whose underlying order is isomorphic to $\mathbb{R}$, and the argument sketched above (using divisibility, completeness, etc.) should get the job done. 


