# Is addition on $\mathbb{R}$ unique up to automorphism?

Consider the set of real numbers $\mathbb{R}$, which we know is the only complete ordered field up to isomorphism.

If we fix the following axioms for the 'addition' operation $\bot:\mathbb{R}^2\to \mathbb{R}$:

1. Associativity
2. Commutativity
3. Identity element 0 (not that there exists some 0 element, but that the identity element is actually the canonical 0).
4. All elements have their own unique inverse

[this defines an abelian group over the set $\mathbb{R}$ with identity 0]

and finally

1. $\bot$ is distributive over multiplication: $\forall a,b,\lambda\in \mathbb{R},$ $\lambda\cdot (a\bot b)=(\lambda \cdot a) \bot (\lambda \cdot b).$

Do these axioms uniquely determine addition on $\mathbb{R}$ up to automorphism?

The example of $a\bot b := (a^3+b^3)^{1/3}$ shows that these axioms do not uniquely determine addition. However, the automorphism $f(x)=x^3$ makes this operation still effectively the same as addition. The question is whether all operations $\bot$ which adhere to these axioms have an automorphism with addition.

And as a secondary question, if it turns out that they do:

Is the fifth axiom necessary, or is the only abelian group with identity 0 on the real numbers addition (up to automorphism)?

EDIT: The second part has been answered, and the answer is yes, it is necessary.

• Clearly the answer to your sceond question is yes, the fifth axiom is necessary. Otherwise, without it you're asking "are there two non isomorphic abelian groups with cardinality $\mathfrak{c}$ ?" to which the answer is obviously yes (you can simply construct a counter-example, for instance $(\Bbb{Z}/2\Bbb{Z})^{\Bbb{N}}$ which is a torsion group and therefore isn't isomorphic to $\Bbb{R}$) – Max Jun 19 '17 at 19:41
• @Max Or even more simply, $(\mathbb{Z}/2\mathbb{Z})\times \mathbb{R}$. – Noah Schweber Jun 19 '17 at 19:42
• @NoahSchweber why make things easy when you can make them complicated haha ... – Max Jun 19 '17 at 19:43
• @MathTrain None of that matters: if I have a group $G$ with cardinality continuum, then there's a bijection between $G$ and $\mathbb{R}$ which sends the identity of $G$ to $0$ - this induces a group structure on $\mathbb{R}$ with the desired properties, isomorphic to $G$. You need some further conditions on the group operation, here. – Noah Schweber Jun 19 '17 at 19:51
• Yes I know but what you're asking (without the distributivity condition) is equivalent to "Do there exist two non isomorphic abelian groups of cardinality $\mathfrak{c}$ ?" . The fact that the two questions are equivalent is relatively easy once you see that having a bijection is simply renaming things. – Max Jun 19 '17 at 19:51

Your axioms do not determine addition uniquely. Note that if $$K$$ is any field and $$f:\mathbb{R}^\times\to K^\times$$ is an isomorphism of the multiplicative groups, then the operation $$(x,y)\mapsto f^{-1}(f(x)+f(y))$$ (and $$(x,0)\mapsto x$$, $$(0,y)\mapsto y$$) will satisfy your axioms, and will only be equivalent to ordinary addition in your sense if $$K$$ is isomorphic to $$\mathbb{R}$$ as a field.
Now note that the abelian group $$\mathbb{R}^\times$$ is not too hard to understand. It is the direct sum of the subgroups $$\{\pm1\}$$ and $$\mathbb{R}_+$$, and $$\mathbb{R}_+$$ is a vector space over $$\mathbb{Q}$$ since every positive real number has an $$n$$th root for any $$n$$. More generally, the same description holds for $$K^\times$$ if $$K$$ is any ordered field in which every positive element has an $$n$$th root for any $$n$$. Since two $$\mathbb{Q}$$-vector spaces of the same uncountable cardinality are isomorphic, it follows that if $$K$$ is any ordered field of the same cardinality as $$\mathbb{R}$$ in which every positive element has an $$n$$th root for any $$n$$, then $$K^\times\cong\mathbb{R}^\times$$.
However, such a field need not be isomorphic to $$\mathbb{R}$$, and so can give rise to a different "addition" on $$\mathbb{R}$$ satisfying your axioms. For instance, you could take $$K$$ to be any non-archimedean real-closed field of cardinality $$2^{\aleph_0}$$ (e.g., the real closure of $$\mathbb{R}(x)$$ where $$x$$ is infinitely large).