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Seen $(\Bbb N,+,\cdot)$ as a semiring, is it possible to extend it to a semiring $(R,+,\cdot)$ so that the additive and multiplicative monoids become isomorphic? This means there is some monoid-isomorphism

$$\varphi:(R,\cdot)\cong(R,+)$$

and $\Bbb N$ is a sub-semiring of $R$. Here, $\Bbb N$ is meant to include $0$. I do not think that there is such an extension, but I cannot find a contradiction. I also wonder if the problem becomes easiert when asking for an isomorphism

$$\varphi:(R\setminus\{0\},\cdot)\cong(R,+)$$

instead?


Observations

The multiplication will be commutative. Also, there will be many new "numbers", e.g. a unique additively absorbing element $\eta:=\varphi(0)$, i.e. $\eta+x=\eta$ for all $x\in R$. For $n\in\Bbb N^+$ we have

$$n\cdot \eta=\underbrace{\eta+\cdots+\eta}_n=\eta.$$

Therefore we have further elements $\tilde\eta=\varphi(\eta)$ that absorbe some numbers when added to them, e.g. $\varphi(n),n\in\Bbb N^+$, but not all (there can be only one universally absorbing element).

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    $\begingroup$ The monoids are almost isomorphic already under logarithm. So add countable many zeros to each monoid, and extend freely to a semiring. Gerhard "Equations Allow Building Your Own" Paseman, 2017.07.16. $\endgroup$ – Gerhard Paseman Jul 16 '17 at 15:28
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    $\begingroup$ It seems to me that we are trying to formally close $\mathbb N$ by both $\log$ and $\exp$, besides $+$ and $\cdot$, but thought of in terms of a generic base (rather than $e$) and only the expressions that simplify because of general properties of $\log$ and $\exp$, regardless of base, do so. The question then would be whether we can run into a contradiction, like 0=1, presumably by enforcing distributivity? $\endgroup$ – Yaakov Baruch Jul 16 '17 at 15:55
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    $\begingroup$ @ZachTeitler For example $[0,1]$ with $a+b=\max(a,b)$ and $a\cdot b=\min(a,b)$. Unfortunatly this example is idempotent and the only one I know of. $\endgroup$ – M. Winter Jul 16 '17 at 19:30
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    $\begingroup$ @M.Winter: Do you assume that substructures share the same 0 and 1 as the whole structure? (So, here, must the 0 and 1 of $\mathbb N$ be the same as the 0 and 1 of $R$?) $\endgroup$ – Keith Kearnes Jul 16 '17 at 20:13
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    $\begingroup$ @M.Winter: No, that argument is not correct. Your example of a semiring [0,1] has unit element 1 and zero element 0, but it contains a subsemiring [.1,.9] whose unit is .9 and whose zero is .1. $\endgroup$ – Keith Kearnes Jul 16 '17 at 20:25
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There is an extension $R$: take the closure of $\mathbb N$ by the operations $\text{L}$ (or $\varphi$ in the OP) and its inverse $\text{E}$, which are the logarithm and exponential in base $1.2$. Notice that the choice of base (see below) implies that all new numbers generated by repeated applications of the 4 operations ($+$, $\cdot$, $\text{L}$ and $\text{E}$) are $\gt 1$, except for infinities, so that there is never a need to apply $\text{L}$ to negative numbers.

For infinities impose the following rules:

$0\cdot x=0$

$\text{L}(0)+x=\text{L}(0)$

$\text{L}(0)\cdot x=\text{L}(0)$ if $x\ne0$

$\text{L}(\text{L}(0))+x=\text{L}(\text{L}(0))$ if $x\ne \text{L}(0)$

$\text{L}(\text{L}(0))\cdot x=\text{L}(\text{L}(0))$ if $x\ne \text{L}(0),0$

$\text{L}(\text{L}(\text{L}(0)))+x=\text{L}(\text{L}(\text{L}(0)))$ if $x\ne \text{L}(\text{L}(0)),\text{L}(0)$

...

These rules obey the semiring axioms and cover all possibilities for mixing reals and infinities. Notice also that $\text{L}(0)$, $\text{L}(\text{L}(0))$, $\text{L}(\text{L}(\text{L}(0)))$ etc. are then the only non-reals in $R$.

As for the choice of a base $\lt\sqrt{2}$, notice that if $a=1.2577...$ is the solution to $1.2^a=a$, then both $\text{L}(x)>a$ and $\text{E}(x)>a$ if $x>a$ and therefore the only finite elements in this model $R$ that can be $<a$ are $0$, $1$, $\text{E}(1)$, $\text{E}(\text{E}(1))$, $\text{E}(\text{E}(\text{E}(1)))$ etc.

Update. A much simpler model is the completion of the one above, but with $\sqrt{2}$ as the base:

$ \text{L(x)}=\log_{\sqrt{2}}(x)$, $\text{E}(x)=\left(\sqrt{2}\right)^{x}$ its inverse, and

$R= \{\dots\ \text{L}^3(0), \text{L}^2(0), \text{L}(0), 0, 1, \sqrt{2}, \sqrt{2}^\sqrt{2},\sqrt{2}^{\sqrt{2}^\sqrt{2}}\dots\}\cup\{x\in\mathbb R| x\ge 2\}$

with the same rules for infinities as already explained.

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  • $\begingroup$ I think this model is what Gerhard meant in one of his comments, with details filled in. $\endgroup$ – Yaakov Baruch Jul 17 '17 at 7:27
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    $\begingroup$ Notice that this same model, if we strip the infinities, also answer the second question in the OP, since $\text{L}$ is then an isomorphism $(R\backslash\{0\},\cdot) \rightarrow (R,+)$. $\endgroup$ – Yaakov Baruch Jul 17 '17 at 16:49
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    $\begingroup$ For the record, I have implemented these rules (for the completed version with base $\sqrt{2}$) in Maple and thereby checked that there are not any hidden problems with the semiring axioms. A very interesting example! $\endgroup$ – Neil Strickland Jul 21 '17 at 17:50
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    $\begingroup$ In fact, if we write $b \uparrow \uparrow n$ for the tower of powers of $b$ of height $n$, and $b_k$ for the (unique) solution to $b_k \uparrow \uparrow k = 2$, then $b_k > \sqrt{2}$ should also be a suitable base so long as the sequence of values $b_k \uparrow \uparrow n$ is bounded (which it is when $b_k < e^{1/e}$, true for $k \ge 5$). $\endgroup$ – Robin Saunders Jul 24 '17 at 5:41
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    $\begingroup$ But surely then $\text{L}^4(2) < b$? A base larger than $\sqrt{2}$ requires $b \uparrow \uparrow k = 2$ for some $k$, does it not? $\endgroup$ – Robin Saunders Jul 24 '17 at 9:46

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