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If you took the familiar Peano Axioms and replaced the axiom

$x \in \mathbb{N} \implies \exists y\in \mathbb{N}(y =S(x))$

with

$x \in \mathbb{M} \implies (\exists y_1\in \mathbb{M})(\exists y_2\in \mathbb{M})(y_1 =S_1(x)\wedge y_2 =S_2(x) \wedge y_1\ne y_2)$

and the other axioms (including the ones defining addition and multiplication) modified accordingly.

The structure here ($\mathbb{M}$) would seem to resemble a tree of numbers, which each "level" $n$ containing $2^n$ elements. I was wondering if there is a name for this set of numbers.

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  • $\begingroup$ How, exactly, do you modify the addition axioms? Normally, $a+S(b)=S(a+b)$. Do you want $a+S_1(b)=S_1(a+b)$ and $a+S_2(b)=S_2(a+b)$? EDIT: This would work, I guess, but we lose $a+b=b+a$. Take $a=S_1(0),b=S_2(0)$. $\endgroup$ Commented Apr 12, 2015 at 15:22
  • $\begingroup$ And for multiplication, normally $aS(b)=ab+a$. Do you want $aS_1(b)=aS_2(b)=ab+a$? $\endgroup$ Commented Apr 12, 2015 at 15:27

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Assuming I have the correct interpretation of other axioms modified accordingly, you're describing an infinite binary tree, or more precisely an infinite, rooted, complete binary tree.

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  • $\begingroup$ Are addition and multiplication normally defined over binary trees? And if so, what form do they take?. (Sorry, I should have made it clear that I meant the complete set of PA; including the definitions of addition and multiplication) $\endgroup$
    – k_g
    Commented Apr 3, 2015 at 2:33
  • $\begingroup$ The usual five basic peano axioms themselves give rise to addition and multiplication. In this case, the axioms would not give rise to the same sort. The infinite binary tree is a non-commutative monoid on the corresponding operation to addition for the natural numbers. Basically, if you tell me two elements $x$ and $y$, i.e. two sequences of $S_1$ and $S_2$ operations, then you can take the element you get by first doing the operations for $x$ and then the operations for $y$, but if you do them in the other order, you may get somewhere else. $\endgroup$
    – aes
    Commented Apr 3, 2015 at 2:38
  • $\begingroup$ Would something like the Hadamard product come into play here? $\endgroup$ Commented Apr 3, 2015 at 2:52
  • $\begingroup$ @martycohen I'm not sure what relationship you're seeing. Basically, the infinite binary tree is a subset of the free group on two elements $S_1$, and $S_2$, consisting of words in $S_1$ and $S_2$ with no negative powers. It's closed under the group operation (i.e. it's a sub-monoid of the free group). This corresponds to addition on the natural numbers in this setup. We could maybe come up with some generalization of multiplication, but I don't see how to get it to satisfy anything useful like the distributive law for example. $\endgroup$
    – aes
    Commented Apr 3, 2015 at 3:05
  • $\begingroup$ @aes I tinkered around a bit with the coordinate representations $\langle a, b\rangle$, where $a$ is the depth (or number of steps from 0) and $b$ is the index, or "which branch", and modified versions of the Peano addition definition to find the definition $\langle a, b \rangle + \langle c, d \rangle = \langle a+c, 2^c b+d \rangle$, which is a unique definition (although it is unfortunately not commutative). $\endgroup$
    – k_g
    Commented Apr 3, 2015 at 19:12

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