Expressing every natural number as a sum of elements of two disjoint subsets in a unique way

Question

I thought of the variation of the following question on AOPS: "Find two infinite subsets $A$ and $B$ in $\mathbb{N}_{0}$ such that every positive integer can be written uniquely as the sum of an element in $A$ and an element in $B$." The variation added by me was that the sets must intersect only at $\{0\}$. Few trivial construction were to write the base $b$ representation of every positive integer (here $b$ is a prime number), and then consider the vector space $\mathcal{P}(b)$ of polynomials of $b$ with coefficients in $\mathbb{Z}_b$ and then find subspaces $U$ and $V$ such that $\mathcal{P}(b)=U\oplus V$. I am looking forward for other constructions which also satisfy this variation of the problem.

Answer 1

$18034677905=10004070905+8030607000$, a sum of two numbers, one having nonzero digits only at the odd places, the other only at the evens. Here, $18034677905$ is a variable. [What I mean is, this construction works for any positive integer $n$.]

EDIT: Note that (formally) $$1+x+x^2+x^3+\cdots=(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots$$ Now break the right side up into two products in any way you like, just making sure that each of the two products has infinitely many terms: $$\prod_{i\in R}(1+x^{2^i})\prod_{i\in S}(1+x^{2^i})$$ where $R$ and $S$, each infinite, form a partition of $\{\,0,1,2,\dots\,\}$. Then define $A$ and $B$ by letting $$\prod_{i\in R}(1+x^{2^i})=\sum_{i\in A}x^i,\qquad\prod_{i\in S}(1+x^{2^i})=\sum_{i\in B}x^i$$ Then $A$ and $B$ have the desired property. But perhaps this is still considered trivial. Still, it says the crucial thing is finding factorizations of $1+x+x^2+x^3+\cdots=(1-x)^{-1}$, preferably factorizations which aren't related to any base $b$ arithmetic. I'm not sure there are any.

MORE EDIT: Thinking about this a little more, here's something we can do: $$1+x+x^2+\cdots=(1+x)A(x)$$ where $$A(x)=1+x^2+x^4+\cdots=(1+x^2+x^4)B(x)$$ where $$B(x)=1+x^6+x^{12}+x^{18}+\cdots=(1+x^6+x^{12}+x^{18})C(x)$$ where $$C(x)=1+x^{24}+x^{48}+\cdots=(1+x^{24}+x^{48}+x^{72}+x^{96})D(x)$$ and so on gives us a factorization of $(1-x)^{-1}$ as a product of an infinity of polynomials which we can split into two infinite sets however we like to get a new pair of sets that isn't based on any one base but more like a "mixed radix" representation. E.g., from $$(1+x)(1+x^6+x^{12}+x^{18})(1+x^{120}+\cdots+x^{600})\cdots$$ $$=1+x+x^6+x^7+x^{12}+x^{13}+x^{18}+x^{19}+x^{120}+\cdots$$ and $$(1+x^2+x^4)(1+x^{24}+x^{48}+x^{72}+x^{96})\cdots$$ $$=1+x^2+x^4+x^{24}+x^{26}+x^{28}+x^{48}+x^{50}+x^{52}+\cdots$$ we get the two sets $$A=\{\,0,1,6,7,12,13,18,19,120,\dots\,\}$$ $$B=\{\,0,2,4,24,26,28,48,50,52,72,74,76,96,98,100,\dots\,\}$$ There is a lot of freedom in this construction, and maybe it gives all possible answers.

Answer 2

Indeed, mixed radix systems are all the solutions. We have, in an obvious notation, that $A \oplus B= \mathbb{N}.$ $0$ is in both $A$ and $B$ and wlog $1\in A$. Let $m$ be the least positive in $B.$ Then $A^*=\{0,1,2,\cdots,m-1\} \subset A.$ If we show $A=A^*\oplus mA'$ and $B=mB'$ where $A' \oplus B'= \mathbb{N}$ we have the result. By induction on $j,$ for $T=\{jm,jm+1,\cdots,jm+m-1\}$ we have $T\cap A=T$ or $\emptyset$ and, in the second case , $T \cap B=\{jm\}$ or $\emptyset.$ The result follows.