Partition of N with ratios close to $2$ a) Prove that given constants $a,b$ with $1<a<2<b$, there is no partition of the set of positive integers into two subsets $A_0$ and $A_1$ such that: if $j \in \{0,1\}$ and $m,n$ are in $A_j$, then either $\frac{n}{m} <a$ or $\frac{n}{m}>b$.
(b) Find all pairs of real numbers $(a,b)$ with $1<a<2<b$ for which the following property holds: there exists a partition of the set of positive integers into three subsets $A_0, A_1, A_2$ such that if $j \in \{0,1,2\}$ and $m,n$ are in $A_j$, then either $\frac{n}{m} <a$ or $\frac{n}{m}>b$.
I thought a partition of a set is to write such a set as a union of disjoint subsets two by two
 A: First question:
Suppose there is such a partition. Let $M = \frac{a + 1}{b - a}$. Then for any integer $m > M$, we have $bm - a(m + 1) = (b - a)m - a > 1$.
This means that there exists at least one integer $n$ such that $n \geq a(m + 1)$ and $n \leq bm$.
Since $am < a(m + 1) \leq n \leq bm$, we have $a \leq \frac{n}{m} \leq b$. Hence $n$ and $m$ are not in the same subset.
Since $a(m + 1)\leq n \leq bm < b(m + 1)$, we have $a \leq \frac{n}{m + 1} \leq b$. Hence $n$ and $m + 1$ are not in the same subset.
But there are only two subsets, so $m$ and $m + 1$ must be in the same subset.
This being true for all $m > M$ leads to a contradiction.
Note that we only used $b > a$, instead of the full inequality $1 < a < 2 < b$.

Second question:
I only have obtained a sufficient condition: $\sqrt2 < a < 2 < b \leq a^2$.
Under this condition, let us define three subsets of $\mathbb R$: for $j = 0, 1, 2$, define $B_j$ as the union $\bigcup_{k \geq 0} [a^{3k + j}, a^{3k + j + 1})$.
Now define $A_j$ as the set of positive integers in $B_j$. Let us verify that this partition indeed satisfies our condition.
Suppose $n > m$ are in one subset $A_j$. Let $k$ (resp. $t$) be the integer such that $n$ (resp. $m$) is in the interval $[a^{3k + j}, a^{3k + j + 1})$ (resp. $[a^{3t + j}, a^{3t + j + 1})$).
If $k = t$, then we have $\frac{n}{m} < \frac{a^{3k + j + 1}}{a^{3k + j}} = a$.
If $k > t$, then $k \geq t + 1$ and we have $\frac{n}{m} > \frac{a^{3(t + 1) + j}}{a^{3t + j + 1}} = a^2 \geq b$.

I don't know whether this condition is also necessary. This seems to be the key point of the question, the rest being somewhat easy.
