Subfield: Algebra


Question: Let $\mathbb{Z}$ and $\mathbb{Q}$ be the sets of integers and rationals respectively.
Here $X+Y$ denotes the set $\{x+y \mid x \in X, y \in Y\}$, for $X, Y \subseteq \mathbb{Z}$ and $X, Y \subseteq \mathbb{Q}$.


a) Does there exist a partition of $\mathbb{Z}$ into three non-empty subsets $A, B, C$ such that the sets $A+B, B+C, C+A$ are disjoint?

b) Does there exist a partition of $\mathbb{Q}$ into three non-empty subsets $A, B, C$ such that the sets $A+B, B+C, C+A$ are disjoint?

Solution:
a) The residue classes modulo 3 yield such a partition:

$$
A=\{3 k \mid k \in \mathbb{Z}\}, \quad B=\{3 k+1 \mid k \in \mathbb{Z}\}, \quad C=\{3 k+2 \mid k \in \mathbb{Z}\}
$$

b) The answer is no. Suppose that $\mathbb{Q}$ can be partitioned into non-empty subsets $A, B, C$ as stated. Note that for all $a \in A, b \in B, c \in C$ one has

$$
a+b-c \in C, \quad b+c-a \in A, \quad c+a-b \in B\tag{1}
$$

Indeed $a+b-c \notin A$ as $(A+B) \cap(A+C)=\emptyset$, and similarly $a+b-c \notin B$, hence $a+b-c \in C$. The other two relations follow by symmetry. Hence $A+B \subset C+C, B+C \subset A+A, C+A \subset B+B$.

The opposite inclusions also hold. Let $a, a^{\prime} \in A$ and $b \in B, c \in C$ be arbitrary. By (1) $a^{\prime}+c-b \in B$, and since $a \in A, c \in C$, we use (1) again to obtain

$$
a+a^{\prime}-b=a+\left(a^{\prime}+c-b\right)-c \in C .
$$

So $A+A \subset B+C$ and likewise $B+B \subset C+A, C+C \subset A+B$. In summary

$$
B+C=A+A, \quad C+A=B+B, \quad A+B=C+C .
$$

Furthermore suppose that $0 \in A$ without loss of generality. Then $B=\{0\}+B \subset A+B$ and $C=\{0\}+C \subset A+C$. So, since $B+C$ is disjoint with $A+B$ and $A+C$, it is also disjoint with $B$ and $C$. Hence $B+C$ is contained in $\mathbb{Z} \backslash(B \cup C)=A$. Because $B+C=A+A$, we obtain $A+A \subset A$. On the other hand $A=\{0\}+A \subset A+A$, implying $A=A+A=B+C$.

Therefore $A+B+C=A+A+A=A$, and now $B+B=C+A$ and $C+C=A+B$ yield $B+B+B=A+B+C=A, C+C+C=A+B+C=A$. In particular if $r \in \mathbb{Q}=A \cup B \cup C$ is arbitrary then $3 r \in A$.

However such a conclusion is impossible. Take any $b \in B(B \neq \emptyset)$ and let $r=b / 3 \in \mathbb{Q}$. Then $b=3 r \in A$ which is a contradiction.
We prove that the example for $\mathbb{Z}$ from the first solution is unique, and then use this fact to solve part $b)$.

Let $\mathbb{Z}=A \cup B \cup C$ be a partition of $\mathbb{Z}$ with $A, B, C \neq \emptyset$ and $A+B, B+C, C+A$ disjoint. We need the relations (1) which clearly hold for $\mathbb{Z}$. Fix two consecutive integers from different sets, say $b \in B$ and $c=b+1 \in C$. For every $a \in A$ we have, in view of (1), $a-1=a+b-c \in C$ and $a+1=a+c-b \in B$. So every $a \in A$ is preceded by a number from $C$ and followed by a number from $B$.

In particular there are pairs of the form $c, c+1$ with $c \in C, c+1 \in A$. For such a pair and any $b \in B$ analogous reasoning shows that each $b \in B$ is preceded by a number from $A$ and followed by a number from $C$. There are also pairs $b, b-1$ with $b \in B, b-1 \in A$. We use them in a similar way to prove that each $c \in C$ is preceded by a number from $B$ and followed by a number from $A$.

By putting the observations together we infer that $A, B, C$ are the three congruence classes modulo 3. Observe that all multiples of 3 are in the set of the partition that contains 0 .



Now we turn to part b). Suppose that there is a partition of $\mathbb{Q}$ with the given properties. Choose three rationals $r_{i}=p_{i} / q_{i}$ from the three sets $A, B, C, i=1,2,3$, and set $N=3 q_{1} q_{2} q_{3}$.

Let $S \subset \mathbb{Q}$ be the set of fractions with denominators $N$ (irreducible or not). It is obtained through multiplication of every integer by the constant $1 / N$, hence closed under sums and differences. Moreover, if we identify each $k \in \mathbb{Z}$ with $k / N \in S$ then $S$ is essentially the set $\mathbb{Z}$ with respect to addition. The numbers $r_{i}$ belong to $S$ because

$$
r_{1}=\frac{3 p_{1} q_{2} q_{3}}{N}, \quad r_{2}=\frac{3 p_{2} q_{3} q_{1}}{N}, \quad r_{3}=\frac{3 p_{3} q_{1} q_{2}}{N}
$$

The partition $\mathbb{Q}=A \cup B \cup C$ of $\mathbb{Q}$ induces a partition $S=A^{\prime} \cup B^{\prime} \cup C^{\prime}$ of $S$, with $A^{\prime}=A \cap S$, $B^{\prime}=B \cap S, C^{\prime}=C \cap S$. Clearly $A^{\prime}+B^{\prime}, B^{\prime}+C^{\prime}, C^{\prime}+A^{\prime}$ are disjoint, so this partition has the properties we consider.

By the uniqueness of the example for $\mathbb{Z}$ the sets $A^{\prime}, B^{\prime}, C^{\prime}$ are the congruence classes modulo 3 , multiplied by $1 / N$. Also all multiples of $3 / N$ are in the same set, $A^{\prime}, B^{\prime}$ or $C^{\prime}$. This holds for $r_{1}, r_{2}, r_{3}$ in particular as they are all multiples of $3 / N$. However $r_{1}, r_{2}, r_{3}$ are in different sets $A^{\prime}, B^{\prime}, C^{\prime}$ since they were chosen from different sets $A, B, C$. The contradiction ends the proof.