# Does {$\Bbb Z_0$,$\Bbb Z_1$, $\Bbb Z_2 ,\cdots$, $\Bbb Z_{m-1}$} form a partition of $\Bbb Z$?

Definition 5. Let $X$ be a nonempty set. By a partition $P$ of $X$ we mean a set of nonempty subsets of $X$ such that

(a) If $A, B \in \mathscr P$ and $A \neq B$, then $A \cap B = \emptyset$,
(b) $\bigcup\limits_{C \in \mathscr P} C=X$

Example 6. Let $m$ be any fixed positive integer. For each integer $j$, $0\le j \lt m$, let $\Bbb Z_j=\{x \in \Bbb Z\,|\, x-j=km \text{ for some } k \in \Bbb Z\}$. Then the set $$\{\Bbb Z_0,\Bbb Z_1, \Bbb Z_2 ,\cdots, \Bbb Z_{m-1}\}$$ forms a partition of $\Bbb Z$.

Source: Set Theory by You-Feng Lin, Shwu-Yeng T. Lin.

$\Bbb Z_0, \Bbb Z_1, \Bbb Z_2 ,\cdots, \Bbb Z_{m-1}$ are subsets of $\Bbb Z$, and each are different, so $\{\Bbb Z_0, \Bbb Z_1, \Bbb Z_2,\cdots, \Bbb Z_{m-1}\}$ satisfies the condition (a) in definition 5, but does it also satisfy the condition (b) $\bigcup\limits_{C \in \mathscr P}=X$? I don't think it does because the union of them is not $\Bbb Z$. That is the finite set $\Bbb Z_0 \bigcup \Bbb Z_1 \bigcup \Bbb Z_2 \bigcup \cdots \bigcup \Bbb Z_{m-1} \neq \Bbb Z$ since $\Bbb Z$ is an infinite set. So why $\{\Bbb Z_0, \Bbb Z_1, \Bbb Z_2,\cdots, \Bbb Z_{m-1}\}$ forms a partition of $\Bbb Z$? Isn't it insufficient to be a partition of $\Bbb Z$?

• $\Bbb Z_0 \bigcup \Bbb Z_1 \bigcup \Bbb Z_2 \bigcup \cdots \bigcup \Bbb Z_{m-1}$ isn't finite - it's a finite union of infinite sets.
– πr8
Jan 11, 2016 at 10:30
• For example, if $m=2$, this describes the partition of $\Bbb Z$ in to odd and even numbers ... There are just two parts, but every integer is (either) odd or even Jan 11, 2016 at 10:41
• @πr8 $m$ is a fixed positive integer. As for $\Bbb Z_0=\{x \in \Bbb Z\,|\, x=km \text{ for some } k \in \Bbb Z\}$, "for some" in "for some k" is an existential quantifier, and m is fixed. So $\Bbb Z_0$ is a finite set, and similarly so are the others. Jan 11, 2016 at 10:44
• $\mathbb{Z}_0$ is the set of multiples of $m$. Why should it be finite?
– πr8
Jan 11, 2016 at 10:54
• @πr8 Shoudn't it be $\Bbb Z_0=\{x \in \Bbb Z\,|\, x=km \text{ for all } k \in \Bbb Z\}$ to be an infinite set? Jan 11, 2016 at 11:05

Given two integers $a$ and $b$, with $b ≠ 0$, there exist unique integers $q$ and $r$ such that $$a = bq + r$$ and $$0 ≤ r < |b|,$$ where $|b|$ denotes the absolute value of $b$.

So given a integer $a$, there exist a $r$ such that $a = mq + r$ with $0 ≤ r < |b|$. By the definition of $Z_i$,$i=0,\dots,m-1$, $a\in Z_r$. So $$\bigcup_{i=0}^{m-1}Z_i = Z$$

Hint. The relation $a=b \mod m$ is an equivalence relation and $\mathbb{Z}_i, i=0 \ldots m-1$ are the equivalence classes with respect to the relations. Then use a theorem that the set of all equivalence classes form a partition of the set $\mathbb{Z}.$

• A one-line hint that's nothing more than a reformulation of the question. Good job. Jan 11, 2016 at 10:02

Condition (b) is also satisfied, since any integer has a remainder in $\;\{0,1,\dots, m-1\}\;$ by the Euclidean division theorem.

Why don't you think it's not a partition?

It's clear that the union is a subset of $\mathbb Z$, so the only way it would not become $\mathbb Z$ is that there's an integer not in the union.

So assume that $z\in \mathbb Z$, then by Euclid division you have $z = qm + r$ where $0\le r< m$. Therefore $z\in \mathbb Z_r$ and therefore in the union.

Note that your motivation that (a) is fulfilled is a bit sloppy. It's not enough that they're different, they have to be pairwise disjoint as well. That too follows from Euclid division (the remainder in the interval is unique).

Their union is $\mathbb{Z}$ as every consecutive point in $\mathbb Z_j$ for some $j$ belonging to $(0,1,...,m-1)$ is equispaced with a distance $m$, so each $\mathbb Z_j$ completes $1/m$th of the $\mathbb Z$ by its $m$ spaced motion along the real line and as they are non intersecting so their union completes $m$ times $1/m$th of $\mathbb Z$, i.e. the whole $\mathbb Z$.