Proving a family of sets is a partition of the set of integers.

Question

I am trying to show that for

$E_n = \{10n, 10n + 1, 10n + 2 . . . , 10n + 9\}$,

$\{E_n\}$, $n ∈Z$ is a partition of $ Z$.

Should this be broken down into cases or is there a more general way to solve this?

Answer 1

First you show $E_m \cap E_n = \emptyset$ for $m \neq n$. Let $k \in E_m$, and also in $E_n$, then $k = 10m+ r = 10n + s$. Thus $10(m-n) = s-r$, and $10\mid |r-s|$, and this cannot happen since $|r-s| < \text{max}(r,s)< 10. $, and each $n \in \mathbb{Z}, n = a_i10^i + a_{i-1}\cdot 10^{i-1} + \cdots + a_1\cdot 10 + a_0 = 10d+a_0 \to n \in E_d$. Done.

Answer 2

A partition of a set is a collection of non-empty subsets such that every element of the set is a member of one and only one subset.

So show that any arbitrary integer $x$ can be a member of one and only one $E_n$. That is:

$$(\forall\, x\in \Bbb Z)(\exists !\, n \in \Bbb Z)( x\in E_n)$$

In other words, that $(\forall\, x\in \Bbb Z)(\exists !\, n \in \Bbb Z)(\exists!\, k\in \{0,1,...,9\}): (x=10n+k)$