Prove $A_r$ is a Partition

Question

Dr. Pinter's "A Book of Abstract Algebra" presents the following exercise:

Prove that each of the following is a partition of the indicated set. Then describe
the equivalence relation associated with that partition.

For $r \in \lbrace 0,1,2,...,9\rbrace$, let $A_r$ be the set of all the integers whose units digit (in decimal notation) is equal to $r$.

Prove: $\lbrace, A_0, A_1, A_2, ...,A_9\rbrace$ is a partition of $\mathbb{Z}$

I believe that the following diagram represents $A_r$:

Where $A_0 = [0], A_1 = [1], ...$.

As a result, it's a partition since:

$A_0 \cap A_1 \cap \text{...} \cap A_9=\emptyset$

$A_0 \cup A_1 \cup \text{...} \cup A_9=A$

Does this proof show a partition in $A_r$?

Also, please give me a hint on how to answer the equivalence relation.

Answer 1

I suppose that you know that, given an equivalence relation on a set $S$ , the set of its equivalence classes is a partition of $S$.

In your case the equivalence relation, defined for $m,n $ integers,is:

$n \sim m $ iff $m$ and $n$ have the same units digit (in decimal notation).

This is the relation that define the sets of the partition, that are the equivalence classes.

Showing that this relation is reflexive, symmetric and transitive ( it's easy) you show that it is an equivalence relation and that its equivalence classes are a partition.

Answer 2

Your general idea is definitely correct - you will, however, need to show or at least state that the sets $A_r$ are pairwise disjoint (consider for example, the sets $\{1,2\}$, $\{1,3\}$ and $\{2,3\}$ - they satisfy $\{1,2\} \cap \{1,3\} \cap \{2,3\} = \emptyset$ and $\{1,2\} \cup \{1,3\} \cup \{2,3\} = \{1,2,3\}$ - however these sets are not a partition of $\{1,2,3\}$ due to not being pairwise disjoint).

As for your second question, partitions and equivalence relations are intimately connected - to help make that connection clearer, you might want to think about the partitions induced by different equivalence classes on the integers and what they look like to work backwards.

Answer 3

You can start by showing it on $\mathbb{N} \ni 0$. Clearly every integer has a decimal representation, so the next step would be to ensure pairwise disjointness, i.e. that if $(a_{m})_{m = 1}^{M}, (b_{n})_{n = 1}^{N}$ are both terminating sequences of the digits $\{ 0, 1, \ldots, 9 \}$, then if $x =\sum_{m = 1}^{M} a_{m} 10^{m} = \sum_{n = 1}^{N} b_{n} 10^{n}$, then $a_{1} = b_{1}, \ldots, a_{M} = b_{N}$. To do this, we see that \begin{align*} \sum_{m = 1}^{M} a_{m} 10^{m} & = a_{1} + 10 \sum_{m = 1}^{M - 1} a_{m + 1} 10^{m} , \\ = \sum_{n = 1}^{N} b_{n} 10^{n} & = b_{1} + 10 \sum_{n = 1}^{N - 1} a_{n + 1} 10^{n} . \end{align*} First we show that $a_{1} = b_{1}$ by assuming otherwise, i.e. suppose for contradiction that $a_{1} \neq b_{1}$. We have that $x \in a_{1} + 10 \mathbb{Z}, x \in b_{1} + 10 \mathbb{Z}$. Thus $a_{1} - b_{1} \in 10 \mathbb{Z}$, but if $a_{1} - b_{1} \neq 0$, then $|a_{1} - b_{1}| \geq 10$, which just couldn't have happened if $0 \leq a_{1}, b_{1} \leq 9$. You can continue this process inductively to show that $a_{2} = b_{2}, \ldots, a_{3} = b_{3}, \ldots, a_{M} = b_{N}$.

As for explicitly defining the equivalence relations, the above expressions are probably rather telling.