Prove: In $\mathbb{Z}$, m ~ n iff $|m|=|n|$

Question

Prof. Pinter's "A Book of Abstract Algebra" presents this exercise:

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

In $\mathbb{Z}$, m ~ n iff $|m|=|n|$.

Here's my attempt at the proof.

Given:

...---(-2)--(-1)---0---1---2---...

$A_{-2}=2\\ A_{-1}=1\\ A_{0}=0\\ A_{1}=1\\ A_{2}=2$

Observe that $A_{-2} \sim A_2$ and $A_{-1} \sim A_1$.

For $n \in \mathbb{Z}$, excluding $0$, $A_{-n} \sim A_n$.

$A_2 \cup A_1 \cup ...=A$ $A_2 \cap A_1 ...=\emptyset$

Is this a valid proof?

Answer 1

Certainly $|m|=|m|$. If $|m|=|n|$ then $|n|=|m|$ by the reflexive property of $=$. Finally, if $|m|=|n|$ and $|n|=|p|$ then we have $|m|=|p|$ because $=$ is transitive.

Since $|-n|=|n|$ we have $-n\sim n$. If $m\neq n$ and $m\neq -n$, then $|m|\neq |n|$ so $m\not\sim n$ (this last observation is not a proof, it is only a claim). This is in line with what you were trying to do.

This is sort of a "technical" thing to prove and you could lose sight of what are you trying to show when you go through the proof. Equivalence relations are marvelous things (one use being the decomposition of an arbitray function on a set).

Answer 2

Let $X$ be a set and $\sim$ a relation amongst the elements of $X$. Then $\sim$ is an equivalence relation if $\sim$ is symmetric, reflexive and transitive.

Symmetric: $ x \sim y \Rightarrow y \sim x$

Reflexive: $\ x \sim x$

Transitive: $\ x \sim y, y \sim z \Rightarrow x \sim z$.

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In your example we take $\ \mathbb{Z}/\sim$ where $x \sim y \iff |x| = |y|$.

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Reflexive: $x \sim x$ since $|x| = |x|$

Symmetric: If $x \sim y$ then $|x|= |y|\ $ so ??

Transitive: ??

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$\textbf{Two Realizations}$

Some equivalence relations have geometric significance and in this case I believe what this is geometrically is an inifinte collection of tangent unit balls.

This relation does partitions $\mathbb{Z}$ into intervals of the form $\{-n,n\}$.

$$ \mathbb{Z} = \bigsqcup_{n} \{-n,n\}$$

Answer 3

Try to describe for any $n \in \mathbb{Z}$ its class of equivalence and this describe the partition of $Z$ in this case is $n, -n$ so the particion is given by this sets.