Do equivalence classes partition all of the elements of the set that are in the relation or does it partition the entire set?

Question

My discrete math textbook says that "The equivalence classes associated with an equivalence relation on a set A form a partition of A." However, I am not sure how this would hold for something like this: A relation R on the set of all integers not including 0 by aRb when ab > 0.

This relation does not hold for a lot of integers like a = -1 and b = 5 (which would make ab = -5), so I don't see how the equivalence classes for this relation would form a partition of Z.

On a side note, I am also a bit confused about the fact that we are given two constants, a and b, but the relation is defined on Z / {0} (all integers except 0) instead of Z^2 / {0,0}. I think this might be because we are considering if ab > 0, but I'm not sure.

Answer 1

The equivalence classes do form a partition of the set $\Bbb{Z}\setminus\{0\}$. Pick any integer $k\in \Bbb{Z}\setminus \{0\}$. If $k$ is positive, then $k$ is in the equivalence class of $1$ and if $k$ is negative, then $k$ is in the equivalence class of $-1$. The intersection of these equivalence classes is $\emptyset$ as a negative integer can't be in the equivalence class of $1$ and a positive integer can't be in the equivalence class of $-1$.

Let $A_1$ denote the equivalence class of $1$ and let $A_2$ denote the equivalence class of $-1$. Then $$A_1\cap A_2=\emptyset \space\text{and } A_1\cup A_2=\Bbb{Z}\setminus \{0\}$$

Thus the equivalence classes form a partition of $\Bbb{Z}\setminus \{0\}$.

Edit: I should also add that this relation only has two equivalence classes. One equivalence class of a positive integer and one equivalence class of a negative integer. If we were to replace $-1$ by another negative integer $-k$, then the equivalence class of $-k$ would be the same as the equivalence class of $-1$. The same holds for positive integers. The equivalence class of a positive integer $k$ is the same as the equivalence class of $1$.

To prove this, it suffices to show that ever integer $k\in \Bbb{Z}^+$ is in the equivalence class of $1$ and every integer $k\in \Bbb{Z}^-$ is in the equivalence class of $-1$. And it just follows from the definition of the relation that the intersection of these two classes is $\emptyset$.

Answer 2

To answer the question asked in the title, remember that the properties of an equivalence relation are:

1. Reflexive ($xRx$)
2. Symmetric ($xRy \iff yRx$)
3. Transitive ($xRy, yRz \implies xRz$)

The reflexivity is important for the partitioning because it means that every element of the set does relate to something (itself), so you will never have $x \in S$ such that $xRy$ doesn't exist.

That's why you have to exclude zero from the relation you've given - since there is no $x \in \mathbb{Z}$ such that $0x > 0$, you can't relate 0 to anything, not even itself, so it can't belong to an equivalence class.