0
$\begingroup$

Consider the equivalence relation defined on the set A = Z \ {0}, where a~b if an only if ab > 0.

I assume this means that A is the set of all integers except 0.

How many equivalence classes are there in the above equivalence relation? Describe each of the equivalence classes.

This is my first equivalence relation assignment, and I am not sure I understand equivalence relations and classes yet. But I assume that to "form" a class, I must select all pairs (a,b) that make ab = 1 (for equivalence class 1), all those that make ab = 2 (for equivalence class 2), etc.

However, would that not mean there are infinite equivalence classes since each integer number must have a class of its own?

$\endgroup$
0
$\begingroup$

You are reading the relation incorrectly. Write $a \sim b$, if $ab > 0$. E.g. $1 \sim 2$ since $1\cdot 2 > 0$, in fact $1 \sim a$, with any $a > 0$, since $1a > 0$. What about $a < 0$?

$\endgroup$
  • $\begingroup$ Thank you for your response. I'm a bit confused about what you mean. I am forming a class for each integer greater than zero that is formed by ab. For example, the class 1 would be {(1,1), (-1,-1)} (one pair of which accounts for a < 0). Doesn't this mean there are infinite classes that can be formed since there are infinite integers greater than 0? But the problem question suggests that there are countable classes, which has me confused. $\endgroup$ – Lorraine Oct 11 '16 at 13:43
  • $\begingroup$ The definition of an equivalence class is as follows: Let $X$ be a set and let $\sim$ be an equivalence relation on $X$. Then the equivalence class of $x \in X$, is $[x] = \left\{ y \in X \: \middle| \: x \sim y \right\}$. Thus, in this exercise, the equivalence class of $1$ is the set of all integers $a$ such that $1a > 0$, since $1 \sim a$ if $1a > 0$. $\endgroup$ – Matias Heikkilä Oct 11 '16 at 13:47
0
$\begingroup$

An equivalence relation on a set $S$ "divides" that set into disjoint subsets, called equivalence classes. Now, suppose $\sim$ is an equivalence relation and $a \in S$. The equivalence class of $a$ under $\sim$, denoted $[a]$ is defined by

$$[a] := \{ b \in S \,|\, a \sim b\} $$

So, in your example, given a non-zero integer $n$, its equivalence class is the set

$$[n]:= \{ m \in \mathbb{Z}\setminus\{0\} \, \mid \, nm > 0\}$$

By definition of an equivalence relation, $\sim$ is reflexive, which means that $a \sim a$ for every $a \in S$, so $a \in [a].$

Using this property for $2 \in \mathbb{Z}\setminus \{0\},$ we see that $ 2 \in [2]$, which is true since $2 \cdot 2 >0$.

$3$ is also a member of $[2]$, since $2 \cdot 3 > 0$, but $-2 \not\in [2]$, since $2 \cdot (-2) < 0,$ so we have at least two distinct equivalence classes.

Hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.