0
$\begingroup$

Are these equivalence relations? If yes, give a corresponding partition. If no show a counterexample.

  1. On the set $Q\setminus \{0\}:a\sim b\iff a\cdot b>0$

  2. On the set $Z:a\approx b\iff b-1\leq a\leq b+1$

  1. Reflexivity:

let $a\in\mathbb{Q}\setminus\{0\}$.

then $a\cdot a>0$, since $a\not=0$ it holds.

Symmetry:

let $a,b\in\mathbb{Q}\setminus\{0\}$.

if $a\cdot b>0$, then $b\cdot a>0$.

since $\cdot$ is commutative, this holds.

Transitivity:

suppose $(a\sim b)\wedge(b\sim c)\wedge(a\not\sim c)$.

case 1: let $a<0$.

then $b<0$ and $c<0$, otherwise the inequalities don't hold.

but then $a\sim c$ holds.

case 2: let $a>0$.

then $b>0$ and $c>0$, otherwise the inequalities don't hold.

but then $a\sim c$ holds.

therefore the assumption is wrong and $\sim$ is transitive.

Therefore $\sim$ is an qquivalence relation on $\mathbb{Q}\setminus\{0\}$.

Let the partition $P=\{\{\mathbb{Q}\setminus\{0\}\} \}$

  1. Transitivity:

let $a=0$, $b=1$ und $c=2$.

$a\approx b$ and $b\approx c$, and $a\not\approx c$.

Therefore it's not an equivalence relation.

How to make a partition in this context? Partition is known to me, I just don't know if mine is that what they are asking for.

$\endgroup$
1
$\begingroup$

There is a shortcut for 1.

Whenever a relation $R$ on some set $X$ can be characterized by: $$xRy\iff f(x)=f(y)$$ where $f$ denotes some function that has $X$ as its domain then this relation is an equivalence relation.

Observe that e.g. the obviously true statement: $$\forall x\in X [f(x)=f(x)]$$ allready guarantees reflexivity of $R$.

Also for symmetry and transitivity there are statements like that.

The equivalence class represented by $x\in X$ takes the form: $$[x]:=\{y\in X\mid f(y)=f(x)\}$$

Then the partition induced by $R$ can be written as:$$\mathcal P=\{[x]\mid x\in X\}$$

If $S:=\{f(x)\mid x\in X\}$ then the partition can also be written as: $$\mathcal P=\{f^{-1}(\{r\})\mid r\in S\}$$

where $f^{-1}(\{r\}):=\{x\in X\mid f(x)=r\}$. Sets like $f^{-1}(\{r\})$ are the fibres of the function and the non-empty fibers of a function always form a partition of the domain of the function.


Now note that $a\sim b\iff f(a)=f(b)$ where $f$ is the function prescribed by $a\mapsto \frac{a}{|a|}$, i.e. it sends $a$ to its sign.

Also note that here $S=\{-1,1\}$.

So the partition turns out to be $\{P,N\}$ where $P$ denotes the subset of positive and $N$ denotes the subset of negative elements of $\mathbb Q\setminus\{0\}$.


In the context of proving that a relation is an equivalence it is very useful to start with looking for such function. Actually if the relation is an equivalence then automatically such a function exists (but can be well covered): the function $x\mapsto[x]$.

$\endgroup$
  • $\begingroup$ are other partitions besides the one you described possible? $\endgroup$ – SAJW Nov 7 '16 at 12:55
  • $\begingroup$ There is a one-to-one relation between equivalences and partitions. If you have an equivalence relation then it is connected with exactly one partition (and vice versa). Of course there are more partitions, but they are not interesting because they do no correspond with the equivalence relation. $\endgroup$ – drhab Nov 7 '16 at 12:59
  • $\begingroup$ so this particular partition means: you either take both elements, a and b, out of P or out of N, but not one out of P and one out of N. (otherwise a*b<0). Are there good excercises to get a feel for that? $\endgroup$ – SAJW Nov 7 '16 at 13:13
  • $\begingroup$ This partition means that $a\sim b$ is true if and only if $a,b\in P$ or $a,b\in N$. $\endgroup$ – drhab Nov 7 '16 at 13:27

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.