Equivalence relation and partitions.

Question

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 $P=\{\{\mathbb{Q}\setminus\{0\}\} \}$

2. 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.

Answer 1

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.

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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 $\mathbb Q\setminus\{0\}$.

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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 $x\mapsto[x]$.