# Prove: R is an equivalence relation.

In $\mathbb{R^2}$ we consider the relation $(x,y)R(a,b)$ if and only if exists $n \in \mathbb{Z}$ such that $n-1<y\leq n \$ and $n-1<b\leq n \$. Prove: $R$ is an equivalence relation.

$R$ is an equivalence relation if it is reflexive, symmetric and transitive .

Reflexive: $\forall (x,y)\in \mathbb{R^2} \Rightarrow (x,y)R(x,y) \Rightarrow n-1<y\leq n \ \wedge n-1<y\leq n$

Symmetric: $\forall (x,y)\in \mathbb{R^2} \Rightarrow (x,y)R(a,b) \Rightarrow n-1<y\leq n \ \wedge n-1<b\leq n\Rightarrow n-1<b\leq n \ \wedge n-1<y\leq n\Rightarrow (a,b)R(x,y)$

Transitive $\forall (x,y)\in \mathbb{R^2}$ \begin{split} (x,y)R(a,b)& \| n-1<y\leq n \ \wedge n-1<b\leq n \\ (a,b)R(c,d)& \|n-1<b\leq n \ \wedge n-1<d\leq n \\ &\Rightarrow n-1<y\leq n \ \wedge n-1<d\leq n\\&\Rightarrow\ (x,y)R(c,d) \end{split}

Is correct my proof ?

• For reflexivity, what is $n$? For transitivity, you are assuming the $n$ given from $(a,b)R(c,d)$ is the same as the $n$ given by $(x,y)R(a,b)$. It's incorrect to make this assumption. Overall the readability of the proof can be improved by using more words and less symbols. – John Griffin Sep 19 '17 at 19:57
• $n$ is an integer number and constant? – B. David Sep 19 '17 at 20:08
• The issue is that in order to show $(x,y)R(x,y)$, you must show that the particular $n$ exists. This means you must prove its existence in some way or describe it explicitly. – John Griffin Sep 19 '17 at 20:11

To expand on my comment, here's how I would adapt your proof.

Fix $(x,y),(a,b),(c,d)\in\mathbb{R}^2$.

For reflexivity, note that $\cup_{n\in\mathbb{Z}}(n-1,n]=\mathbb{R}$ and hence there exists $n\in\mathbb{Z}$ such that $n-1 < y \le n$. (Alternatively you can define $n:=\lceil y \rceil$ if you are familiar with the ceiling function.) Thus $(x,y)R(x,y)$.

For symmetry, suppose $(x,y)R(a,b)$. Then there exists $n\in\mathbb{Z}$ satisfying $$n-1 < y \le n \quad \text{and} \quad n-1 < b \le n,$$ which by the symmetry of "and", is equivalent to $$n-1 < b \le n \quad \text{and} \quad n-1 < y \le n.$$ Consequently $(a,b)R(x,y)$.

For transitivity, suppose $(x,y)R(a,b)$ and $(a,b)R(c,d)$. This means there are $n,m\in\mathbb{Z}$ such that $$n-1 < y \le n \quad \text{and} \quad n-1 < b \le n$$ and $$m-1 < b \le m \quad \text{and} \quad m-1 < d \le m.$$ Since $(n-1,n]$ and $(m-1,m]$ are disjoint whenever $n\ne m$ and $b\in(n-1,n]\cap(m-1,m]$, we infer $n=m$. Hence $n-1=m-1 < d \le m = n$ so that we have $$n-1 < y \le n \quad \text{and} \quad n-1 < d \le n.$$ Therefore $(x,y)R(c,d)$.

This completes the proof that $R$ is an equivalence relation.

(first evaluation is in $b$, not in $y$)
Also now will need to prove same $n$ for $y,b$ and $d$...
Another way to verify this is to make sure that the seeming equivalence relation generates well-defined equivalence classes that partition the underlying set. This particular relation appears to partition the plane into horizontal stripes: $\{(x,y)|n-1<y\leq n\}$ for each integer $n$. Thus, it appears to be an equivalence relation.