# Relational proof

• Mar 12th 2009, 08:27 PM
thehollow89
Relational proof
If R is reflexive and transitive, then R U R^U is an equivalence. I need to prove this using algebraic-relation calculations or disprove by counter example. I'm really not understanding this stuff. (Worried)
• Mar 13th 2009, 03:42 AM
Plato
Quote:

Originally Posted by thehollow89
If R is reflexive and transitive, then R U R^U is an equivalence. I need to prove this using algebraic-relation calculations or disprove by counter example. I'm really not understanding this stuff. (Worried)

What is the meaning of $R^U$?
• Mar 13th 2009, 07:50 AM
thehollow89
R converse I believe.
• Mar 14th 2009, 12:25 AM
Equivalence Relation
Hello thehollow89

$R^U$ is the converse, or the inverse, of $R$, means that if the element $(x, y)$ in R, the element $(y,x)$ is in $R^U$ - because that's what the inverse relation is: the relation you get by switching the order of the elements in the relation. (It's usually denoted by $R^{-1}$, as you might expect).

So, if $R$ is reflexive and transitive, all you'll need to prove in order for $R\cup R^U$ to be an equivalence relation is:

• $R^U$ is also reflexive and transitive, and therefore $R \cup R^U$ is reflexive and transitive
• $R \cup R^U$ is symmetric; in other words, $(x, y) \in R \cup R^U \Rightarrow (y,x) \in R \cup R^U$

Can you do that, or do you need further help?

• Mar 14th 2009, 12:43 PM
thehollow89
Quote:

Hello thehollow89

$R^U$ is the converse, or the inverse, of $R$, means that if the element $(x, y)$ in R, the element $(y,x)$ is in $R^U$ - because that's what the inverse relation is: the relation you get by switching the order of the elements in the relation. (It's usually denoted by $R^{-1}$, as you might expect).

So, if $R$ is reflexive and transitive, all you'll need to prove in order for $R\cup R^U$ to be an equivalence relation is:

• $R^U$ is also reflexive and transitive, and therefore $R \cup R^U$ is reflexive and transitive
• $R \cup R^U$ is symmetric; in other words, $(x, y) \in R \cup R^U \Rightarrow (y,x) \in R \cup R^U$

Can you do that, or do you need further help?

I mean I understand the question and what I need to prove, I'm just not sure how I go about proving it. The prof isn't really all that clear. He'll just teach us about the kinds of relations and then ask us to prove something and most kids end up stumped on what method of proof writing he wants...
• Mar 14th 2009, 04:32 PM
Equivalence Relation
Quote:

So, if $R$ is reflexive and transitive, all you'll need to prove in order for $R\cup R^U$ to be an equivalence relation is:

• $R^U$ is also reflexive and transitive, and therefore $R \cup R^U$ is reflexive and transitive
• $R \cup R^U$ is symmetric; in other words, $(x, y) \in R \cup R^U \Rightarrow (y,x) \in R \cup R^U$

First we prove that $R^U$ is reflexive.

Assume that
$R$ and $R^U$ are defined on the set $S$.

Then
$R^U$ is reflexive $\iff (x, x) \in R^U, \forall x \in S$.

Now
$(x, y) \in R^U \Rightarrow (y, x) \in R$. Therefore $(x, x) \in R^U, \forall x\in S$, since $(x,x) \in R$, because $R$ is reflexive.

$\Rightarrow R^U$ is reflexive.

Next, we prove that
$R^U$ is transitive.

$R^U$ is transitive $\iff ((x,y) \in R^U \wedge (y,z) \in R^U) \Rightarrow (x,z) \in R^U$

Now
$(x,y) \in R^U \Rightarrow (y,x) \in R$, and $(y,z) \in R^U \Rightarrow (z,y) \in R$

$\Rightarrow (z, x) \in R$, since $R$ is transitive

$\Rightarrow (x, z) \in R^U$

$\Rightarrow R^U$ is transitive

Finally, we prove that
$R \cup R^U$ is symmetric.

$(x,y) \in R \cup R^U \Rightarrow (x,y) \in R$ or $(x,y) \in R^U$

Now if
$(x,y) \in R$, then $(y,x) \in R^U$, and if $(x,y) \in R^U$, then $(y,x) \in R$.

Either way,
$(x,y) \in R\cup R^U \Rightarrow (y,x) \in R\cup R^U$

So
$R\cup R^U$ is symmetric.

Finally, then,
$R$ and $R^U$ are both reflexive and transitive, and $R\cup R^U$ is symmetric. So $R\cup R^U$ is an equivalence relation.

• Mar 14th 2009, 05:59 PM
thehollow89
Quote:

First we prove that $R^U$ is reflexive.

Assume that
$R$ and $R^U$ are defined on the set $S$.

Then
$R^U$ is reflexive $\iff (x, x) \in R^U, \forall x \in S$.

Now
$(x, y) \in R^U \Rightarrow (y, x) \in R$. Therefore $(x, x) \in R^U, \forall x\in S$, since $(x,x) \in R$, because $R$ is reflexive.

$\Rightarrow R^U$ is reflexive.

Next, we prove that
$R^U$ is transitive.

$R^U$ is transitive $\iff ((x,y) \in R^U \wedge (y,z) \in R^U) \Rightarrow (x,z) \in R^U$

Now
$(x,y) \in R^U \Rightarrow (y,x) \in R$, and $(y,z) \in R^U \Rightarrow (z,y) \in R$

$\Rightarrow (z, x) \in R$, since $R$ is transitive

$\Rightarrow (x, z) \in R^U$

$\Rightarrow R^U$ is transitive

Finally, we prove that
$R \cup R^U$ is symmetric.

$(x,y) \in R \cup R^U \Rightarrow (x,y) \in R$ or $(x,y) \in R^U$

Now if
$(x,y) \in R$, then $(y,x) \in R^U$, and if $(x,y) \in R^U$, then $(y,x) \in R$.

Either way,
$(x,y) \in R\cup R^U \Rightarrow (y,x) \in R\cup R^U$

So
$R\cup R^U$ is symmetric.

Finally, then,
$R$ and $R^U$ are both reflexive and transitive, and $R\cup R^U$ is symmetric. So $R\cup R^U$ is an equivalence relation.