# Thread: What kind of relation is that?

1. ## What kind of relation is that?

(exist z) x=yz

What kind of relation is that?

***************
I did it this way in my exam:

For z = 1 the equation is correct, because x=y*1 is x=y

The relation is symmetric
***************

2. Originally Posted by aRTx
(exist z) x=yz

What kind of relation is that?

***************
I did it this way in my exam:

For z = 1 the equation is correct, because x=y*1 is x=y

The relation is symmetric
***************
We appear to be missing some information. For example, how did you know that x = y?

-Dan

3. Originally Posted by topsquark
We appear to be missing some information. For example, how did you know that x = y?

-Dan
If he says that $x=yz$ and $z =1$ than $x=y*1$ --> $x=y$

though that's how I interpreted it...

4. Originally Posted by shinhidora
If he says that $x=yz$ and $z =1$ than $x=y*1$ --> $x=y$

though that's how I interpreted it...
Hmmm... It looks more to me that the problem is written
$\exists z \text{ such that }x = yz$
So the relation would be between sets of x and y. However the problem I see is that any two x and y are equivalent if we are talking $x, y, z \in \mathbb{R}$. Possibly this would be some sort of "maximal" relation? (My ignorance here is probably going to make a fool out of me, so I'm likely done trying to help out on this one.)

-Dan

5. Originally Posted by topsquark
Hmmm... It looks more to me that the problem is written
$\exists z \text{ such that }x = yz$
So the relation would be between sets of x and y. However the problem I see is that any two x and y are equivalent if we are talking $x, y, z \in \mathbb{R}$. Possibly this would be some sort of "maximal" relation? (My ignorance here is probably going to make a fool out of me, so I'm likely done trying to help out on this one.)

-Dan
I thought he meant:

He knows for sure that:

$x=yz$ and $z = 1$ and then it's obvious that $x=y$

Edit: I would believe what you, Topsquark and Isomorphism say, If I'd understand it I just don't, so I stick with the stuff I understand xD

6. Originally Posted by topsquark
Hmmm... It looks more to me that the problem is written
$\exists z \text{ such that }x = yz$
So the relation would be between sets of x and y. However the problem I see is that any two x and y are equivalent if we are talking $x, y, z \in \mathbb{R}$.
-Dan
I, too, believe this. Isnt it an equivalence relation?

$\text{xRy if}\, \exists z \text{ such that }x = yz$

A) Reflexive: xRx by choosing z=1.

B) Symmetric: xRy implies there exists z such that x = yz. But x=yz => y = x(1/z) . Thus yRx.

A) Transitive: xRy , yRw => there exists a and b such that x =ay and y = bw. But this means x = ay = a(bw) = (ab)w. Thus xRw.

7. We really do need to know the domain of this relation.
If the domain contains 0 then it is not symmetric.
For example: $\left( {0,2} \right) \in R \quad \mbox{but} \quad \left( {2,0} \right) \notin R$.