(exist z) x=yz
What kind of relation is that?
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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
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Hmmm... It looks more to me that the problem is written
$\displaystyle \exists z \text{ such that }x = yz$
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 $\displaystyle x, y, z \in \mathbb{R}$. 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:
$\displaystyle x=yz$ and $\displaystyle z = 1$ and then it's obvious that $\displaystyle x=y$
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
I, too, believe this. Isnt it an equivalence relation?
$\displaystyle \text{xRy if}\, \exists z \text{ such that }x = yz$
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.