Order Relation

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• Sep 18th 2012, 03:46 PM
keynes
Order Relation
Q. Define an order relation

(x,y)<(j,k) if and only if x+k<y+j

I do not any idea where to start?
• Sep 18th 2012, 03:59 PM
Plato
Re: Order Relation
Quote:

Originally Posted by keynes
Q. Define an order relation
(x,y)<(j,k) if and only if x+k<y+j I do not any idea where to start?

BUT what is the question?
• Sep 18th 2012, 04:25 PM
keynes
Re: Order Relation
The whole question is:

We defined Z to be the set of equivalence classes of tuples (x,y) with a,b is in Z with respect to the equivalence relation
(x,y)~(x',y') if and only if x+y'=x'+y
Define an order relation by
(x,y)<(j,k) if and only if x+k<y+j
Show that if (x,y)~(x',y') and (x,y)<(j,k) then (x',y')<(j,k). {{I can solve this part if the above part is defined}}
Dont we have to show that the order relation is defined. Like, when asked to define a given equivalence relation, we showed that the relation is transitive, reflexive and symmetric?
:-)
• Sep 18th 2012, 04:44 PM
Plato
Re: Order Relation
Quote:

Originally Posted by keynes
The whole question is:
We defined Z to be the set of equivalence classes of tuples (x,y) with a,b is in Z with respect to the equivalence relation
(x,y)~(x',y') if and only if x+y'=x'+y
Define an order relation by
(x,y)<(j,k) if and only if x+k<y+j
Show that if (x,y)~(x',y') and (x,y)<(j,k) then (x',y')<(j,k). {{I can solve this part if the above part is defined}}
Dont we have to show that the order relation is defined. Like, when asked to define a given equivalence relation, we showed that the relation is transitive, reflexive and symmetric?

Sorry to say that still makes no sense for me. It may be a translation problem.
What is the 'underlying' set?
• Sep 18th 2012, 04:49 PM
keynes
Re: Order Relation
There is no underlying set except Z, the set of integers. if thats what you mean.
we defined the set of integers to be the set of equivalence classes of tuples... means that's what we proved in class.
and order relation means greater than or smaller than order, not ordered pair type order.
• Sep 19th 2012, 06:52 AM
emakarov
Re: Order Relation
When one says, "Define an order relation," it may mean "Define a relation about which we will later prove that it is an order." The phrase "Define an order relation by (x,y)<(j,k) if and only if x+k<y+j" defines a relation. First, one needs to show that < is well-defined. This means that the result of comparison does not depend on the representative of the equivalence class we use. This is exactly what the problem asks to show. Later presumably one is supposed to prove that < is an order.