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?

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- September 18th 2012, 02:46 PMkeynesOrder 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? - September 18th 2012, 02:59 PMPlatoRe: Order Relation
- September 18th 2012, 03:25 PMkeynesRe: 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?

:-) - September 18th 2012, 03:44 PMPlatoRe: Order Relation
- September 18th 2012, 03:49 PMkeynesRe: 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. - September 19th 2012, 05:52 AMemakarovRe: 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.