1. Equivalence Relation

in RxR, let (x,y)R(u,v) if a(x^2)+b(y^2)=a(u^2)+b(v^2) , where a,b >0. Determine if the relation R is an equivalence relation.

I know it has to satify the Reflexive, Symmetry, & Transitive properties but i only did relations with xRy... any hints please

2. Are you able to write down what those three properties would translate to for this specific relation?

3. Originally Posted by nikie1o2
in RxR, let (x,y)R(u,v) if a(x^2)+b(y^2)=a(u^2)+b(v^2) , where a,b >0. Determine if the relation R is an equivalence relation.

I know it has to satify the Reflexive, Symmetry, & Transitive properties but i only did relations with xRy... any hints please
This is the exact same thing. $(x,y)\sim (x,y)$ because $1x^2+1y^2=1x^2+1y^2$...etc.

4. Drexel28 is doing the case a= b= 1. I read the problem as saying that a and b can be any positive numbers.

To have reflexivity, we must have (x, y)R(x, y) for any numbers x and y: we must have $ax^2+ by^2= ax^2+ by^2$. Well, that's certainly true so this is reflexive. Notice that this is true be the two sides of that equation are true. This relation is reflexive because "=" is reflexive.

To have symmetry, we must have that if (x, y)R(u, v) then (u, v)R(x, y).
The first says that $ax^2+ by^2= au^2+ by^2$. Because "=" is symmetric, that is the same as $au^2+ bv^2= ax^2+ by^2$.

To have transitivity, we must have that if (x, y)R(u, v) and (u, v)R(p, q), then (x, y)R(p, q). I'll leave it to you to write down the equations and use the fact that "=" is transitive to show this is true.

5. Thank you all very much..i was just confused on if i had to show (x,y)R(x,y) for reflexive and if i had to show (u,v)R(u,v) and so on... Have a greay day!