Is R an Equivalence Relation in R x R Based on a Quadratic Equation?

[nikie1o2]

In R x R , ley (x,y) R (u,v) if ax^2 +by^2=au^2 + bv^2, where a,b >0. Determine the relation R is an equivalnce relation. Prove or give a counter example

 

[tiny-tim]

Welcome to PF!

Hi nikie1o2! Welcome to PF!

(try using the X² tag just above the Reply box )

Tell us how far you've got, and where you're stuck, and then we'll know how to help!

 

[nikie1o2]

Hi nikie1o2! Welcome to PF!

(try using the X² tag just above the Reply box )

Tell us how far you've got, and where you're stuck, and then we'll know how to help!

Hello, thank you for the warm welcome.

I have done equivalence relations before but with just two variables not 4. So i was confused on how to prove the reflexive, symmetric & transitive properties.

For reflexive i was thinking if (x,y)R(x,y) then ax^2+by^2=ax^2+by^2- so that is true
Symmetry: (x,y)R(u,v) then (u,v)R(x,y) is true

For Transitive i knoe if(x,y)R(u,v) and (u,v)R(a,b), then (x,y)R(a,b). I am just confused on how to show the equations for that and that it's true...

 

[Redbelly98]

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[tiny-tim]

Hello nikie1o2!

(Please use the X² tag just above the Reply box )

For Transitive i knoe if(x,y)R(u,v) and (u,v)R(a,b), then (x,y)R(a,b). I am just confused on how to show the equations for that and that it's true...

Just write out the definitions of (x,y)R(u,v) and (u,v)R(a,b) … then it should be obvious!

(btw, the equivalence classes are a well-known geometrical shape … can you se which?)

 

[Redbelly98]

Just a comment: using "a" and "b" is a bad choice of variable names in this problem. May I suggest using "s" and "t" instead? I.e., use

(u,v)R(s,t)​

instead of

(u,v)R(a,b)​

when working out the transitive property.