Equivalence Relation: Is (x,y) \propto (a,b) ∝ x^2 + y^2 = a^2 + b^2?

[gtfitzpatrick]

Homework Statement

is the relation \propto deifined on RxR defined by (x,y) \propto (a,b) \Leftrightarrow x^{2} + y^{2} = a^{2} + b^{2} an equivalence relation?

The Attempt at a Solution

I know i have to show that they hold true for the 3 properties
1 reflexive
2 symetric
3 transiative

but I am unsure to go about this?

 

[LeonhardEuler]

Can you test whether each of the 3 properties hold? For instance, the reflexive property means that (x,y)\propto (x,y) for all x and y. Is that true?

 

[gtfitzpatrick]

it think it is true because x=x and y=y and x^2=x^2 and y^2=y^2

 

[LeonhardEuler]

Exactly, continue this process for the other two properties, and you will have an answer.

 

[gtfitzpatrick]

it it holds true for reflexive because if x\equiv a and y\equiv c so x+y \equiv a+b and hence x^{2} + y^{2} \equiv a^{2} + b^{2} ?is this right?

 

[LeonhardEuler]

Yes, except I assume you mean y=b, instead of y=c.

 

[gtfitzpatrick]

oh yes sorry, thanks a million

 

[gtfitzpatrick]

same again for symmetric

if x\equiva and y\equivb then a\equivx and b\equivy

and so it follow that a^{2} + b^{2} \equiv x^{2} + y^{2} and follows that it is symetric

 

[gtfitzpatrick]

but for transitive i need to show if a\equivb and b\equivc then a\equivc but in my relation which values do i use?

 

[LeonhardEuler]

same again for symmetric

if x\equiva and y\equivb then a\equivx and b\equivy

and so it follow that a^{2} + b^{2} \equiv x^{2} + y^{2} and follows that it is symetric

This is not quite right. The symmetric property requires that if (x,y)\propto (a,b) then (a,b)\propto (x,y). The fact that (x,y)\propto (a,b) does not imply that both x=a and y=b, merely that x^2 + y^2 = a^2 + b^2. But it is still extremely easy to show that the symmetric property holds anyway.

For the transitive property, there are three pairs of numbers involved. You can call them (x,y), (a,b) and (u,v) and you will find that the test is straightfoward.