- #1
hitmeoff
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Homework Statement
Let S := ([tex]\Re[/tex] x [tex]\Re[/tex] \ {(0,0)}. For (x,y), (x',y') [tex]\in[/tex] S, let us say (x,y) ~ (x',y') if there exists a real number [tex]\lambda[/tex] > 0 such that (x,y) = ([tex]\lambda[/tex]x',[tex]\lambda[/tex]y'). Show that ~ is an equivalence relation; moreover, show that each equivalence class contains a unique representative that lies on the unit circle (i.e., the set of points (x,y) such that x2 + y2 = 1).
Homework Equations
I know in order to show that something is an equivalence relation if the following 3 properties hold
reflexive: a ~ a for all a [tex]\in[/tex] S
symmetric: a ~ b implies b ~ a for all a,b [tex]\in[/tex] S
transitive: a ~ b and b ~ c implies a ~ c for all a, b, c [tex]\in[/tex] S
and for an equivalence relation ~ the equivalence class as the set {x [tex]\in[/tex] : x ~ a}
The Attempt at a Solution
What I don't get is, if there is no operation defined how do we show equivalence? Or is the operation scalar multiplication?
Im not sure where to go from here, especially showing that the solution to the unit circle is in the equivalent class.