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While I understand equivalence classes, I can't seem to grasp this problem. This is what I am working with:

Let’s define a relation ∼ on $\Bbb R^2$ by $(x, y) ∼ (p, q)$ if and only if $(x, y) = (λp, λq)$ for some positive real number $λ$ Now define an addition operation on the equivalence classes of this relation: $$[(x, y)] + [(p, q)] = [(xp − yq, xq + yp)].$$ Notice that our definition involved choosing specific representative elements $(x, y)$ and $(p, q)$ for the two input equivalence classes. Show that this addition operation is “well-defined.”

I know I need to show that whatever elements I pick, I will get the same defined output (the definition of well-defined), however whatever test variables I use, I can't seem to get that to work. For example, if I choose $x =8, y= 6, p= 4, q =3$, both sides of the equation are different after their respective equivalence classes. If I can get that to work, I might be able to come up with something that shows that the same output will always be achieved.

Any sort of direction with this problem would be awesome, thank you in advance!

Ivo Terek
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user225265
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1 Answers1

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To show that the operation is well defined, you have to check that: $$\begin{cases} (x,y)∼(x',y') \\ \text{and }(p,q)∼(p',q')\end{cases} \implies (xp-yq,xq+yp) ∼ (x'p'-y'q',x'q'+y'p').$$

This ensures that, in the language of classes, if $[(x,y)] = [(x',y')]$ and $[(p,q)] = [(p',q')]$, then $$[(x,y)]+[(p,q)] = [(x',y')]+[(p',q')],$$ that is, the operation $+$ makes sense.

I think you misunderstood the definition of $+$ in the exercise. The operation $[\cdot]$ does not seems to be linear with respect to $+$, that is, $$[(x,y)+(p,q)]\neq [(x,y)]+[(p,q)],$$ in general, and that is what you have perceived with your example $x = 8$, etc. I think that with these ideas you can go on. If not, give me a yell in the comments and we'll go into the gory details of proving that $+$ is well-defined.

Ivo Terek
  • 77,665