# Equivalence Relation and classes

On $\mathbb{R}^2$, define $(a_1, a_2)\sim(b_1, b_2)$ if $a_1^2 + a_2^2 = b_1^2 + b_2^2$. Check that this defines an equivalence relation. What are the equivalence classes?

My work:

Let $a_1=b_1 = 1$ and $a_2=b_2 = 2$. We have $(1,2)\sim(1,2)\implies 5 = 5$ hence it is reflexive.

For symmetry, let $(a_1,a_2)\sim(b_1,b_2)$ then we should have $(b_1,b_2)\sim(a_1,a_2)$ to be satisfied the condition of symmetry. Like I did for reflexive, it is also symmetry.

Also this holds transitivity: let $c_1=1,c_2=2$.

I feel like I'm not on the right track and I don't know how to find the equivalence classes....

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The equivalence classes correspond to circles centered at the origin of varying radii. This much is obvious. – oldrinb Sep 5 '13 at 1:24

Most of what's presented in the question amounts to proofs by example; it's a logically fallacy, since one can prove untrue statements using proofs by example.

To prove reflexivity, your task is to show that $(a,b) \sim (a,b)$ for all $(a,b) \in \mathbb{R}^2$. You need to show that $$a^2 + b^2 = c^2 + d^2$$ holds whenever $(a,b)=(c,d)$.

To prove symmetry, you need to show that if $(a,b) \sim (c,d)$ then $(c,d) \sim (a,b)$ for all $(a,b),(c,d) \in \mathbb{R}^2$.

Try completing this: If $(a,b) \sim (c,d)$ then [some equation] holds, this implies [some other equation] holds, which implies that $(c,d) \sim (a,b)$.

To prove transitivity, you need to show that if $(a,b) \sim (c,d)$ and $(c,d) \sim (e,f)$ then $(a,b) \sim (e,f)$ for all $(a,b),(c,d),(e,f) \in \mathbb{R}^2$.

Try completing this: If $(a,b) \sim (c,d)$ then [some equation] holds. Further, if $(c,d) \sim (e,f)$, then [some other equation] holds. Together, these imply [some other equation] holds, which implies that $(a,b) \sim (e,f)$.

Finding a succinct way of describing the equivalence classes is often not straightforward. In this case, oldrinb's comment is quite helpful here:

The equivalence classes correspond to circles centered at the origin of varying radii.

Formally, the equivalence class containing $(a,b) \in \mathbb{R}^2$ is the set $$\{(c,d) \in \mathbb{R}^2:(a,b) \sim (c,d)\}$$ but this doesn't give much insight into what's in this class, and to why it's interesting. We can rewrite it $$\{(c,d) \in \mathbb{R}^2:c^2+d^2=k\}$$ where $k=a^2+b^2$, which is where oldrinb's hint comes from. (Note also that $(0,0)$ needs to be treated separately.)

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So I don't need to use norm? – therexists Sep 5 '13 at 3:40
You need to recognize them as circles. You can do that in your preferred way. I'd do it by recognizing that $x^2+y^2=k$ is the equation of a circle. – Rebecca J. Stones Sep 5 '13 at 3:53
Your helps totally helped me out. Thanks Rebecca. : ) – therexists Sep 6 '13 at 1:50
My pleasure! Thank you. – Rebecca J. Stones Sep 6 '13 at 1:50

Careful! The relation will be reflexive if it is true that $(a,b)\sim (a,b)$ for any pair $(a,b)$ you choose. Proving this is true for just one doesn't do it. But you know equality is already reflexive, right?

Think about complex numbers. If $z=a+bi$, then $|z|^2=\cdots$? What geometrical figure is $C_r=\{z\in \Bbb C:|z|=r\}$?

If you don't know about complex numbers, pick any $(a,b)$; and set $r^2=a^2+b^2$. Define $\lVert (a',b')\rVert=\sqrt{a'^2+b'^2}$. We're then saying that $(a,b)\sim (a',b')$ iff $\lVert (a,b)\rVert^2 =\lVert (a',b')\rVert^2$. So, the equivalence classes are of the form $$\widehat{(a,b)}=\left\{(x,y):\lVert (x,y)\rVert^2 =x^2+y^2=r^2\right\}$$

What does $\sqrt{x^2+y^2}$ measure for a point in $\Bbb R^2$?

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it might be more 'enlightening' for those in lower-level math to instead consider $\{(x,y)\in\mathbb{R}^2:x^2+y^2=r^2\}$ ;-) complex numbers may confuse him... +1 though – oldrinb Sep 5 '13 at 1:25
I found a website mathcs.org/analysis/reals/logic/answers/projspc.html – therexists Sep 5 '13 at 1:28
This relates to my question, right? – therexists Sep 5 '13 at 1:28
@therexists I think you can answer that! =D – Pedro Tamaroff Sep 5 '13 at 1:28
@PeterTamaroff well of course I do not argue -- you are correct in both your points. My point was only that it may confuse the OP who may not immediately see the connection between the problems – oldrinb Sep 5 '13 at 1:29