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Prove the relation define on $\mathbb{R} \!\,^2$ by $$(x_1,y_1) \sim(x_2,y_2) \Leftrightarrow x_1^2+y_1^2=x_2^2+y_2^2$$ is an equivalence relation

Ok, so I know what an equivalence relation is. It must be: reflexive symmetric transitive

But I don't get how to prove the relation defined on $\mathbb{R} \!\,^2$. I think I am just over thinking this. Any tips on how to get started?

Reflexive: $x^2+y^2=x^2+y^2$

Symmetric: $x^2+y^2=y^2+x^2$

Transitive: $x^2+y^2=y^2+z^2=x^2+z^2$

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Tol prove it's reflexive, let $(x, y)$ be any point of $R^2$,. You want to show that $(x, y) \sim (x, y)$. Write that out: it says that

$$x^2 + y^2 = x^2 + y^2$$

which is evidently true. So you've proved reflexivity. Now...you give symmetry a try,...

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  • $\begingroup$ I made an attempt! Let me know how I did $\endgroup$ – Math Major Jan 29 '15 at 19:36
  • $\begingroup$ Not good. You need to suppose that $(x_1, y_1) \sim (x_2, y_2)$, and write down what that means; then you need to show that this implies that $(x_2, y_2) \sim (x_1, y_1)$. $\endgroup$ – John Hughes Jan 29 '15 at 19:47
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Hint: Start with this

$1-$ $(x,y) \sim (x,y)$

$2 -$ $(x,y) \sim (z,w) \Rightarrow (z,w) \sim (x,y)$

$3 - $ $(x,y) \sim (x',y')$ and $(x',y') \sim (x'',y'')$ then $(x,y) \sim (x'',y'')$

Edit:

$2 -$ $x^2 +y^2 = z^2 + w^2 \Rightarrow z^2 + w^2 = x^2 + y^2$ by the symmetry of $"="$

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    $\begingroup$ Feel free to ask any questions. $\endgroup$ – Aaron Maroja Jan 29 '15 at 19:39
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An alternate approach is to use the fact an equivalence relation is the same thing as a partition. The definition here is that two points are in the same equivalence class if and only if they have the same distance to the origin. So the relation divides the plane into all the circles centred at the origin and we have obviously have a partition.

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