So this may be a really simple obvious question, but this is something that kind of trips me up. I'm a beginner at this type of stuff, and still learning.

In my past experience if I see the relation defined, it makes complete sense on how it works, but if a question asks me to find a defined relation I get confused.

Now the question starts with Let R be the relation defined on the set of integer pairs by (x1, y1)R(x2, y2) whenever

$x_{1}^{2}$ + $y_{1}^{2}$ = $x_{2}^{2}$ + $y_{2}^{2}$

What's really throwing me off is the integer pairs.

Now besides this example, a general explanation of defining relations would be nice. Thank you!


closed as unclear what you're asking by Morgan Rodgers, Dbchatto67, Alexander Gruber Apr 29 at 1:48

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    $\begingroup$ I don't really understand what you are asking. What type of general explanation are you looking for? $\endgroup$ – Derek Luna Apr 21 at 5:31
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    $\begingroup$ This relation is an equivalence relation. $\endgroup$ – Dbchatto67 Apr 21 at 5:34
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    $\begingroup$ What do you mean by "finding" the relation? You are given the relation. This formula tells you how to decide if two integer pairs are related under $R$. $\endgroup$ – Morgan Rodgers Apr 21 at 5:35
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    $\begingroup$ It is a relation on $\Bbb Z \times \Bbb Z.$ So in this case $$R \subseteq (\Bbb Z \times \Bbb Z) \times (\Bbb Z \times \Bbb Z).$$ $\endgroup$ – Dbchatto67 Apr 21 at 5:37
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    $\begingroup$ Obviously it is reflexive because $x^2+y^2 = x^2 + y^2.$ So $(x,y)\ R\ (x,y).$ It is symmetric because $x^2 + y^2 = y^2 + x^2.$ Try to see that $R$ is also transitive. Try yourself! I give you $5$ minutes. $\endgroup$ – Dbchatto67 Apr 21 at 5:39

Two elements, in this case integer pairs, are related when the sum of the squares of the elements of each pair are equal. To check properties of the relation consider the elements you are already given: $(x_1,y_1)$ and $(x_2,y_2)$. To check if R is reflexive we check if $(x_1,y_1)R(x_1,y_1)$ which is clearly true.

  • $\begingroup$ Please don't help him further @Derek Luna. I earnestly request you. Spoon-feeding something is strongly condemned in this site. $\endgroup$ – Dbchatto67 Apr 21 at 5:45
  • $\begingroup$ Yes, that is for the best. $\endgroup$ – Derek Luna Apr 21 at 5:48

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