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On the set $\mathbb{R}^2$, define $(x,y) R (a,b)$ if and only if $x^2-y =a^2-b$. Show that $R$ is an equivalence relation.

1) $\forall (x,y)\in\mathbb{R^2}$ , we have $(x,y) R (x,y)$ since $x^2-y = x^2-y$, which shows that $R$ is reflexive

2) If $(x,y) R (a, b)$, then $x^2-y=a^2-b$ , and so $a^2-b=x^2-y$ , which implies that $(a,b)R(x,y)$, and so $R$ is symmetric

3) If $(x,y)R(a,b)$ and $(a,b)R(c,d)$ , then $x^2-y=a^2-b$ and $a^2-b=c^2-d$ The transitive law for equality implies that $x^2-y=c^2-d$, and therefore $(x,y)R(c,d)$, so $R$ is transitive.

We conclude that $R$ is an equivalence relation.

Is correct my proof?

B. David
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    Yes, except for a typo in 2): "$(a,b)R(c,d)$" should be "$(a,b)R(x,y)$". – zipirovich Oct 21 '17 at 16:37
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    at 2) you wrote at the end $(a,b)R(c,d)$ and not $(a,b)R(x,y)$ and at 3) you said that The transitive law for equality implies that $x^2-y=a^2-b$ instead of $x^2-y=c^2-d$ – ℋolo Oct 21 '17 at 16:44
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    In general, any function $f:A \to B$ induces an equivalence relation on $A$ where $a_1 \sim a_2 \iff f(a_1)=f(a_2) \in B$. Yours is just the case where $A=\mathbb{R}^2$, $B=\mathbb{R}$, and $f(a_1,a_2)=a_1^2-a_2$. – Geoffrey Trang Jun 26 '20 at 19:24

1 Answers1

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Your proof is correct; good job! This indeed mostly boils down to utilizing the fact that equality itself is an equivalence relation.

Mostly posting this so this question can be finally considered to have an answer. Made it Community Wiki since I have nothing of substance to add.

PrincessEev
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