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If we have a equivalence relation $R=_{def} \{((x_1,y_1),(x_2,y_2)) ~~| ~~x_1-y_1=x_2-y_2 \} \subseteq \mathbb{R}^2 \times \mathbb{R}^2 $

What is the equivalence class $[(0,1)]_R$?

I thought it could be $[(0,1)]_R = \{ x_2 \in \mathbb{R}, y_2 \in \mathbb{R} ~~|~~ x_2-y_2 = -1\} $ But I'm pretty sure that this is mathematically incorrect, isn't it?

Second question: Is $[(0,1)]_R$ a rep system (Don't know the correct word in english) of the relation?

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Your attempt is correct. By definition, $$[(0,1)]_R=\{(x,y)\in\mathbb{R}^2:(0,1),(x,y)\in R\}=\{(x,y)\in\mathbb{R}^2:0-1=x-y=-1\}$$and this correspond to the set of points in the line $y=x+1$!

However, $[(0,1)]_R$ is far from being a set of representatives of the equivalence relation $R$. In fact, by definition, a set of representatives is a set $M$ that contains exactly one element of each class. Thus, the set $[a]_R$ will never be a set of representatives of an equivalence relation $R$ on $X$ (except for the extremely trivial case when $X=\{a\}$!), because it contains only elements in one of the classes.

Notice that the equivalence class $[(x,y)]_R$ is characterized in this case by the real number $x-y=b$, that is, if $x-y=b$ then $(x_1,y_1)R(x,y)$ if and only if $x_1-y_1=b$. Therefore, a set of representatives $M$ will be a set such that every real number appears as the difference of the components of a pair in $M$, and all the possible differences are different.

You can check somewhat easily that the $x$-axis, $M=\{(x,0):x\in\mathbb{R}\}\subseteq \mathbb{R}^2\}$ is a set of representatives. Can you think about other examples?

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All points on on lines parallel to the line x=y are equivalent to each other.

so {(x,y)| x-y = -1} would indeed be the equivalence class containing (0,1)

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