Hello,

Here's a relation \sim defined on set \mathbb{R}:
x \sim y \Leftrightarrow $\lfloor x \rfloor = $\lfloor y \rfloor

So the number of equivalence classes is the same as the number of integers, am I right? Because that's how many differet "floors" we have. Here I have another relation R defined on the set of these classes, meaning R \subseteq \mathbb{R}/\sim \times \mathbb{R}/\sim

\left[ x \right] _{\sim} R \left[ y \right] _{\sim} \Leftrightarrow x \sim y \vee x < y

And a few questions to answer.

I. Prove that the definition of the relation is correct.
This means I have to prove that no matter which member of the equivalence class I choose, the relation between equivalence classes will stand, meaning that for x,y \in \mathbb{R} if \left[ x \right] _{\sim} R \left[ y \right] _{\sim} then for any x_{1} \in \left[ x \right] _{\sim} ,y_{1} \in \left[ y \right] _{\sim} \mathbb{R} are too in relation \left[ x_{1} \right] _{\sim} R \left[ y_{1} \right] _{\sim}.
We know that $\lfloor x_{1} \rfloor=$\lfloor x \rfloor and $\lfloor y_{1} \rfloor=$\lfloor y \rfloor. So there's two cases to analyze, when two equivalence classes are related by the first part of the logical disjunction, then $\lfloor x_{1} \rfloor=$\lfloor x \rfloor=$\lfloor y\rfloor=$\lfloor y_{1} \rfloor, so the definiton is correct. For the second part of the logical disjunction we start by stating that x<$\lfloor y \rfloor, because otherwise x would be in the equivalence class of y and fulfill the first part of the disjunction, any by extensions, x_{1}<$\lfloor y_{1} \rfloor whic means x_{1} < y_{1}, so the definition is correct in this case also.

II. Prove that this relation is a linear order, meaning that for any two equivalence classes either [x]_{\sim} R [y]_{\sim} or [y]_{\sim} R [x]_{\sim} . It suffices to say that if two numbers have the same floor then their equivalence classes are in relation, if not than one must be larger than the other so it would still fulfill the definition.

III. Show, that \left\langle \mathbb{R}/\sim,R \right\rangle is isomorphic with \left\langle \mathbb{Z},\leq \right\rangle, which is an order of integers by normal greater-equal relation. Well the bijection f:\mathbb{R}/\sim \to \mathbb{Z} in this case to a equivalence class \left[ x \right] _{\sim} assigns an integer $\lfloor x \rfloor.

Is this right?