Then give information about the equivalence classes as specified for

The relation $R$ on $ℝ$ given by $xRy$ iff $x-y∈ℚ$. Give the equivalence class of $0$; of $1$ ; $\sqrt{2}$.

First In order to solve this problem one must list what he/she knows in order to solve this problem.

Reflexive : R is reflexive on $A$ iff for all $x∈ A$, $xRx.$

Symmetric : R is symmetric iff for all $x$ and $y∈ A$ if $xRy,$ then $yRx.$

Transitive: R is transitive on A if for all $x$,$y$ $z ∈A$ if $xRy$, and $yRz$ then $xRz$.

An equivalence relation is defined if something is reflexive,symmetric, and transitive on A.

My question is how does one determine how this is a Reflexive relationship ?Does one have to substitute real numbers or rational numbers into the equation in order to find the relation. Does this also apply to Symmetric and transitive?

I also know how to determine an equivalence class, $\frac{x}{R} =$ {$ y∈A: xRy$}. I am just perplexed on how one proceeds.

This is what I have deduced so far.

$[0]$ = { $x ∈ ℝ$ : $0$Rx}

Any advice on how to proceed and solve

this, would be appreciated.


The relation is reflexive since for all $x \in \Bbb R$, we have $x-x = 0 \in \Bbb Q$. To write the classes, use the definition of the relation. For the other properties, you have to check that if $x,y,z \in \Bbb R$ are such that $x-y,y-z\in \Bbb Q$, then $x-z \in \Bbb Q$, and finally that if $x-y \in \Bbb Q$, then $y-x \in \Bbb Q$.

About the classes, for instance, we have $$[0] = \{x \in \Bbb R \mid xR0\} = \{x \in \Bbb R \mid x - 0 \in \Bbb Q\} = \Bbb Q.$$In general, the classes will be "copies" of $\Bbb Q$ shifted by the representant, as in $$[\sqrt{2}] = \sqrt{2}+\Bbb Q = \{ \sqrt{2}+q \mid q \in \Bbb Q \},$$ etc.

  • $\begingroup$ So for this problem does one not need to show a scenario where it is transitive and symmetric? So since x is real numbers y is rational numbers is that why it is symmetric? $\endgroup$ – Jon Jul 1 '16 at 20:32
  • $\begingroup$ Why shouldn't you have to prove the other properties? Of course you have to. I said it in the answer, to prove symmetry you take $x,y \in \Bbb R$. Then assume that $x-y \in \Bbb Q$. And check that $y-x \in \Bbb Q$. $\endgroup$ – Ivo Terek Jul 1 '16 at 20:34
  • $\begingroup$ You misunderstand, I just wanted to know that what you said above is enough to justify whether it be symmetric and transitive now I know more work is required. Sorry for the discrepancies. $\endgroup$ – Jon Jul 1 '16 at 20:37
  • $\begingroup$ Yes, it is the definition of these properties. What does $xRy$ mean? Look at the definition of the relation we're given. $\endgroup$ – Ivo Terek Jul 1 '16 at 20:37
  • $\begingroup$ X is a relationship to y. $\endgroup$ – Jon Jul 1 '16 at 20:42

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