Let $R$ be the relation on all real numbers defined by: $$ \forall x,y \in \Bbb R , xRy \iff x-y \in \Bbb Z. $$ Is it true that for all $a \in \Bbb R$; there exists an irrational number $b$ so that $b \in [a]$ ?

I think this is FALSE. Why?

Let $a = 2$ and $b$ be an irrational number. Now, $[2] = \{ x \in \Bbb Z | x \}$ which is a set of all integers. So $b \notin [2]$ because $b$ is irrational and all $[2]$ is a set of integers which are rational.

Is this proof logically correct? I just started learning about equivalence. Thanks :) !!

  • 1
    $\begingroup$ Relax. It's just fine. $\endgroup$ – José Carlos Santos Mar 30 '18 at 20:28
  • $\begingroup$ Thanks im glad its correct ! :) $\endgroup$ – FabolousPotato Mar 30 '18 at 20:30

Your proof is basically correct.

Maybe you should just prove that $[2]$ is actually the set of all integers (it is). I think you know how to do this and you didn't write all passages here to not overextend the question. But if you don't know, don't hesitate to ask.


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