Is This a Valid Equivalence Relation on ℚ?

[RJLiberator]

Homework Statement

For each of the relations defined on ℚ, either prove that it is an equivalence relation or show which properties it fails.

x ~ y whenever xy ∈ Z

Homework Equations

The Attempt at a Solution

Here's my problem: I am starting off the proof with the first condition of reflexivity.
Now, do I let x ∈ ℚ ? I would think so, if that is the case, then x can be 2/3.
So xx is thus 4/9 which does not exist in the integers.
Thus, the proof would fail.

But I think I am missing something here. Did I do this right or am I making a fatal error by suggesting that x can be any ℚ?

 

[PeroK]

I'm not sure why you doubt what you've done.

 

[RJLiberator]

I search for many proofs of equivalence relations online and just feel like something might be off here.

Let me give you an example.

The first part of this question, part a is
a) x~y whenever x-y∈ℤ
I proved the reflective part by showing x-x = 0 always.
II symetric part by stating x-y must be an integer so y-x = -(x-y) thus that must be an integer.
III transitive, by showing that (x-y) + (y-z) = x-z and since the first two are integers, this must be an integer also.

I guess, in this part I did not need to suggest a rational number anywhere. Hm...

Okay, back to the original question.

We say x~y whenever x*y ∈ℤ
An example would be x = 1/9 and y = 9/1 since x*y = 9 and that is an integer.

So if we do 1. reflexive, does it have to be an x such that x*x is an integer by the rule? Or can I pick any x value I want out of the rational numbers. This is my problem.
We say x~y whenever xy ∈ℤ.
Must we also say that x~x whenever xx ∈ℤ ?

 

[PeroK]

The reflexive rule must apply for all $x$ (not just for some $x$). So, you only need to find one counterexample, as you did, to show that the rule does not, in general, hold.

 

[RJLiberator]

Excellent. Thank you for your help.