Can All Real Numbers Be Multiplied to Create an Integer?

[iamsmooth]

Homework Statement

Prove or disprove: ∀x ∈ R ∃y ∈ R so that xy ∈ Z.

(R denotes set of all real nuimbers, Z denotes set of all integers)

Homework Equations

The Attempt at a Solution

I'm not sure how to attack this question. It seems false, but I can't think of a good counterexample.

Like If I say take pi, I don't think there is any other number you could multiply pi by to make an integer, but I don't know how to formulate this into a proof that makes sense. I also thought of playing with irrationals since maybe they could provide a counter example, but I'm not sure how to prove something like this. Maybe it's just something I'm not seeing.

Can anyone help get me started? Thanks!

 

[Hurkyl]

Sometimes, starting simpler is helps. Since you want to prove something for all integers (namely, that there exists an x such that for all y, xy is not that integer), you could first try to get ideas by proving it for one specific integer.

 

[l'Hôpital]

Well, try two cases.

Case 1: x = 0.
Trivial

Case 2: x =/= 0.

What if y = 1/x?

 

[iamsmooth]

well if you give me any integer, multiplying it by any integer will produce an integer

if you give me an irrational, how can you prove that no number can make an integer.

it's defined on reals, not integers. i don't think that 1/x example works since it says there exists, as long as there exists one number that can be multiplied with x to make an integer, it's true.

 

[iamsmooth]

i'm so stupid, i figured it out, given any real number, multiply it by 0 and 0 is an integer...

i wasn't thinking simple enough...

 

[Hurkyl]

If you're given a nonzero, can you multiply it by something to get, say, 1?

 

[iamsmooth]

like l'Hopital said, 1/x * x = 1, so yeah I guess that works.

Thanks a lot guys :D