Does there exist a nonzero rational number $x$ for every irrational number $y$ such that the product of $x$ and $y$ is rational?

Question

I came across this statement in my math textbook: $$\exists x\in\mathbb{Q}\ :\ \forall y\in\mathbb{R}\setminus\mathbb{Q},\ xy\in\mathbb{Q}$$ The only number $x$ for which it's true (that I can think of) is $0,$ as in: $$xy=0\times y=0$$ I know that this is enough to prove the statement true, however, I wonder if there is another number which satisfies this statement. Because if there isn't, $x$ could well be denoted as belonging to the set of natural numbers.

So here's my question: can you think of any other rational number $x$ besides $0$ for which this statement would be true, if there is one at all? And if not, could the statement be rewritten to $\exists x\in\mathbb{N}\ \forall y\in\mathbb{R}\setminus\mathbb{Q}:xy\in\mathbb{N}$ and still keep the same meaning?

For what it's worth, you've mistranslated the statement. Verbally, your title better fits $$\forall y\in\Bbb R\setminus\Bbb Q,\exists x\in\Bbb Q:xy\in\Bbb Q.$$ While this is also true, the statement you've proved is much stronger: "There exists a rational number $x$ such that for any irrational number $y,$ the number $xy$ is rational." — Cameron Buie, Feb 05 '22 at 15:16

score 2 · Accepted Answer · answered Feb 05 '22 at 14:25

2

If $x\in\Bbb Q\setminus\{0\}$, then $\sqrt2\,x\notin\Bbb Q$. In fact, if $\sqrt 2\,x=q\in\Bbb Q$, then $\sqrt2=\frac qx\in\Bbb Q$. So, yes, $0$ is the only rational number with that property.

answered Feb 05 '22 at 14:25

José Carlos Santos

427,504

score 2 · Answer 2 · answered Feb 05 '22 at 14:25

2

$0$ is the only rational that satisfies the statement, as you say. Your version is also true, but it is a slightly different statement.

answered Feb 05 '22 at 14:25

Ross Millikan

374,822

Does there exist a nonzero rational number $x$ for every irrational number $y$ such that the product of $x$ and $y$ is rational?

2 Answers2