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Like I stated in the title, I am interested in whether the following always holds true:

$\forall x\in\mathbb{Q}( \forall y\in\mathbb{Q} (x\mathbb{Z} \cap y\mathbb{Z} \neq \emptyset))$

and how to go about solving this question. Note that I am not a mathematician and I am also too old to become one. Still I am eager to learn though.

Andrés E. Caicedo
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apriori
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3 Answers3

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Notice that for any $x \in \mathbb{Q}$ we have $x \mathbb{Z} \ni 0$. Thus, $x \mathbb{Z} \cap y \mathbb{Z} \supset \{0\}$. The answers below give more elements that must be in this intersection.

D. Ungaretti
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  • I think the OP might have been happier with a proof that $xZ^+ \cap yZ^+ \neq\emptyset$. Your proof works even lightening the restrictions on $x$ and $y$ to $x, y \in \Bbb C$. Of course, then the problem should have said $\Bbb z^+$ rather than $\Bbb Z$. – Mark Fischler Aug 16 '19 at 19:09
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Yes, the statement is valid in ordinary ZF arithmetic.

The proof is sketched as follows:

For arbitrary $x \in \Bbb Q$ there exists $p,q \in \Bbb Z$ such that $x=p/q$. Similarly, there exists $r,z \in \Bbb Z$ such that $y=r/s$.

Now consider the least common multiple of $q$ and $s$; call that $t$. Since $$xt = \frac{t}{q} p $$ and by the definition of least common multiple $\frac{t}q \in \Bbb Z$, we have $xt \in \Bbb Z$. Similarly, $yt \in \Bbb Z$.

Now consider $xtyt = x (tyt) = (xtt) y$. Since $ty$ is an integer, so is $tyt$, so $x(tyt) \in x\Bbb Z$. Since $xt$ is an integer, so is $xtt$, so $(xtt)y \in x\Bbb Z$. Therefore, $xyt^2$ is in both $ x\Bbb Z$ and $y\Bbb Z$ so their intersection is non-empty.

Mark Fischler
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let $x = \frac {p}{q}$ let $y = \frac{r}{s}$

$rqx = pr = spy$

$pr \in x\mathbb Z\cap y\mathbb Z$

Doug M
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