Is $\mathbb Q-\mathbb N$ dense in $\mathbb R$? Is $\mathbb Q-\mathbb N$ dense in $\mathbb R$? I believe that the solution is about two relatively prime integers. However, I do not know how to proceed.
 A: Let's try to find a noninteger rational between any two reals. For any real numbers $a <b$, since $\mathbb Q$ is dense in $\mathbb R$, there is $p\in(a,b)\cap \mathbb Q$. If $p\notin \mathbb N$, it is our desired rational. Otherwise, there is $q\in(p,b)\cap\mathbb Q$. If $q\notin \mathbb N$, it is our desired rational. Otherwise, since $p < q$ and $p,q\in\mathbb N$, we have $p < p + 1/2 < q$ and $p + 1/2 \notin \mathbb N$. Since $a < p < p + 1/2 < q < b$,  $p + 1/2$ is our desired rational.
A: Here is a topological perspective:
If $X,Y \subset Z$ and $X$ is discrete in $Z$ while $Y$ is dense in $Z$, then $Y-X$ is dense in $Z$.
Indeed, assume the contrary, i.e. $Y-X$ is not dense. Then there exists $a \in Z$ and a neighborhood $U$ of $a$ such that $U \cap (Y-X) = \emptyset$. Since $X$ is discrete, you can shrink $U$ so that $U \cap X = \emptyset$. But then $U \cap Y = \emptyset$, being the union of two empty sets. This would imply that $Y$ is not dense, contradiction.
EDIT: this answer is incorrect, the shrinking argument is false.
A: Assuming that you have proven that $\mathbb Q$ is dense in $\mathbb R$, we know that if $x<y$ then there is a $q\in\mathbb Q$ such that $x<q<y$. However, we can actually establish a stronger result: if $x<y$ then there are infinitely many $q\in\mathbb Q$ such that $x<q<y$. To prove this, suppose for the sake of contradiction that there are only finitely many rational numbers lying strictly between $x$ and $y$. This means that there is a least rational $q_0$ such that $x<q_0<y$. Therefore, there is no rational number $q$ such that $x<q<q_0$, contradicting the density of $\mathbb Q$ in $\mathbb R$.
Can you see how this implies that $\mathbb Q-\mathbb N$ is dense in $\mathbb R$? Hint: between any two real numbers, there are only finitely many natural numbers.
A: Let $r\in\mathbb{R}$ and $\varepsilon>0$. Then

*

*$(r-\varepsilon,r+\varepsilon)\cap\mathbb{Q}$ is infinite, because $\mathbb{Q}$ is dense in $\mathbb{R}$;

*$(r-\varepsilon,r+\varepsilon)\cap\mathbb{N}$ is finite.

Therefore
$$
(r-\varepsilon,r+\varepsilon)\cap(\mathbb{Q}-\mathbb{N})=
\bigl((r-\varepsilon,r+\varepsilon)\cap\mathbb{Q}\bigr)-
\bigl((r-\varepsilon,r+\varepsilon)\cap\mathbb{N}\bigr)
$$
is infinite, in particular not empty.
We use that $A\cap(B-C)=(A\cap B)-(A\cap C)$.
A: You can see that $\mathbb{Q}-\mathbb{N}$ is dense in $\mathbb{Q}$.(If $q\in \mathbb{N}$, consider $q_n=q+1/(n+1)$.
$\mathbb{Q}$ is dense in $\mathbb{R}$ and hence...
