# Using density of $\mathbb{Q}$ to show density of $\mathbb{R}\setminus\mathbb{Q}$

See a previous answer given here.

Basically, by the density of rationals, for $x, y \in \mathbb{R}$, there exists an $r \in \mathbb{Q} \setminus \{ 0 \}$, $$\dfrac{x}{\sqrt{2}} < r < \dfrac{y}{\sqrt{2}} \Longleftrightarrow x < r\sqrt{2} < y\text{.}$$

The product of a non-zero rational ($r$) and an irrational ($\sqrt{2}$) is irrational (which is easy to show), so we've proven the density of irrationals.

The statement of the density of rationals I have is:

If $x, y \in \mathbb{R}$, there exists an $r \in \mathbb{Q}$ such that $x < r < y$.

The problem I see with this is that $r$ could very well be $0$. How are we allowed to assume that there exists an $r\in \mathbb{Q} \setminus \{ 0 \}$? I would guess intuitively that we could work with a sequence of nested intervals of some sort (perhaps?) to find a non-zero rational, but sequences have not been covered in this course yet.

• If $x=0$ or $y=0$, then you do not need to worry about $r=0$. If neither $x$ nor $y$ is zero, and you get $r=0$, then repeat the procedure to find some rational $r'$ such that $x<r'<r=0$ or $0=r<r'<y$. – angryavian May 20 '16 at 18:53

Well, if $x<0$ and $y < 0$, then since $x< r< y$, we would have $r < 0$. Similarly, if $x > 0$ and $y > 0$, then since $x < r < y$, $r > 0$.
So in the above two cases, $r \neq 0$. The only other case to consider is if $x < 0$ and $y > 0$. If so, by the density of $\Bbb Q$ in $\Bbb R$, find $r \in \Bbb Q$ such that $0 < r < y$ (and consequently, we have $r \neq 0$). Then this same $r$ satisfies $x < r < y$, because $x < 0$. So we can find a nonzero $r$.
• So, in other words, if your two values in $\mathbb{R}$ are of the same sign, the proof is trivial. If they are of different sign, consider the maximum of the two, which will be greater than $0$. Apply the density of $\mathbb{Q}$ in $\mathbb{R}$ to $0$ and the maximum. Done. – Clarinetist May 20 '16 at 18:57