The Density of the Rational/Irrational Numbers

# The Density of the Rational/Irrational Numbers

We will now look at a theorem regarding the density of rational numbers in the real numbers, namely that between any two real numbers there exists a rational number.

Theorem 1 (The Density of the Rational Numbers): Let $x, y \in \mathbb{R}$ be any two real numbers where $x < y$. Then there exists a rational number $r \in \mathbb{Q}$ such that $x < r < y$. |

**Proof:**Suppose that $x > 0$. Since $x < y$ we have that $y > 0$ and furthermore we have that $y - x > 0$. Now we know by the Archimedean properties that since $y - x > 0$, then there exists a natural number $n \in \mathbb{N}$ such that $\frac{1}{n} < y - x$.

- If we multiply this out we get that $1 < ny - nx$ or rather $nx + 1 < ny$. Now we know that since $n > 0$ and since $x > 0$, and by the Archimedean properties that since $nx > 0$ then there exists a natural number, call it $A \in \mathbb{N}$ such that $A - 1 ≤ nx < A$ or equivalently $A ≤ nx + 1 ≤ A + 1$.

- Therefore $nx ≤ A ≤ nx + 1 ≤ ny$ and so $nx < A < ny$ and so the rational number $r = \frac{A}{n}$ works for $x < r < y$. $\blacksquare$

Our next question might be to ask whether between any two real numbers there always exists an irrational number. The answer, once again, is yes, as we will see in this following corollary.

Theorem 2 (The Density of the Irrational Numbers): Let $x, y \in \mathbb{R}$ be any two real numbers where $x < y$. Then there exists an irrational number $q \in \mathbb{R} \setminus \mathbb{Q}$ such that $x < q < y$. |

**Proof:**Consider the real numbers $\frac{x}{\sqrt{2}}$ and $\frac{y}{\sqrt{2}}$. By the theorem above there exists a rational number $r$ such that:

\begin{align} \quad \frac{x}{\sqrt{2}} < r < \frac{y}{\sqrt{2}} \quad (*) \end{align}

- We can assume that $r \neq 0$ for if $r = 0$ we can apply theorem 1 again to get an $r^* \in \mathbb{Q}$ such that:

\begin{align} \frac{x}{\sqrt{2}} < r^* < r < \frac{y}{\sqrt{2}} \Rightarrow \frac{x}{\sqrt{2}} < r^* < \frac{y}{\sqrt{2}} \end{align}

- And we can choose the $r^*$ instead of $r$. So we assume that $r \neq 0$. We multiply the inequality at $(*)$ by $\sqrt{2}$ to get $x < r\sqrt{2} < y$, and let $q = r\sqrt{2}$. Since $r \neq 0$ and $r \in \mathbb{Q}$ we have that $q$ is an irrational number. $\blacksquare$