(Theorem) Between any two real numbers is a rational number

Theorem: Density of the rationals in \mathbb R

(i) For all real numbers $x,y\in \mathbb{R}$ with $x,y>0$, there exists $n\in \mathbb{N}$ such that $nx>y$. (ii) For all $x,y\in \mathbb{R}$ with $x<y$, there exists a rational number $a\in \mathbb{Q}$ such that $x<a<y$.

(Nearly) direct proof of (ii) §

From Honors Mathematics B, Homework 1.

Lemma ( MATH-UN1208 1.6)

If $x\in \mathbb{R}$ has $x\ge 0$, the set of natural numbers $N_{\le x}=\{n\in \mathbb{N}|n\le x\}$ is finite.

Sketch.

1. Prove that for any real number $\delta >0$, there exists a natural number $n$ so that $1/n<\delta$.
2. If $z$ is a real number, let’s denote ${\mathbb{Q}}_{<z}=\{a\in \mathbb{Q}\mid a<z\}$. Prove that $\mathsf{lub}\left({\mathbb{Q}}_{<z}\right)=z.$
3. If $x<y$ are two real numbers, prove that there exists a rational $a$ with $x<a<y$.

Proof of 1.

We can prove there exists some $n\in \mathbb{N}$ with $1/n<\delta$ by showing that the set of all such $n$ is nonempty. Using the equivalent relation $1/\delta <n$, we define this set to be

$$\{n\in \mathbb{N}|1/n<\delta \}=\{n\in \mathbb{N}|1/\delta <n\}.$$

By trichotomy, this set is empty if the set $\{n\in \mathbb{N}|1/\delta \ge n\}$ is all of $\mathbb{N}$. But by Lemma 1.6, we know that given the real number $1/\delta$, the set of $n\in \mathbb{N}$ with $n\le 1/\delta$ is finite. Thus, the set of all $n\in \mathbb{N}$ with $1/n<\delta$ must be nonempty, which proves the existence of some $n$ that satisfies the relation.

Proof of 2.

First, let $\ell =\mathsf{lub}({\mathbb{Q}}_{<z})$. By definition of the least upper bound, we have $\ell \le z$ with precisely $\ell =z$ if $z$ is indeed the least upper bound, as well as $0\le z-\ell$.

If $\ell <z$, then the difference $0<z-\ell$ is a positive real number. By part (1), there exists some $n\in \mathbb{N}$ with $1/n<z-\ell$. We can also construct an even smaller term $1/2n<1/n$. Putting these expressions together, we have

$$0<1/2n<1/n<z-\ell ,$$ $$\ell +0=\ell <\ell +1/2n<\ell +1/n<\ell +(z-\ell )=z.$$

Now since $\ell$ is the least upper bound, by (Theorem) Elements in a subset of the reals can be arbitrarily close to the least upper bound, for all real $ϵ>0$ there exists some $q$ such that $\ell \le q+ϵ$. By definition of an upper bound, we can also guarantee $q\le \ell$. Choose $ϵ=1/2n$, and let $q$ an element which satisfies $\ell \le q+1/2n<q+1/n$. Using the definition $q\le \ell$ and the first set of relations we showed, we have

$$q+1/n\le \ell +1/n<z,$$

which means in particular that $q+1/n$ is a rational less than $z$. If both $\ell <q+1/n$ and $q+1/n\in {\mathbb{Q}}_{<z}$ are true, then this contradicts our assumption that $\ell =\mathsf{lub}({\mathbb{Q}}_{<z})$. Therefore, any real number $\ell <z$ cannot be an upper bound for ${\mathbb{Q}}_{<z}$, and we must have $\ell =z=\mathsf{lub}({\mathbb{Q}}_{<z})$.

Proof of 3.

First let $c=\frac{x+y}{2}$ denote the real average of $x$ and $y$, so we have $x<c<y$. By part (2), we have $c=\mathsf{lub}({\mathbb{Q}}_{<c})$.

Fix $ϵ=\frac{c-x}{2}$. Since $c$ is the least upper bound, we can use (Theorem) Elements in a subset of the reals can be arbitrarily close to the least upper bound to choose an $a\in {\mathbb{Q}}_{<c}$ so that $a<c\le a+ϵ$. First, notice that $a\le c<y$ for all $a\in {\mathbb{Q}}_{<c}$. To see $x<a$ as well, we can use the property $c\le a+ϵ$ and the fact that $0<\frac{c-x}{2}<c-x$ to get

$$\begin{array}{rl}x<c\le a+ϵ& ⟹x+\frac{c-x}{2}<x+c-x\le a+\frac{c-x}{2}\\ & ⟺x+ϵ<c\le a+ϵ\\ & ⟺x<a.\end{array}$$

Explicitly, this means we can fix $ϵ$ so that there is a rational $a$, based our choice of $ϵ$, which satisfies $x<a<y$ for any real numbers $x,y$. This completes the proof that between any two real numbers is a rational number.

Proof of (i) and (ii) by contradiction §

#wip See notes from Modern Analysis I.