Proof of The Rational Numbers are Dense in the Real Numbers

Let $\mathbb{N}$ be the natural numbers, $\iota$ the canonical map of $\mathbb{R}$, and $\mathbb{Z}$ the integers. Every step below in which the same quantity is added to both sides of an inequality, or both sides are multiplied by a positive quantity, is justified by the elementary order arithmetic in an ordered field.

Claim 1. Let $x<y$ in $\mathbb{R}$, so $0<y-x$. By claim 3 of the Archimedean property there is $N\in\mathbb{N}$ such that $\iota(N)^{-1}$ exists and

$$0<\iota(N)^{-1}<y-x .$$

By claim 3 of the properties of the canonical map, $0<\iota(N)$.

Let $n=\lfloor\iota(N)\,x\rfloor$ be the integer part of $\iota(N)x$, so that

$$n\le\iota(N)\,x<n+1 ,$$

and put $q=(n+1)\,\iota(N)^{-1}$. Both $n+1$ and $\iota(N)$ are integers, by claim 2 of the arithmetic lemma for the integers and by the definition of $\mathbb{Z}$ respectively, and $\iota(N)\ne0$; hence $q\in\mathbb{Q}$ by the definition of the rational numbers.

Multiplying the strict inequality $\iota(N)x<n+1$ by the positive number $\iota(N)^{-1}$ preserves it, and $\iota(N)^{-1}\bigl(\iota(N)x\bigr)=x$, so

$$x<(n+1)\,\iota(N)^{-1}=q .$$

Multiplying $n\le\iota(N)x$ by the positive $\iota(N)^{-1}$ gives $n\,\iota(N)^{-1}\le x$, so

$$q=(n+1)\,\iota(N)^{-1}=n\,\iota(N)^{-1}+\iota(N)^{-1}\le x+\iota(N)^{-1}<x+(y-x)=y .$$

Hence $x<q<y$.

Claim 2. Let $x\in\mathbb{R}$ and let $\varepsilon>0$ be real. Then $x<x+\varepsilon$, so claim 1 provides $q\in\mathbb{Q}$ with $x<q<x+\varepsilon$. Then $0<q-x<\varepsilon$, so $|x-q|=|q-x|=q-x<\varepsilon$ by the definition of the absolute value.

Claim 3. By the identification of the Euclidean distance on the real line with the absolute value metric, $d(s,t)=|s-t|$ for all $s,t\in\mathbb{R}$, so for $x\in\mathbb{R}$ and real $\varepsilon>0$ the open ball of centre $x$ and radius $\varepsilon$ is $\{s\in\mathbb{R}:|s-x|<\varepsilon\}$. By the theorem that metric open sets form a topology, the collection $\mathcal{T}_{d}$ of metric open subsets of $\mathbb{R}$ is a topology, so the closure of a subset of $\mathbb{R}$ is defined.

By the characterization of the closure in a metric space by open balls, a point $x\in\mathbb{R}$ belongs to the closure of $\mathbb{Q}$ if and only if every open ball of centre $x$ and positive radius contains a point of $\mathbb{Q}$. Let $x\in\mathbb{R}$ and let $\varepsilon>0$ be real; claim 2 supplies $q\in\mathbb{Q}$ with $|x-q|<\varepsilon$, and this $q$ lies in the open ball of centre $x$ and radius $\varepsilon$. Hence every $x\in\mathbb{R}$ belongs to the closure of $\mathbb{Q}$, so that closure is $\mathbb{R}$ itself, which is precisely the assertion that $\mathbb{Q}$ is dense in $(\mathbb{R},\mathcal{T}_{d})$.