[proofplan] We construct an explicit sequence of rational decimal truncations converging to $\sqrt{2}$ in $\mathbb{R}$. This sequence is Cauchy for the absolute-value metric, hence also Cauchy in the rational [metric space](/page/Metric%20Space) $(\mathbb{Q},d)$. If it converged to a rational number in $\mathbb{Q}$, the same estimates would force that rational number to equal $\sqrt{2}$, contradicting the irrationality of $\sqrt{2}$. [/proofplan] [step:Construct rational truncations of $\sqrt{2}$] We use the ambient sets declared in the theorem statement: $\mathbb{R}$ is the real line with its usual absolute value, $\mathbb{Q}\subset \mathbb{R}$ is the set of rational numbers, $\mathbb{Z}\subset \mathbb{Q}$ is the set of integers, and $\mathbb{N}=\{1,2,3,\dots\}$. For each $n \in \mathbb{N}$, apply the integer-part property of the real line to the real number $10^n\sqrt{2}$, obtaining the unique integer $r_n \in \mathbb{Z}$ satisfying \begin{align*}r_n \leq 10^n\sqrt{2} < r_n + 1.\end{align*} Define the sequence \begin{align*}a: \mathbb{N} \to \mathbb{Q}\end{align*} by \begin{align*}a_n := \frac{r_n}{10^n}.\end{align*} Since $r_n \in \mathbb{Z}$ and $10^n \in \mathbb{N}$, each $a_n$ belongs to $\mathbb{Q}$. Dividing the defining inequality for $r_n$ by $10^n$ gives \begin{align*}a_n \leq \sqrt{2} < a_n + 10^{-n}.\end{align*} Hence \begin{align*}0 \leq \sqrt{2}-a_n < 10^{-n}.\end{align*} Therefore $a_n \to \sqrt{2}$ in $\mathbb{R}$ with respect to the absolute-value metric. [guided] We need a [Cauchy sequence](/page/Cauchy%20Sequence) whose natural limit lies outside $\mathbb{Q}$. The number $\sqrt{2}$ is the standard target, but we must build the rational approximating sequence explicitly. For each $n \in \mathbb{N}$, apply the integer-part property of the real line to the real number $10^n\sqrt{2}$. This gives a unique integer $r_n \in \mathbb{Z}$ satisfying \begin{align*}r_n \leq 10^n\sqrt{2} < r_n + 1.\end{align*} This integer $r_n$ is the integer part of $10^n\sqrt{2}$. Now define a sequence \begin{align*}a: \mathbb{N} \to \mathbb{Q}\end{align*} by \begin{align*}a_n := \frac{r_n}{10^n}.\end{align*} Each $a_n$ is rational because it is a quotient of two integers and $10^n \neq 0$. Dividing the inequality \begin{align*}r_n \leq 10^n\sqrt{2} < r_n + 1\end{align*} by the positive number $10^n$ gives \begin{align*}\frac{r_n}{10^n} \leq \sqrt{2} < \frac{r_n}{10^n} + \frac{1}{10^n}.\end{align*} Using the definition of $a_n$, this becomes \begin{align*}a_n \leq \sqrt{2} < a_n + 10^{-n}.\end{align*} Subtracting $a_n$ from all parts of the inequality gives \begin{align*}0 \leq \sqrt{2}-a_n < 10^{-n}.\end{align*} Since $10^{-n} \to 0$ in $\mathbb{R}$, the squeeze estimate shows that $a_n \to \sqrt{2}$ in $\mathbb{R}$. [/guided] [/step] [step:Show the rational sequence is Cauchy in $(\mathbb{Q},d)$] Let $\varepsilon > 0$. Choose $N \in \mathbb{N}$ such that $2\cdot 10^{-N} < \varepsilon$. If $m,n \geq N$, then the triangle inequality in $\mathbb{R}$ and the estimate from the previous step give \begin{align*}d(a_m,a_n)=|a_m-a_n| \leq |a_m-\sqrt{2}|+|\sqrt{2}-a_n|.\end{align*} Using $|a_k-\sqrt{2}|<10^{-k}$ for $k \in \mathbb{N}$, we obtain \begin{align*}d(a_m,a_n) < 10^{-m}+10^{-n}.\end{align*} Since $m,n \geq N$, we have $10^{-m} \leq 10^{-N}$ and $10^{-n} \leq 10^{-N}$, so \begin{align*}d(a_m,a_n) < 2\cdot 10^{-N} < \varepsilon.\end{align*} Thus $(a_n)_{n=1}^{\infty}$ is a Cauchy sequence in $(\mathbb{Q},d)$. [/step] [step:Prove that $\sqrt{2}$ is irrational] For an integer $s \in \mathbb{Z}$, say that $s$ is even if there exists an integer $t \in \mathbb{Z}$ such that $s=2t$. Suppose, for contradiction, that $\sqrt{2} \in \mathbb{Q}$. Then there exist integers $p,q \in \mathbb{Z}$ with $q \neq 0$, $\gcd(p,q)=1$, and \begin{align*}\sqrt{2}=\frac{p}{q}.\end{align*} Squaring both sides gives \begin{align*}p^2 = 2q^2.\end{align*} Thus $p^2$ is even. If $p$ were not even, then $p=2\ell+1$ for some $\ell \in \mathbb{Z}$, and then $p^2=4\ell^2+4\ell+1=2(2\ell^2+2\ell)+1$, which is not even. Hence $p$ is even. Write $p=2k$ for some $k \in \mathbb{Z}$. Substituting this into $p^2=2q^2$ gives \begin{align*}4k^2 = 2q^2.\end{align*} Dividing by $2$ gives \begin{align*}q^2 = 2k^2.\end{align*} Thus $q^2$ is even. The same parity argument applied to $q$ shows that $q$ is even. Hence $p$ and $q$ are both divisible by $2$, contradicting $\gcd(p,q)=1$. Therefore $\sqrt{2}\notin \mathbb{Q}$. [/step] [step:Rule out convergence to any rational number] Assume, for contradiction, that $(a_n)_{n=1}^{\infty}$ converges in $(\mathbb{Q},d)$. Then there exists $q \in \mathbb{Q}$ such that \begin{align*}d(a_n,q) \to 0.\end{align*} By the definition of $d$, this means \begin{align*}|a_n-q| \to 0\end{align*} in $\mathbb{R}$. We show that this forces $q=\sqrt{2}$. Let $\varepsilon > 0$. Choose $N_1 \in \mathbb{N}$ such that $|a_n-q|<\varepsilon/2$ whenever $n \geq N_1$, and choose $N_2 \in \mathbb{N}$ such that $|a_n-\sqrt{2}|<\varepsilon/2$ whenever $n \geq N_2$. For $n \geq \max\{N_1,N_2\}$, the triangle inequality gives \begin{align*}|q-\sqrt{2}| \leq |q-a_n|+|a_n-\sqrt{2}| < \varepsilon.\end{align*} Since $\varepsilon>0$ was arbitrary and $|q-\sqrt{2}|$ is a fixed nonnegative real number, we have $|q-\sqrt{2}|=0$. Hence $q=\sqrt{2}$, contradicting $\sqrt{2}\notin \mathbb{Q}$. Therefore the Cauchy sequence $(a_n)_{n=1}^{\infty}$ in $(\mathbb{Q},d)$ has no limit in $\mathbb{Q}$. Hence $(\mathbb{Q},d)$ is not complete. [/step]