[proofplan] We must verify two facts: that the termwise sum and the termwise product of rational Cauchy sequences are themselves Cauchy (so the operations land in the right set), and that they respect the equivalence relation $(a_n) \sim (b_n) \iff |a_n - b_n| \to 0$. Additivity follows immediately from the triangle inequality. Multiplicativity uses the standard add-and-subtract trick combined with the fact, established earlier in the chapter, that every rational Cauchy sequence is bounded; the boundedness controls the "cross terms" that appear when comparing $a_n b_n$ with $a_n' b_n'$. [/proofplan]

[step:Fix Cauchy representatives and recall that they are bounded] Let $(a_n), (a_n'), (b_n), (b_n') \subset \mathbb{Q}$ be rational Cauchy sequences with $(a_n) \sim (a_n')$ and $(b_n) \sim (b_n')$, meaning \begin{align*} |a_n - a_n'| \to 0 \quad \text{and} \quad |b_n - b_n'| \to 0 \quad \text{as } n \to \infty. \end{align*} By the Boundedness of Cauchy Sequences in $\mathbb{Q}$ (every rational Cauchy sequence is bounded in absolute value by some positive rational), there exist rational constants $M_a, M_a', M_b, M_b' > 0$ such that \begin{align*} |a_n| \le M_a, \quad |a_n'| \le M_a', \quad |b_n| \le M_b, \quad |b_n'| \le M_b' \quad \text{for all } n \in \mathbb{N}. \end{align*} Define $M := \max\{M_a, M_a', M_b, M_b'\} > 0$, so that all four sequences are bounded by $M$ in absolute value. [/step]

[step:Show that the termwise sum of two rational Cauchy sequences is Cauchy] We verify that $(a_n + b_n)$ is a rational Cauchy sequence. Fix $\varepsilon \in \mathbb{Q}_{>0}$. Since $(a_n)$ is Cauchy, there exists $N_1 \in \mathbb{N}$ such that $|a_n - a_m| < \varepsilon/2$ for all $n, m \ge N_1$. Similarly there exists $N_2 \in \mathbb{N}$ with $|b_n - b_m| < \varepsilon/2$ for all $n, m \ge N_2$. Set $N := \max\{N_1, N_2\}$. For $n, m \ge N$, by the triangle inequality on $\mathbb{Q}$, \begin{align*} |(a_n + b_n) - (a_m + b_m)| \le |a_n - a_m| + |b_n - b_m| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Hence $(a_n + b_n)$ is Cauchy. The same argument applied to $(a_n', b_n')$ shows $(a_n' + b_n')$ is Cauchy. [/step]

[step:Show that the termwise product of two rational Cauchy sequences is Cauchy] Fix $\varepsilon \in \mathbb{Q}_{>0}$. Using the bound $M$ from Step 1 (so $|a_n|, |b_m| \le M$ for all $n, m$), we estimate via the add-and-subtract identity \begin{align*} a_n b_n - a_m b_m = a_n(b_n - b_m) + b_m(a_n - a_m). \end{align*} By the triangle inequality on $\mathbb{Q}$ and the bound $M$, \begin{align*} |a_n b_n - a_m b_m| \le |a_n|\,|b_n - b_m| + |b_m|\,|a_n - a_m| \le M\bigl(|b_n - b_m| + |a_n - a_m|\bigr). \end{align*} Since $(a_n)$ and $(b_n)$ are Cauchy, choose $N \in \mathbb{N}$ so that for $n, m \ge N$ both $|a_n - a_m| < \varepsilon/(2M)$ and $|b_n - b_m| < \varepsilon/(2M)$. Then for all $n, m \ge N$, \begin{align*} |a_n b_n - a_m b_m| \le M\left(\frac{\varepsilon}{2M} + \frac{\varepsilon}{2M}\right) = \varepsilon. \end{align*} Thus $(a_n b_n)$ is Cauchy. The same estimate applied to $(a_n' b_n')$ shows it is Cauchy as well. [/step]

[step:Verify that the sum respects the equivalence relation] We show $(a_n + b_n) \sim (a_n' + b_n')$, i.e. $|(a_n + b_n) - (a_n' + b_n')| \to 0$. By the triangle inequality on $\mathbb{Q}$, \begin{align*} \bigl|(a_n + b_n) - (a_n' + b_n')\bigr| = \bigl|(a_n - a_n') + (b_n - b_n')\bigr| \le |a_n - a_n'| + |b_n - b_n'|. \end{align*} Fix $\varepsilon \in \mathbb{Q}_{>0}$. Since $|a_n - a_n'| \to 0$, there exists $N_1$ with $|a_n - a_n'| < \varepsilon/2$ for $n \ge N_1$; since $|b_n - b_n'| \to 0$, there exists $N_2$ with $|b_n - b_n'| < \varepsilon/2$ for $n \ge N_2$. For $n \ge \max\{N_1, N_2\}$, the displayed inequality gives \begin{align*} \bigl|(a_n + b_n) - (a_n' + b_n')\bigr| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Therefore $(a_n + b_n) \sim (a_n' + b_n')$, so the sum $x + y = [(a_n + b_n)]$ does not depend on the choice of representatives. [/step]

[step:Verify that the product respects the equivalence relation via the cross-term decomposition]We show $(a_n b_n) \sim (a_n' b_n')$. Apply the add-and-subtract identity \begin{align*} a_n b_n - a_n' b_n' = a_n(b_n - b_n') + b_n'(a_n - a_n'). \end{align*} By the triangle inequality on $\mathbb{Q}$ and the uniform bound $|a_n| \le M$, $|b_n'| \le M$ from Step 1, \begin{align*} |a_n b_n - a_n' b_n'| \le |a_n|\,|b_n - b_n'| + |b_n'|\,|a_n - a_n'| \le M\bigl(|b_n - b_n'| + |a_n - a_n'|\bigr). \end{align*} Fix $\varepsilon \in \mathbb{Q}_{>0}$. Since $|a_n - a_n'| \to 0$ and $|b_n - b_n'| \to 0$, there exists $N \in \mathbb{N}$ such that \begin{align*} |a_n - a_n'| < \frac{\varepsilon}{2M} \quad \text{and} \quad |b_n - b_n'| < \frac{\varepsilon}{2M} \quad \text{for all } n \ge N. \end{align*} For such $n$, \begin{align*} |a_n b_n - a_n' b_n'| \le M\left(\frac{\varepsilon}{2M} + \frac{\varepsilon}{2M}\right) = \varepsilon. \end{align*} Thus $|a_n b_n - a_n' b_n'| \to 0$, i.e. $(a_n b_n) \sim (a_n' b_n')$, so the product $x \cdot y = [(a_n b_n)]$ does not depend on the choice of representatives.[/step]

[guided]We want to show $(a_n b_n) \sim (a_n' b_n')$, i.e. $|a_n b_n - a_n' b_n'| \to 0$ as $n \to \infty$. Unlike the sum, the difference $a_n b_n - a_n' b_n'$ does not factor through a single difference like $a_n - a_n'$ or $b_n - b_n'$ — both factors change at once. The standard remedy is to insert a hybrid term and split the difference: \begin{align*} a_n b_n - a_n' b_n' = (a_n b_n - a_n b_n') + (a_n b_n' - a_n' b_n') = a_n(b_n - b_n') + b_n'(a_n - a_n'). \end{align*} Each summand is now a product of a "bounded" factor ($a_n$ or $b_n'$) and a "small" difference ($b_n - b_n'$ or $a_n - a_n'$, both of which tend to $0$). This is the same trick used to prove that the product of two convergent sequences converges, and it works for the same reason: small times bounded is small. Applying the triangle inequality on $\mathbb{Q}$ to this decomposition, \begin{align*} |a_n b_n - a_n' b_n'| \le |a_n|\,|b_n - b_n'| + |b_n'|\,|a_n - a_n'|. \end{align*} We now use the uniform bound $M$ from Step 1: by the Boundedness of Cauchy Sequences in $\mathbb{Q}$ each of the four sequences $(a_n), (a_n'), (b_n), (b_n')$ is bounded, and we set $M := \max\{M_a, M_a', M_b, M_b'\}$, so in particular $|a_n| \le M$ and $|b_n'| \le M$ for every $n$. Substituting these bounds, \begin{align*} |a_n b_n - a_n' b_n'| \le M\,|b_n - b_n'| + M\,|a_n - a_n'| = M\bigl(|b_n - b_n'| + |a_n - a_n'|\bigr). \end{align*} Why does this estimate close the argument? The right-hand side is $M$ times a sum of two quantities that tend to $0$. The constant $M$ is fixed (chosen once and for all in Step 1, independent of $n$), so multiplication by $M$ does not destroy convergence to $0$: if $c_n \to 0$ then $M c_n \to 0$. The crucial input is precisely that Cauchy sequences in $\mathbb{Q}$ are bounded — without this, the factor multiplying $|b_n - b_n'|$ could blow up and we could not conclude. To convert this into an explicit $\varepsilon$-statement, fix $\varepsilon \in \mathbb{Q}_{>0}$. We want each of the two terms on the right to be at most $\varepsilon/2$. Since $|a_n - a_n'| \to 0$, there is $N_1 \in \mathbb{N}$ with $|a_n - a_n'| < \varepsilon/(2M)$ for all $n \ge N_1$; since $|b_n - b_n'| \to 0$, there is $N_2 \in \mathbb{N}$ with $|b_n - b_n'| < \varepsilon/(2M)$ for all $n \ge N_2$. Set $N := \max\{N_1, N_2\}$. Then for every $n \ge N$, \begin{align*} |a_n b_n - a_n' b_n'| \le M\left(\frac{\varepsilon}{2M} + \frac{\varepsilon}{2M}\right) = \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Since $\varepsilon \in \mathbb{Q}_{>0}$ was arbitrary, $|a_n b_n - a_n' b_n'| \to 0$, which is exactly the statement $(a_n b_n) \sim (a_n' b_n')$. Therefore the product $x \cdot y = [(a_n b_n)]$ does not depend on the choice of representatives.[/guided]

[step:Combine the steps to conclude well-definedness] By Steps 2 and 3, the termwise sum and termwise product of two rational Cauchy sequences are again rational Cauchy sequences, so the equivalence classes $[(a_n + b_n)]$ and $[(a_n b_n)]$ are well-defined elements of $\mathbb{R}$. By Steps 4 and 5, these classes are independent of the chosen representatives. Therefore \begin{align*} x + y := [(a_n + b_n)] \quad \text{and} \quad x \cdot y := [(a_n \cdot b_n)] \end{align*} define operations $+, \cdot : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$, completing the proof. [/step]