Rational Sequences as Real Numbers
The rational numbers are dense in the real line, but they are not complete: a Cauchy sequence of rationals need not converge to a rational number. The completion construction addresses this by treating a Cauchy sequence as a point and identifying sequences that become arbitrarily close to one another. For \(\mathbb{Q}\), these points can be identified exactly: each is the real limit of its rational sequence.
Give \(\mathbb{Q}\) the metric \(d_{\mathbb{Q}}(p,q)=|p-q|\). A sequence \((q_n)\) in \(\mathbb{Q}\) is Cauchy if for every \(\varepsilon>0\) there is \(N\in\mathbb{N}\) such that \(|q_n-q_m|<\varepsilon\) whenever \(n,m\ge N\). Let \(\mathcal{C}\) be the set of all such sequences. For \((q_n),(r_n)\in\mathcal{C}\), write \((q_n)\sim(r_n)\) when \(|q_n-r_n|\to0\). The completion constructed from Cauchy sequences consists of the equivalence classes \([(q_n)]\) under this relation.
The formula makes intuitive sense: two sequences represent the same point precisely when their distance tends to zero. The key observation is that every rational Cauchy sequence also converges in \(\mathbb{R}\), since the real line with its usual metric is complete. We now prove that taking this real limit identifies the entire completion, not just some of its points.
The Limit Map Identifies the Completion
Proof. First, because \((q_n)\) is Cauchy in \(\mathbb{Q}\), it is Cauchy in \(\mathbb{R}\) with the same metric. Completeness of \(\mathbb{R}\) gives a real limit; denote it by \(x\). If \((q_n)\sim(r_n)\), let \(x=\lim q_n\) and \(y=\lim r_n\). The triangle inequality gives \(|x-y|\le |x-q_n|+|q_n-r_n|+|r_n-y|\) for every \(n\). Each term on the right tends to zero, so \(x=y\). Thus \(\Phi\) does not depend on the chosen representative.
For any two representatives, let \(x=\lim q_n\) and \(y=\lim r_n\). The inequality \(\big||q_n-r_n|-|x-y|\big|\le |q_n-x|+|r_n-y|\) shows that \(|q_n-r_n|\to |x-y|\). Consequently \(d_{\mathcal{C}}([(q_n)],[(r_n)])=|x-y|=|\Phi([(q_n)])-\Phi([(r_n)])|\). This proves that \(\Phi\) preserves distances, and in particular is injective: if its values agree, the distance between the classes is zero, so the classes are equal.
To prove surjectivity, take any \(x\in\mathbb{R}\). For each \(n\in\mathbb{N}\), choose an integer \(k_n\) with \(k_n\le 10^n x<k_n+1\), and put \(q_n=k_n/10^n\). The Archimedean property of the real numbers gives such an integer, including when \(x\) is negative. Then \(q_n\in\mathbb{Q}\) and \(0\le x-q_n<10^{-n}\), so \(q_n\to x\). Every convergent sequence is Cauchy, so \((q_n)\in\mathcal{C}\) and \(\Phi([(q_n)])=x\). Finally, for a constant sequence \(q_n=q\), its real limit is \(q\). This verifies the assertion about the canonical embedding and completes the proof. \(\square\)
This identification says more than that \(\mathbb{Q}\) sits densely inside \(\mathbb{R}\). It describes every point added by completion: it is represented by a rational Cauchy sequence, and its position on the real line is exactly that sequence’s real limit. Different sequences can represent the same point; what matters is their distance tending to zero, not term-by-term equality.
Constructing a Rational Sequence for \(\sqrt{2}\)
Worked Example: Bisection Approximations to \(\sqrt{2}\)
We construct rational numbers \(q_n\) by repeatedly bisecting an interval whose endpoint squares lie on opposite sides of \(2\). Start with \(I_0=[1,2]\). Given \(I_{n-1}=[a,b]\), take its midpoint \(q_n=(a+b)/2\). If \(q_n^2<2\), set \(I_n=[q_n,b]\); if \(q_n^2>2\), set \(I_n=[a,q_n]\). The comparisons are strict here; if equality occurred, the midpoint itself would be a rational square root and the process could stop.
The first steps are exact. For \(I_0=[1,2]\), the midpoint is \(q_1=3/2\), and \(q_1^2=9/4>2\), so \(I_1=[1,3/2]\). Its midpoint is \(q_2=5/4\), and \(q_2^2=25/16<2=32/16\), so \(I_2=[5/4,3/2]\). The next midpoint is \(q_3=11/8\), with \(q_3^2=121/64<2=128/64\), giving \(I_3=[11/8,3/2]\). Then \(q_4=23/16\), and \(q_4^2=529/256>2=512/256\), giving \(I_4=[11/8,23/16]\).
Each interval contains the root and has width \(2^{-n}\): the starting width is \(1\), and each bisection halves it. To see that \((q_n)\) is Cauchy, take \(m,n\ge N+1\). Both \(q_m\) and \(q_n\) lie in \(I_N\), so \(|q_m-q_n|\le 2^{-N}\). Given \(\varepsilon>0\), choose \(N\) with \(2^{-N}<\varepsilon\). Hence the sequence is Cauchy and defines a point in the completion.
In \(\mathbb{R}\), the sequence has a limit \(r\). For each fixed \(N\), all later midpoints lie in the closed interval \(I_N\), so \(r\in I_N\). The endpoints of \(I_N\) have squares on opposite sides of \(2\). As \(N\to\infty\), both endpoints converge to \(r\); continuity of squaring therefore gives \(r^2=2\). Also \(r\ge1\), so \(r\) is the positive square root \(\sqrt{2}\).
Check: the first four terms are \(3/2,5/4,11/8,23/16\). Their squares are \(9/4,25/16,121/64,529/256\), respectively; the first and fourth exceed \(2\), while the second and third are less than \(2\), matching the interval updates. Thus \((3/2,5/4,11/8,23/16,\ldots)\) represents the real number \(\sqrt{2}\).
A Second Irrational Limit
Worked Example: Bisection Approximations to \(\sqrt{3}\)
Use the same midpoint rule, now starting from \(J_0=[1,2]\) and comparing midpoint squares with \(3\). The initial midpoint is \(q_1=3/2\), and \(q_1^2=9/4<3\), so \(J_1=[3/2,2]\). Next \(q_2=7/4\), with \(q_2^2=49/16>3=48/16\), so \(J_2=[3/2,7/4]\). The next midpoint is \(q_3=13/8\), and \(q_3^2=169/64<3=192/64\), giving \(J_3=[13/8,7/4]\).
Then \(q_4=27/16\), and \(q_4^2=729/256<3=768/256\), so \(J_4=[27/16,7/4]\). The next midpoint is \(q_5=55/32\), since \((27/16+7/4)/2=(27/16+28/16)/2=55/32\). Its square is \(3025/1024<3=3072/1024\), giving \(J_5=[55/32,7/4]\).
Each \(J_n\) has width \(2^{-n}\), and the midpoint \(q_n\) lies in \(J_{n-1}\). For \(m,n\ge N+1\), both terms lie in \(J_N\), hence \(|q_m-q_n|\le2^{-N}\). Choosing \(N\) with \(2^{-N}<\varepsilon\) proves the sequence is Cauchy. Its real limit \(s\) lies in every \(J_N\). The endpoints converge to \(s\), while their squares bracket \(3\); continuity of squaring gives \(s^2=3\), and \(s\ge1\). Thus this rational Cauchy sequence represents \(\sqrt{3}\).
Check: the five midpoint squares are \(9/4,49/16,169/64,729/256,3025/1024\). Comparing with \(3\) gives, in order, less, greater, less, less, less, exactly the comparisons used to retain each interval.
These two limits are irrational. For \(\sqrt{2}\), if \(\sqrt{2}=a/b\) with relatively prime positive integers \(a,b\), then \(a^2=2b^2\), so \(a\) is even. Writing \(a=2k\) gives \(b^2=2k^2\), so \(b\) is even as well, contradicting relative primality. For \(\sqrt{3}=a/b\) in lowest terms, \(a^2=3b^2\) implies \(3\) divides \(a\); writing \(a=3k\) then implies \(b^2=3k^2\), so \(3\) also divides \(b\), again a contradiction. Thus these sequences converge in \(\mathbb{R}\) but not to any rational number. Their classes are points of the completion not in the canonical image of \(\mathbb{Q}\).
Worked Example: Two Representatives of the Same Rational Point
For \(n\in\mathbb{N}\), let \(q_n=1+1/n\) and \(r_n=1+1/(2n)\). Both sequences are rational. They converge in \(\mathbb{R}\) to \(1\), so the limit map assigns both classes to \(1\). Directly, \(|q_n-r_n|=|1/n-1/(2n)|=1/(2n)\to0\), so \((q_n)\sim(r_n)\). Neither sequence is the constant sequence \(1\): for every \(n\ge1\), \(1/n>0\) and \(1/(2n)>0\). Nevertheless, both represent the same completion point as that constant sequence.
Check: the distance from \(q_n\) to the constant sequence \(1\) is \(|q_n-1|=1/n\to0\); the distance from \(r_n\) to \(1\) is \(|r_n-1|=1/(2n)\to0\). Each class is therefore the canonical image of \(1\).
A Common Mistake
Common pitfall. It is wrong to require equivalent sequences to have equal terms, or even to agree from some point onward. For example, \(q_n=1+1/n\) is never equal to \(1\), but \(|q_n-1|=1/n\to0\). The sequences are equivalent by definition. The correct test is whether the distance between corresponding terms tends to zero.
Another tempting error is to suppose that every rational Cauchy sequence must converge to a rational number. The bisection sequences above disprove this: their real limits are \(\sqrt{2}\) and \(\sqrt{3}\), which are irrational. They are Cauchy as sequences in \(\mathbb{Q}\) because their terms become arbitrarily close to one another; completeness of \(\mathbb{Q}\) would additionally require a limit in \(\mathbb{Q}\), and that is precisely what fails.
Check Your Understanding
Use the definitions and the limit map to answer the following.
- Show that the sequence \(q_n=3+1/n\) represents the same completion point as the constant sequence \(3\).
- For the \(\sqrt{2}\) bisection, explain why \(q_2=5/4\) is retained as the lower endpoint after its square is compared with \(2\).
- Why does the estimate \(|q_m-q_n|\le2^{-N}\) for \(m,n\ge N+1\) prove that a bisection sequence is Cauchy?
- Explain why the limit map is injective, using the completion metric.
- For a given \(x\in\mathbb{R}\), why does choosing \(k_n\le10^n x<k_n+1\) produce a rational sequence converging to \(x\)?