Prove completeness of the constructed field

Try

A rational interval can locate a number that no rational equals. Start with [1,2] and ask for a number whose square is 2. Compare the midpoint square with 2 using integer arithmetic. Since (3/2)²=9/4 is too large, keep [1,3/2]. Since (5/4)²=25/16 is too small, keep [5/4,3/2]. Nothing here uses a square-root function or assumes the missing number already exists.

The preceding lessons built an ordered field E from rational Cauchy sequences. Two such sequences represent the same element when their difference tends to zero using rational tolerances. Write ι(q) for the constant-sequence class of a rational q. We now prove that this field has the completeness promised by the construction. A finite interval display illustrates the proof; it cannot replace its infinite argument.

Predict

Predict the next bracket before computing (11/8)². Then decide whether endlessly narrowing brackets prove existence merely because their widths get small. In Q they do not: the rational square-root obstruction remains. In E we can explicitly use the rational endpoint sequence as a new field element. The crucial difference is the definition of the ambient field, not a more accurate display.

Also predict what can go wrong with approximating every term xₙ of a Cauchy sequence by a rational within distance 2. Even for xₙ=0, the allowed rational choices (−1)ⁿ fail to be Cauchy. Approximation errors must tend to zero.

Experiment

Build a root from rational brackets

Find a positive number whose square is N using only rational arithmetic. Keep lower² < N < upper², test the midpoint, and retain the correct half. The integer comparisons below do not call a square-root function.

Target square N2357 Bisect once Restart brackets Reset

After 0 bisections: lower = 1/1; upper = 2/1. Exact width = 1/1.

Initial check: 1² < 2 < 2². Predict which half the first midpoint test will keep.

The graph is a rounded rendering of exact rational endpoints. It does not establish an infinite limit. The lesson proves the lower endpoints are rational Cauchy, constructs their class r, and proves r²=N. Given that proof, these brackets bound r with error at most 1/1; increasing the display count alone is not a completeness proof.

Inspect every rational bracket and square comparison

Integer arithmetic; fractions deliberately retain their common denominator

Step	Lower	Upper	Width	Midpoint test
0	1/1	2/1	1/1	Initial bracket

Predict the next midpoint decision before clicking. Explain why both endpoint sequences represent the same class. Then change N and check which parts of that proof are unchanged. All arithmetic comparisons stay within exact safe integers; controls are temporary and award no completion.

Use the square target 2. Record three midpoint comparisons, and explain which strict endpoint inequality each decision preserves. Compare your record with the exact table. For a changed experiment choose target 3, keeping the rational comparisons separate from any familiar decimal approximation.

Before asserting that bisection continues with strict inequalities, rule out an exact rational hit. If a rational p/q in lowest terms had square 2, then p²=2q² would force p and q both even. Thus no rational midpoint has square 2. This divisibility argument is about rational numbers alone; it does not assume the existence of r.

If lₙ,uₙ are these endpoints after n steps, all later endpoints lie in [lₙ,uₙ]. Their pairwise distances are at most 2⁻ⁿ, proving each endpoint sequence is rational Cauchy. Their mutual difference also tends to zero, so both define one class r. This is a construction of r, not a reference to an existing real limit.

Understand

First construct limits of sequences of classes. Let xₙ be Cauchy in E. The preceding order lesson gives rational density and proves directly that a class [a] is approximated by its late representative entries ι(aₙ). Choose rational qₙ with |xₙ−ι(qₙ)|<2⁻ⁿ. For positive rational ε, choose a common tail where each approximation error and the distance |xₙ−xₘ| are below ε/3. The triangle inequality then gives |qₙ−qₘ|<ε. Thus q is a rational Cauchy sequence and defines x=[q].

The representative-approximation lemma now gives |x−ι(qₙ)| tending to zero. Combining it with |xₙ−ι(qₙ)|<2⁻ⁿ proves xₙ tends to x. This does not invoke the result being proved: the representative lemma concerns a single rational Cauchy sequence and was obtained from its rational tail inequalities. Rational ε suffice for convergence in E because the Archimedean property gives a smaller positive rational below any positive field tolerance.

Selection need not hide a new choice assumption: enumerate Q and select the first rational meeting each approximation condition. Density guarantees a selection. This specifies a mathematical sequence; it does not assert that comparisons involving arbitrary input classes are computationally decidable. Another vanishing-error choice gives the same class by the triangle inequality. Limits are unique because two proposed limits can be put within arbitrarily small rational tolerances of the same late term.

Now derive least upper bounds. Let S be nonempty and bounded above in E. Choose s∈S and an upper bound B. Archimedean order supplies rational l₀<s and u₀>B. Maintain two different invariants: lₙ is not an upper bound of S, whereas uₙ is an upper bound. At the rational midpoint m, replace u by m if m is an upper bound; otherwise replace l by m. Classical excluded middle makes this a legitimate existence construction even when that predicate is not algorithmically decidable.

The nested rational intervals have width (u₀−l₀)2⁻ⁿ. Their endpoints are rational Cauchy and define the same class c. For fixed n, every later endpoint remains between lₙ and uₙ; preservation of non-strict order under these rational class limits gives lₙ≤c≤uₙ. Consequently both endpoint classes converge to c.

Prove the two supremum properties separately. If some s∈S exceeded c, choose a late uₙ closer to c than half that positive gap. Then uₙ<s, contradicting its upper-bound invariant. If v<c were an upper bound, choose a late lₙ>v. Since lₙ is not an upper bound, some s∈S exceeds lₙ and therefore v, again a contradiction. Hence c is an upper bound and every smaller candidate fails. This proves least-upper-bound completeness, not membership of c in S.

Finish the square-root construction algebraically. Its class r lies in [1,2]. The residual satisfies

0≤2−lₙ²≤uₙ²−lₙ²=(uₙ−lₙ)(uₙ+lₙ)≤4·2⁻ⁿ.

So lₙ²−2 is a null rational sequence. The already defined multiplication of classes gives r²=2. Two positive roots must agree because (r−s)(r+s)=0 and r+s>0. Finally, if r were a rational p/q in lowest terms, p²=2q² would force p even and then q even, a contradiction. This also shows no rational midpoint can meet the target exactly in this example. Other targets may terminate at an exact rational root, which is a valid distinct case.

Apply

Construct a positive cube root of 5 on paper, starting at [1,2]. The first two brackets are [3/2,2] and [3/2,7/4], since 27/8<5<343/64. Explain every later decision using rational cubes. An exact rational hit is impossible: p³=5q³ in lowest terms forces p divisible by 5, then q³=25(p/5)³ forces q divisible by 5 as well. The residual bound is now u³−l³=(u−l)(u²+ul+l²)≤12(u−l). This proves the limiting class has cube 5. Uniqueness follows from factoring a difference of cubes, with a strictly positive second factor. Neither derivatives nor the intermediate-value theorem are needed.

Repair an argument that begins “Bolzano–Weierstrass gives a real limit of every rational Cauchy sequence.” That theorem in the existing analysis course depends on real completeness. It is useful after construction but circular as its premise. Likewise, do not define the new supremum by taking a supremum in an already complete real field.

Continue

Return to ordering the constructed field if the rational-approximation lemma needs repair. Compare the result with the axiomatic completeness lesson: the earlier course assumes exactly the property now established for E.

Record separately which statements are definitions, rational estimates, ordered-field consequences and completeness conclusions. The new construction validates an ambient field for later sequence spaces; it does not replace the separate work of proving those spaces complete. Nor does a finite bisection trace establish the general construction on its own.