Try
Can the sign of a number be read from the signs of its approximating terms? Consider the rational sequences aₙ=1/n and bₙ=−1/n, starting at n=1. Every term of a is positive and every term of b is negative. Yet their difference 2/n is null, so they represent the same class: zero. A rule that declares a class positive whenever its representative has positive terms is not well-defined.
The preceding lesson built a field F of rational Cauchy sequences modulo null differences. We now construct its order. Write [a] for a class and ι(q) for the class of the constant rational sequence q. All tolerances used in this construction are positive rational numbers. We do not assume the completeness, supremum property or Bolzano–Weierstrass theorem of the real numbers we are constructing.
The repair is a persistent gap: declare [a]>0 if some fixed rational r>0 satisfies aₙ≥r for all sufficiently large n. The gap must not shrink with n. This distinguishes 2−1/n, whose tail stays at least 1, from 1/n, whose positive terms shrink below every positive rational.
Predict
Predict which of these rational Cauchy representatives define positive, zero or negative classes. Give a tail estimate, not a graph-based guess.
| Representative, n≥1 | Tail certificate | Class sign |
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| 1/n | For each rational ε>0, n>1/ε gives |aₙ|<ε | Zero |
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| 2−1/n | aₙ≥1 for all n≥1 | Positive |
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| −3+1/n | −aₙ≥2 for all n≥1 | Negative |
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| (−1)ⁿ/n | |aₙ|=1/n is null | Zero despite alternating signs |
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Now replace a positive representative a by an equivalent b. If aₙ≥r eventually, nullity of a−b gives |aₙ−bₙ|<r/2 eventually. On their common tail bₙ>r/2. Thus positivity survives a change of representative. Reversing the roles proves the converse. Different representatives may require different tail indices; their class sign agrees.
Experiment
Work with x=[a], where aₙ=2+(−1)ⁿ/n. To compare x with the rational a₁₀=21/10, first prove x=ι(2): the difference aₙ−2 is null. Consequently |x−ι(a₁₀)|=1/10. At N=20 the error is 1/20. The pattern suggests that representative terms approach their own class, but this example alone is not the general proof.
Try a finite alteration: replace a₁ by 1000 and a₂ by −1000 while retaining the remaining terms. The class and every eventual sign certificate are unchanged. A proposed order based on the first term or the maximum displayed term would change, revealing why it cannot define the class order.
For a quantitative placement action, suppose two unknown classes satisfy y−x≥3/5 and |x−ι(7/4)|<1/20. Draw rational marks at 7/4−1/20, 7/4+1/20 and 41/20. Check that q=41/20 lies more than 1/4 above x and more than 1/4 below y. This is interval reasoning from proved inequalities, not a claim that x or y has a known decimal expansion.
Understand
Every nonzero class has exactly one sign. If a is not null, the negation of nullity supplies a positive rational ε₀ such that arbitrarily late indices j satisfy |aⱼ|≥ε₀. Choose a rational Cauchy tail where |aₙ−aₘ|<ε₀/2. For each n on that tail, choose a late witness j. The rational triangle inequality gives |aₙ|>ε₀/2. Put r=ε₀/2. On a later tail, require pairwise distances less than r. Two terms of opposite signs would instead differ by more than 2r, so all terms there have the same sign. Hence [a]>0 or −[a]>0.
A null sequence cannot stay above a positive rational gap. Both positive and negative class signs cannot hold together, since representative independence would force one representative above r and below −s on a common tail. These observations prove trichotomy: exactly one of x=0, x>0 and −x>0 holds. No real limit was used to select the sign.
Construct the ordered-field rules. The sum of representatives eventually bounded below by r and s is bounded below by r+s. Their product is bounded below by rs. Thus positive classes are closed under addition and multiplication. Define x<y when y−x>0. Adding positive differences proves transitivity; translation cancels in the difference; multiplication by a positive class preserves positivity of the difference. Together with trichotomy these are the strict total order and ordered-field compatibility laws. Define x≤y to mean x<y or x=y.
The constant class ι(q) is positive exactly when the rational q is positive. Therefore the existing rational field embedding also preserves and reflects order. If x≠0, either x or −x is positive, so x²>0. If x>0, its inverse is nonzero and cannot be negative: multiplying that proposed negative inverse by x would make the rational constant 1 negative. Thus x⁻¹>0.
Absolute values and non-strict inequalities. Define |[a]|=[|aₙ|]. The rational inequality ||r|−|s||≤|r−s| proves that absolute values preserve both the Cauchy property and equivalence. If a representative is eventually nonnegative, its class is nonnegative: a negative class would force the same representative below a fixed negative bound eventually. This one-way rule lets rational tail inequalities pass to classes. Its converse for every representative is false, as −1/n represents the nonnegative class zero.
For a positive class, the representative is eventually positive, so |x|=x; for a negative class, |x|=−x; for zero, absolute values remain null. Applying the rational triangle and reverse triangle inequalities term by term now proves |x+y|≤|x|+|y| and ||x|−|y||≤|x−y|. These order statements have been derived rather than borrowed from an already complete real line.
Archimedean bounds. A rational Cauchy sequence has a rational absolute bound B: choose a tail within 1 of one fixed term, then enlarge its rational bound to cover the finitely many earlier terms. Choose an integer N>B using the Archimedean property of the rationals. The sequence N−|aₙ| is bounded below by the positive rational N−B, so |[a]|<ι(N). Thus no class exceeds every integer in magnitude.
Representative terms approximate their class. Given rational ε>0, choose N₀ so |aₘ−a(N)|<ε/4 for all m,N≥N₀. Fix any N≥N₀. The difference x−ι(a(N)) is represented, with index m, by aₘ−a(N). Consequently ι(ε)−|x−ι(a(N))| has a representative eventually greater than 3ε/4, so it is positive. We have proved
|x−ι(a(N))|<ι(ε) for every N≥N₀.
Notice the two indices: N chooses one rational approximation; m runs through a representative of the difference class. Confusing them would replace a uniform tail argument with a single equality at m=N. This lemma constructs approximation inside F without presupposing that rational Cauchy sequences converge in a pre-existing real field.
Apply
Finish the density proof. If x<y, positivity gives a positive rational δ and a representative of y−x eventually at least δ. The nonnegative-tail rule yields y−x≥ι(δ). Choose a representative term p of x with |x−ι(p)|<δ/4, and set q=p+δ/2. Then q−x lies strictly between δ/4 and 3δ/4. Subtract from the gap y−x≥δ to get y−q>δ/4 too. Hence x<ι(q)<y.
Check every inequality for the numerical placement in Experiment: q−x is strictly between 1/4 and 7/20, while y−q>3/5−7/20=1/4. Both strict margins are required; merely landing on an endpoint would not prove density.
Apply density to 0<z. It supplies a positive rational below z. Therefore rational tolerances suffice to test arbitrary positive class gaps: if a nonnegative class is smaller than every positive rational, it must be zero. Later convergence arguments can use this consequence after proving their estimates, instead of silently importing a metric or order completeness theorem.
Continue
We now have an Archimedean ordered field containing an order-preserving copy of the rationals, together with an explicit approximation lemma. We have not yet proved that every Cauchy sequence of classes has a limit, or that every bounded set has a supremum. Those are the next construction steps. Keep the representative index distinct from the approximation index when building a diagonal rational sequence; the proof above provides exactly the rational error control that step requires.
