Rational Cauchy classes and arithmetic

Try

Compare the rational sequences aₙ=2+1/n and bₙ=2+2/n, indexed by positive integers n. They are not the same sequence, but their difference has magnitude 1/n, which eventually falls below every positive rational tolerance. We want them to describe the same number. That requires a construction, not an appeal to an already existing real limit.

The constant sequence 2 should be a third description of that number. Predict whether changing just the first term of a to zero should change its class. It should not: arbitrarily late comparisons ignore any fixed finite prefix. But the zero first term blocks the naive instruction “take the reciprocal of every term.” We will repair that instruction only after proving it is legitimate on a tail.

For tolerance 1/100, choose an integer N>100. Then 1/n<1/100 whenever n≥N. This is a rational inequality justified by the unboundedness of the positive integers, not a real-limit theorem.

Predict

A rational sequence a is Cauchy if for every rational ε>0 there exists an integer N such that |aₘ−aₙ|<ε whenever m,n≥N. A rational sequence z is null if for every rational ε>0 there exists N such that |zₙ|<ε whenever n≥N. All entries, bounds and tolerances in this lesson belong to the rational numbers.

Define a∼b when a−b is null. A class [a] is the set of Cauchy sequences equivalent to a. Predict which proof is needed before writing [a][b]=[ab]. First, the product must remain Cauchy. Second, replacing a or b by an equivalent sequence must not change the product class. Checking only one of these obligations would not define multiplication on classes.

Also compare 1/n and (−1)ⁿ. The first has no zero term but represents the zero class. The second stays away from zero but is not Cauchy. Neither licenses the inverse construction for a nonzero Cauchy class.

Experiment

Modify the first term of 2+1/n to zero. Define the reciprocal sequence to be zero at that first index and n/(2n+1) afterward. Check that the termwise product equals 1 from the second index onward. Its difference from the constant sequence 1 is eventually zero, hence null. The repaired first term need not itself be an inverse.

Next verify, using rational algebra,

|n/(2n+1)−1/2|=1/(4n+2).

This identifies the reciprocal class with the constant rational 1/2. Try a different rational first value for the reciprocal. Explain why the class remains unchanged even though the finite table changes.

Finally multiply the null sequence 1/n by the unbounded sequence n. The product is constantly 1. This shows why the boundedness step in the product proof is essential. The unbounded sequence n is not Cauchy, so this example is a warning about a missing hypothesis, not a contradiction to the construction.

Understand

Boundedness and equivalence come first

Every rational Cauchy sequence is rationally bounded. Choose N for tolerance 1. For n≥N, |aₙ|<|a_N|+1. A positive rational bound larger than this number and the finitely many earlier magnitudes controls every term. No subsequence or limiting value has been used.

A null sequence is Cauchy: eventually each of |zₘ| and |zₙ| is less than ε/2, so |zₘ−zₙ|<ε. The relation ∼ is reflexive because a−a is zero, symmetric because negation preserves absolute values, and transitive by the triangle inequality with two ε/2 bounds. Finite changes preserve both Cauchyness and the equivalence class, because beyond a fixed index the altered sequences coincide.

Close the operations and remove dependence on representatives

Sums of Cauchy sequences are Cauchy by adding two ε/2 estimates. For products choose positive rational bounds A,B≥1 for the magnitudes of a,b. Then

|aₘbₘ−aₙbₙ|≤A|bₘ−bₙ|+B|aₘ−aₙ|.

Use tolerance ε/(2A) for b and ε/(2B) for a. Beyond the larger cutoff, the product difference is less than ε. Thus the termwise product is an allowed representative.

A null sequence multiplied by a bounded rational sequence is null: if |cₙ|≤C with rational C≥1, the eventual estimate |zₙ|<ε/C gives |cₙzₙ|<ε. If a∼a′ and b∼b′, the sum differences are null and

ab−a′b′=a(b−b′)+b′(a−a′)

is a sum of null sequences by this bounded-product lemma. Therefore [a]+[b]=[a+b] and [a][b]=[ab] are independent of representative choice. Likewise −[a]=[−a] is well-defined.

The constant sequences define 0, 1 and the rational embedding i(r)=[(r,r,…)]. If r≠s, their constant difference cannot be null: the rational tolerance |r−s|/2 already fails. Thus the embedding is injective and 0≠1. It preserves rational addition and multiplication. Associativity, commutativity, distributivity, additive inverses and identity laws hold at every sequence index by rational arithmetic, and therefore hold for the classes. Multiplicative inverses remain the one field obligation that these identities alone do not supply.

Nonzero classes have reciprocal tails

If [a]≠0, then a is not null. Negating the null definition gives a positive rational ε₀ such that every tail contains some term with magnitude at least ε₀. Cauchyness gives a cutoff N₀ after which any two terms differ by less than ε₀/2. Fix any n≥N₀ and choose a witness m≥N₀ with |aₘ|≥ε₀. Then

|aₙ|≥|aₘ|−|aₘ−aₙ|>ε₀/2.

Thus all sufficiently late terms are bounded away from zero by a positive rational r=ε₀/2. Non-nullity alone supplied recurring large terms; the Cauchy property converted that into an eventual bound for every term.

Define cₙ=1/aₙ on this tail and choose arbitrary rational values at the finitely many earlier indices. For late m,n,

|cₘ−cₙ|≤|aₘ−aₙ|/r².

The Cauchy cutoff for a with tolerance εr² proves that c is Cauchy. Its product with a equals 1 eventually, so [a][c]=1. Finite-prefix choices differ by an eventually zero sequence and hence do not affect the class.

If b∼a is another representative of the same nonzero class, it also has an eventual rational lower bound s>0. Their reciprocal difference is bounded by |aₙ−bₙ|/(rs), which is null. Thus the inverse is independent of representative choice too. We have constructed a field containing an injective copy of the rationals. Order and completeness still require their own proofs; no real limit was smuggled into this arithmetic construction.

Apply

Repair the false claim “any sequence with no zero terms has an inverse class.” The sequence zₙ=1/n is null, so its class is zero. Its termwise reciprocal n is not Cauchy: for every cutoff there are later indices with values differing by at least 1. The eventual lower-bound proof correctly refuses this case.

Repair a second claim: “being bounded away from zero is sufficient.” The sequence (−1)ⁿ has magnitude 1, but opposite-parity terms differ by 2 arbitrarily far out. It fails the Cauchy condition with rational tolerance 1 and is not an object in the construction.

For a fresh calculation, let d₁=d₂=0 and dₙ=3−1/n for n≥3. On the tail, its reciprocal is n/(3n−1). The exact difference from 1/3 is 1/[3(3n−1)], a null sequence; arbitrary rational repairs at indices 1 and 2 give the same reciprocal class. Thus [d]=i(3) and its inverse is i(1/3). Write the rational cutoff argument instead of appealing to a previously known real limit.

Continue

Keep the dependency order visible: rational Cauchy bounds, null equivalence, arithmetic closure, representative independence, rational embedding, and finally nonzero inverses. Field identities are inherited pointwise only after the operations have been proved well-defined.

The next lesson will define positivity and order on these classes. A representative that is eventually positive but approaches zero cannot define a positive nonzero class merely from its signs; a fixed positive rational separation is needed. Completeness will later establish that this constructed ordered field fills the gaps that rational arithmetic alone leaves. Those are subsequent arguments, not assumptions used above.