CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-07-26

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The truncated decimal approximations of $\sqrt{2}$ form a Cauchy sequence of rationals with no rational limit

Statement refuted

Refuted claim: every Cauchy sequence of rationals converges to a rational; equivalently, $Q$ is complete (FALSE: the rationals are complete, The rationals as equivalence classes of pairs of integers).

The witness is the sequence of truncated decimal approximations of $\sqrt{2}$ , $s_{0} = 1$ , $s_{1} = 1.4$ , $s_{2} = 1.41$ , $s_{3} = 1.414$ , and so on: $s_{n} = k_{n} / 10^{n}$ where $k_{n}$ is the largest natural number with $k_{n}^{2} \leq 2 \cdot 10^{2 n}$ . That this sequence is Cauchy in $Q$ and has no rational limit is proved in full in FALSE: the rationals are complete and is not repeated here.

What this item adds is the view from $R$ , which is what makes the witness informative rather than merely negative: the same sequence converges in $R$ , and its limit is $\sqrt{2}$ . So the defect is not in the sequence but in $Q$ , and the contrast is exactly The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges, which says that a Cauchy sequence of reals never behaves this way.

Facts & Assumptions

Given: For $n \in N$ the rational $s_{n} = k_{n} / 10^{n}$ , where $k_{n}$ is the largest natural with $k_{n}^{2} \leq 2 \cdot 10^{2 n}$ , together with the properties established for it in FALSE: the rationals are complete; and the real $\sqrt{2}$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ). Rationals are identified with their images in $R$ under the embedding $q \mapsto \hat{q}$ (The rationals embed densely in the reals), so $(s_{n})$ is also a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

The construction and its properties: $s_{n} \geq 0$ , $s_{n}^{2} \leq 2 < (s_{n} + 10^{- n})^{2}$ , and $(s_{n})$ is a Cauchy sequence of rationals with no rational limit (FALSE: the rationals are complete).

[L2]

The embedding $q \mapsto \hat{q}$ is an injective, order-preserving field homomorphism of $Q$ into $R$ , so every identity and inequality between rationals holds between their images and conversely (The rationals embed densely in the reals, The rationals as equivalence classes of pairs of integers).

[L3]

Square roots: $\sqrt{2} \geq 0$ is the unique nonnegative real with $(\sqrt{2})^{2} = 2$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Integer powers $a^{m}$ ).

[L4]

Powers and order: for $a, b \geq 0$ , $a \leq b$ exactly when $a^{2} \leq b^{2}$ , and $a < b$ exactly when $a^{2} < b^{2}$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ); and $(1 / t)^{n} = 1 / t^{n} = t^{- n}$ for $t \neq 0$ (Laws of integer exponents).

[L5]

Order: $10 > 1 > 0$ , so $0 < 1 / 10 < 1$ (Reciprocals and order: $1 / r$ against $1$ , Inverses of positives are positive, and reciprocation reverses order, Ordered field, Complete ordered field (least-upper-bound property)); and $∣ t ∣ = t$ for $t \geq 0$ (Basic properties of the absolute value).

[L6]

For $∣ r ∣ < 1$ the sequence $(r^{n})$ converges to $0$ (For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ ).

[L7]

Squeeze theorem (The squeeze theorem) and the algebra of limits (Algebra of limits: sums, scalar multiples, products and quotients); a constant sequence converges to its value (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

[L8]

Every convergent sequence of reals is Cauchy (Every convergent sequence is Cauchy); every Cauchy sequence of reals converges (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges); limits are unique (A sequence has at most one limit).

[L9]

No rational squares to $2$ (FALSE: some rational number squares to 2).

Counterexample

technique · direct

1.1

The inequalities of [L1] hold verbatim in $R$ , since the embedding preserves the order and the field operations.

givenL1L2

1.2

$0 < 1 / 10 < 1$ and $(1 / 10)^{n} = 10^{- n}$ for every $n$ .

givenL4L5

2.1

From $s_{n} \geq 0$ , $\sqrt{2} \geq 0$ and $s_{n}^{2} \leq 2 = (\sqrt{2})^{2}$ we get $s_{n} \leq \sqrt{2}$ ; from $(\sqrt{2})^{2} = 2 < (s_{n} + 10^{- n})^{2}$ with both bases $\geq 0$ we get $\sqrt{2} < s_{n} + 10^{- n}$ . Hence $0 \leq \sqrt{2} - s_{n} < 10^{- n}$ for every $n$ .

step 1.1L3L4

2.2

The sequence $(10^{- n}) = ((1 / 10)^{n})$ converges to $0$ .

step 1.2L6

3.1

The constant sequence $0$ and the sequence $(10^{- n})$ both converge to $0$ , and $0 \leq \sqrt{2} - s_{n} \leq 10^{- n}$ at every index, so the squeeze theorem gives $\sqrt{2} - s_{n} \to 0$ ; by the algebra of limits $s_{n} = \sqrt{2} - (\sqrt{2} - s_{n}) \to \sqrt{2}$ .

step 2.1step 2.2L7

4.1

In particular $(s_{n})$ converges in $R$ and is therefore Cauchy as a sequence of reals; this is the behaviour The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges guarantees for every Cauchy sequence of reals, and it is what fails in $Q$ .

step 3.1L8

4.2

Suppose $(s_{n})$ converged to a rational $q$ . Then in $R$ it converges to $\hat{q}$ , so $\hat{q} = \sqrt{2}$ by uniqueness of limits, hence ${\hat{q}}^{2} = 2$ ; the embedding is injective and preserves squaring, so $q^{2} = 2$ in $Q$ .

step 3.1L2L3L8

5.1

No rational squares to $2$ , so no such $q$ exists: $(s_{n})$ is a Cauchy sequence of rationals with no rational limit, and the claim that $Q$ is complete is refuted.

step 4.2L1L9∎

Remarks

The limit exists; it is merely not rational. That is the entire content of the counterexample and the reason the construction of $R$ is worth doing. The sequence is Cauchy in $Q$ , so $Q$ "should" have a limit for it, and the point at which it converges lies outside $Q$ .
Decimal truncation is a convenience, not the mechanism. Any sequence of rationals converging to any irrational does the same job, for instance the Babylonian iterates of The Babylonian sequence $x_{1} = 2$ , $x_{k + 1} = (x_{k} + 2 / x_{k}) / 2$ decreases to $\sqrt{2}$ started at $2$ , which are all rational and converge to $\sqrt{2}$ . Truncated decimals are chosen because the two-sided estimate $s_{n} \leq \sqrt{2} < s_{n} + 10^{- n}$ is immediate from the definition of $k_{n}$ and turns into convergence with one application of the squeeze theorem.
Note which completeness is which. FALSE: the rationals are complete refutes Cauchy completeness of $Q$ . $Q$ also fails the least-upper-bound property, on the same underlying fact that $\sqrt{2} \notin Q$ , and the two failures are not the same statement: Cauchy completeness and Dedekind completeness differ in general, and coincide only in the presence of the Archimedean property. Two independent proofs that $R$ is Cauchy complete, and why the library records both records where this library stands on that.

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Sources

Completeness of the real numbers (Wikipedia) (standard reference, not scraped)
Square root of 2 (Wikipedia) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §5.1 and §5.4 (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 and Ch. 3 (standard reference, not scraped)

The truncated decimal approximations of 22​ form a Cauchy sequence of rationals with no rational limit