CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck 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, $\mathbb{Q}$ 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 $k_n^2 \le 2 \cdot 10^{2n}$ . That this sequence is Cauchy in $\mathbb{Q}$ 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 $\mathbb{R}$ , which is what makes the witness informative rather than merely negative: the same sequence converges in $\mathbb{R}$ , and its limit is $\sqrt 2$ . So the defect is not in the sequence but in $\mathbb{Q}$ , 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 $n \in \mathbb{N}$ the rational $s_n = k_n/10^n$ , where $k_n$ is the largest natural with $k_n^2 \le 2 \cdot 10^{2n}$ , together with the properties established for it in FALSE: the rationals are complete; and the real $\sqrt 2$ (Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$ ; the positives are $\{x^2 : x \neq 0\}$ ). Rationals are identified with their images in $\mathbb{R}$ 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: $s_n \ge 0$ , $s_n^2 \le 2 < (s_n + 10^{-n})^2$ , 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 $\mathbb{Q}$ into $\mathbb{R}$ , 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: $\sqrt 2 \ge 0$ is the unique nonnegative real with $(\sqrt 2)^2 = 2$ (Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$ ; the positives are $\{x^2 : x \neq 0\}$ , Integer powers $a^m$ ).

[L4]

Powers and order: for $a, b \ge 0$ , $a \le b$ exactly when $a^2 \le b^2$ , 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 $t \ne 0$ (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 $t \ge 0$ (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 $\mathbb{R}$ , 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 $s_n \ge 0$ , $\sqrt 2 \ge 0$ and $s_n^2 \le 2 = (\sqrt 2)^2$ we get $s_n \le \sqrt 2$ ; from $(\sqrt 2)^2 = 2 < (s_n + 10^{-n})^2$ with both bases $\ge 0$ we get $\sqrt 2 < s_n + 10^{-n}$ . Hence $0 \le \sqrt 2 - s_n < 10^{-n}$ 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 $0 \le \sqrt 2 - s_n \le 10^{-n}$ 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 $\mathbb{R}$ 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 $\mathbb{Q}$ .

step 3.1L8

4.2

Suppose $(s_n)$ converged to a rational $q$ . Then in $\mathbb{R}$ 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 $\mathbb{Q}$ .

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 $\mathbb{Q}$ 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 $\mathbb{R}$ is worth doing. The sequence is Cauchy in $\mathbb{Q}$ , so $\mathbb{Q}$ "should" have a limit for it, and the point at which it converges lies outside $\mathbb{Q}$ .
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 $s_n \le \sqrt 2 < s_n + 10^{-n}$ 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 $\mathbb{Q}$ . $\mathbb{Q}$ also fails the least-upper-bound property, on the same underlying fact that $\sqrt 2 \notin \mathbb{Q}$ , 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 $\mathbb{R}$ 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 2\sqrt{2}2​ form a Cauchy sequence of rationals with no rational limit