ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (openai/gpt-5.4)audited 2026-07-25

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$\sqrt{2}$ exists in every complete ordered field, and is irrational

Example

In any complete ordered field $F$ , the element $2 = 1 + 1$ is positive, so by Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$ ; the positives are $\{x^2 : x \neq 0\}$ applied to $a = 2$ it has a unique $s > 0$ with $s^2 = 2$ : this is $\sqrt{2}$ . Moreover $s$ is not the image of any rational under the embedding $\iota : \mathbb{Q} \to F$ , because no rational squares to $2$ . Thus every complete ordered field contains $\sqrt{2}$ , the canonical gap that $\mathbb{Q}$ lacks, now filled by completeness.

Facts & Assumptions

Given: A complete ordered field $F$ (Complete ordered field (least-upper-bound property)) with unit $1$ ; write $2 := 1 + 1$ . In any ordered field $1 > 0$ , hence $2 = 1 + 1 > 0$ .

[L1]

Every $a \ge 0$ in $F$ has a unique $s \ge 0$ with $s^2 = a$ (Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$ ; the positives are $\{x^2 : x \neq 0\}$ ).

[L2]

There is a unique field homomorphism $\iota : \mathbb{Q} \to F$ ; it is injective and order-preserving, and satisfies $\iota(1) = 1$ (The unique embedding of ℚ into an ordered field).

[L3]

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

Verification

technique · direct

1.1

In $F$ we have $2 = 1 + 1 > 0$ , so in particular $2 \ge 0$ .

given

2.1

Apply Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$ ; the positives are $\{x^2 : x \neq 0\}$ [L1] with $a = 2$ : there is a unique $s \ge 0$ with $s^2 = 2$ , and $s \neq 0$ since $s^2 = 2 > 0$ , so $s > 0$ ; write $\sqrt{2} := s$ .

L1step 1.1

3.1

The element $s$ is not rational: if $s = \iota(q)$ for some $q \in \mathbb{Q}$ , then $\iota(q^2) = \iota(q)^2 = s^2 = 2 = \iota(1) + \iota(1) = \iota(2)$ , so injectivity of $\iota$ [L2] forces $q^2 = 2$ , which is impossible by [L3]; hence $s$ lies outside $\iota(\mathbb{Q})$ .

L2L3step 2.1

4.1

Therefore every complete ordered field contains a unique positive $s = \sqrt{2}$ with $s^2 = 2$ , and this $s$ is irrational: it is exactly the gap in $\mathbb{Q}$ that completeness fills.

step 2.1step 3.1∎

Depends on

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Direct dependencies and their dependencies through the next three levels: 29 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (Thm 1.21, Cor) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §5.5 (standard reference, not scraped)
University of Colorado analysis notes: The real numbers (standard reference, not scraped)
Elias Zakon, Mathematical Analysis: Irrational numbers (standard reference, not scraped)

2\sqrt{2}2​ exists in every complete ordered field, and is irrational