False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

FALSE: every real number has a real square root

Statement

False claim: every real number has a real square root, that is, for every $x \in R$ there exists $y \in R$ with $y^{2} = x$ .

The true statement is Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , which produces a square root only for $x \geq 0$ , and its generalisation Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , which produces an $n$ -th root only for $x \geq 0$ . The nonnegativity hypothesis in both is load bearing, not decoration.

Facts & Assumptions

Given: The complete ordered field $R$ (Complete ordered field (least-upper-bound property), Ordered field), with integer powers as in Integer powers $a^{m}$ .

[A1]

Every square is nonnegative: $y^{2} > 0$ for $y \neq 0$ (Squares of nonzero elements are positive, which states this and only this), and $0^{2} = 0 \cdot 0 = 0$ because a product with a zero factor vanishes (Multiplication by zero: $0 \cdot a = 0$ ); so $y^{2} \geq 0$ for every $y \in R$ .

[A2]

$1 > 0$ , hence $- 1 < 0$ ; and by trichotomy no element satisfies both $z \geq 0$ and $z < 0$ (The multiplicative identity is positive, Ordered field).

[A3]

Nonnegative reals do have roots: for $a \geq 0$ and $n \geq 1$ there is a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

Refutation

technique · contradiction

1.1

Assume, for contradiction, that every real has a real square root; applying this to $- 1 \in R$ produces $y \in R$ with $y^{2} = - 1$ .

assume-contragiven

2.1

By [A1] the element $y^{2}$ is nonnegative, so $- 1 = y^{2} \geq 0$ ; but $- 1 < 0$ by [A2], and no element is both $\geq 0$ and $< 0$ .

step 1.1A1A2

3.1

The obstruction is exactly the order, and it applies in every ordered field, not only in $R$ : completeness is never used, so no ordered field contains a square root of a negative element, and adjoining one, as happens in $C$ , necessarily destroys the order.

step 2.1A1A2

4.1

The assumption of step 1.1 therefore fails: there is no real $y$ with $y^{2} = - 1$ , so the claim that every real has a real square root is false, and the correct statements are [A3] with its hypothesis $a \geq 0$ kept.

step 2.1step 3.1step 1.1A3discharge-contradiction∎

Remarks

What fails is evenness of the exponent, not the taking of roots. Odd roots of negative numbers do exist in $R$ . The library has no general theory of parity, so fix the local abbreviation: call a natural $n$ odd when $n = 2 k + 1$ for some natural $k$ . Every odd $n$ satisfies $n \geq 1$ .
First, $(- 1)^{n} = - 1$ for odd $n$ , which is a computation and not an unstated induction: $(- 1)^{2 k + 1} = ((- 1)^{2})^{k} \cdot (- 1)$ by the addition and iterated-power laws for natural exponents (Laws of integer exponents, claim 1), $(- 1)^{2} = (- 1) (- 1) = 1$ ( $(- 1) (- 1) = 1$ ), and $1^{k} = 1$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , claim 4); so the product is $1 \cdot (- 1) = - 1$ . Consequently $(- u)^{n} = (- 1)^{n} u^{n} = - u^{n}$ for every $u \in R$ and odd $n$ (Laws of integer exponents, Sign rules for products: $(- a) b = - (a b)$ and $(- a) (- b) = a b$ ).
Every real is an $n$ -th power, for odd $n$ . For $a \geq 0$ take $x = a^{1 / n}$ ([A3], Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ). For $a < 0$ take $x : = - (- a)^{1 / n}$ , which is legitimate because $- a > 0$ (Sign rules for products and monotonicity of multiplication), and then $x^{n} = - ((- a)^{1 / n})^{n} = - (- a) = a$ by the previous item.
And the $n$ -th power map is injective for odd $n$ , which the strict increase on ${x \geq 0}$ alone does not give, since a sign statement is not a monotonicity statement. The map is strictly increasing on the whole line. On ${x \geq 0}$ that is Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , claim 2. If $x < y \leq 0$ then $0 \leq - y < - x$ , so $(- y)^{n} < (- x)^{n}$ by that same claim, and negating gives $x^{n} = - (- x)^{n} < - (- y)^{n} = y^{n}$ (Sign rules for products and monotonicity of multiplication). If $x < 0 \leq y$ then $- x > 0$ gives $(- x)^{n} > 0$ , so $x^{n} = - (- x)^{n} < 0$ , while $y^{n} \geq 0$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , claim 1). So $x < y$ always implies $x^{n} < y^{n}$ ; the map is injective, and with the surjectivity above it is a bijection of $R$ onto itself for every odd $n$ .
None of this rescues the even case, and that is the point of the item: for even $n$ , meaning $n = 2 k$ , the same computation gives $(- 1)^{n} = ((- 1)^{2})^{k} = 1^{k} = 1$ , so $x^{n} = (- x)^{n} \geq 0$ for every $x$ , powers of both signs land in ${x \geq 0}$ , and the refutation above applies verbatim with $n = 2$ .

Depends on

Used by

FALSE: a^m/n := (a^1/n)ᵐ extends to negative bases False statement

Dependency tree · two levels

35 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis I (standard reference, not scraped)
Radicals and rational exponents (Emory University) (standard reference, not scraped)
Nth root (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (standard reference, not scraped)
M. Spivak, Calculus, 4th ed., Ch. 1 (standard reference, not scraped)