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

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FALSE: $a^{m / n} : = (a^{1 / n})^{m}$ extends to negative bases

Statement

False claim: the definition $a^{m / n} : = (a^{1 / n})^{m}$ of Rational powers $a^{r}$ of a positive base extends to negative bases, that is, the same formula assigns to every $a < 0$ and every $r \in Q$ a real number $a^{r}$ , depending only on $a$ and on the rational $r$ .

This is the claim that Rational powers $a^{r}$ of a positive base rules out by insisting on $a > 0$ , and this item is the reason for that restriction.

Facts & Assumptions

Given: The base $a = - 8$ and the rational $r = 1 / 3$ ; the formula under test is $a^{m / n} = (a^{1 / n})^{m}$ , in which $a^{1 / n}$ has to denote a real number whose $n$ -th power is $a$ (Rational powers $a^{r}$ of a positive base, Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ).

[A1]

Numerals denote canonical naturals. For a natural $k$ the symbol $k$ inside $R$ means $ι (k) = k \cdot 1_{R}$ , where $ι$ is the canonical order-preserving field embedding; $ι (k) > 0$ for $k \geq 1$ , and $ι$ preserves products (Canonical naturals are positive and strictly increasing, The unique embedding of ℚ into an ordered field). So $8 > 0$ , and therefore $- 8 < 0$ , since $x > 0$ means $x \in P$ and $0 - (- x) = x \in P$ says $- x < 0$ (Ordered field; none of the items just named states this passage from a positive element to its negative). Also $2^{3} = ι (2) ι (2) ι (2) = ι (8) = 8$ (Integer powers $a^{m}$ ). This is where the numerals of this item enter $R$ ; the order on $Q$ (Order on the rationals) is not what is being used when we write $- 8 < 0$ in $R$ .

[A2]

The same rational has many representatives: $1 / 3 = 2 / 6$ in $Q$ , since $1 \cdot 6 = 3 \cdot 2$ (The rationals as equivalence classes of pairs of integers). For a formula in $m$ and $n$ to define a function of $r$ , all representatives must give the same value, which for positive bases is Rational powers do not depend on the representative.

[A3]

No real has sixth power $- 8$ : for every $x \in R$ , $x^{6} = (x^{3})^{2}$ (Laws of integer exponents, claim 1, Integer powers $a^{m}$ ), and a square is nonnegative because a nonzero one is positive (Squares of nonzero elements are positive) while $0^{2} = 0 \cdot 0 = 0$ (Multiplication by zero: $0 \cdot a = 0$ ); so $x^{6} \geq 0$ , whereas $- 8 < 0$ in $R$ by [A1].

[A4]

Exactly one real has cube $- 8$ , namely $- 2$ : if $x^{3} = - 8 < 0$ then $x < 0$ , since $x \geq 0$ would give $x^{3} \geq 0$ ; and then $y : = - x > 0$ satisfies $y^{3} = - x^{3} = 8$ , so $y = 8^{1 / 3} = 2$ by uniqueness of the nonnegative cube root, whence $x = - 2$ (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , Sign rules for products and monotonicity of multiplication, Sign rules for products: $(- a) b = - (a b)$ and $(- a) (- b) = a b$ ).

Refutation

technique · contradiction

1.1

Assume, for contradiction, that the formula does define $a^{r}$ for negative $a$ and every rational $r$ , depending only on $a$ and $r$ ; then in particular $(- 8)^{1 / 3}$ is a real number, and the value obtained from any representative $m / n$ of the rational $1 / 3$ is that same number.

assume-contragivenA2

2.1

Read through the representative $1 / 3$ : the formula gives $(- 8)^{1 / 3} = ((- 8)^{1 / 3})^{1}$ , where $(- 8)^{1 / 3}$ is a real cube root of $- 8$ , and there is exactly one such real, namely $- 2$ ; so the value is $- 2$ .

step 1.1A1A4

2.2

Read through the representative $2 / 6$ : the formula gives $(- 8)^{2 / 6} = ((- 8)^{1 / 6})^{2}$ , and $(- 8)^{1 / 6}$ must be a real sixth root of $- 8$ , of which there is none.

step 1.1A1A3

3.1

The two readings are incompatible: by the assumption the rational $1 / 3 = 2 / 6$ has a single value, which step 2.1 computes to be $- 2$ , while step 2.2 shows that the very expression the formula prescribes for the representative $2 / 6$ names nothing at all in $R$ .

step 1.1step 2.1step 2.2A2

4.1

The assumption therefore fails, and the failure is not an artefact of the chosen numbers: every rational $r$ has representatives with even denominator, and a negative base has no real root of even order by the argument of [A3], so for a negative base the formula depends on the representative and Rational powers do not depend on the representative genuinely breaks down; this is exactly why Rational powers $a^{r}$ of a positive base requires $a > 0$ .

step 3.1step 1.1A2A3discharge-contradiction∎

Remarks

Precisely what fails. For a negative base the formula does not produce two different numbers; it produces a number from some representatives and nothing at all from others. That is still a failure of well-definedness: a definition of $a^{r}$ must depend only on the rational $r$ , and this one depends on how $r$ is written.
The odd-denominator repair, and why it is not adopted. If one restricts to rationals admitting a representative $m / n$ with $n$ odd, and always uses such a representative, the formula is consistent, because odd roots of negatives exist and are unique (FALSE: every real number has a real square root records this). What one gets is a partial operation, defined on the proper subset of $Q$ of rationals with odd denominator in lowest terms, not on $Q$ . The library does not adopt it: it is not the operation of Rational powers $a^{r}$ of a positive base, its exponents form only a proper subring of $Q$ (every rational whose lowest-terms denominator is even is missing, $1 / 2$ among them, so the square root that Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ supplies for nonnegative bases has no counterpart here), and every later use on this page, from Weighted AM-GM inequality with rational weights to Minkowski's inequality for finite sums (rational exponent), needs arbitrary rational exponents on a base that is nonnegative anyway.
The restriction to $a > 0$ is therefore not squeamishness about signs. It is the exact condition under which $a^{1 / n}$ exists for every $n \geq 1$ , which is what makes the value independent of the representative.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

45 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)
Exponentiation (Wikipedia) (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)

FALSE: am/n:=(a1/n)mam/n:=(a1/n)m extends to negative bases