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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 $R$ denotes the image $k \cdot 1_{R}$ , where $ι : Q \to R$ is the canonical order-preserving field embedding; these field images are positive for natural $k \geq 1$ , 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 $2^{3} = 2 \cdot 2 \cdot 2 = 8$ (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

1.1assume-contragivenA2

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.

2.1step 1.1A1A4

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 $- 2$ .

2.2step 1.1A1A3

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.

3.1step 1.1step 2.1step 2.2A2

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 $R$ .

4.1step 3.1step 1.1A2A3discharge-contradiction∎

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 $a > 0$ .

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.
Nonnegativity $a \geq 0$ is the exact condition under which real $n$ -th roots exist for every $n \geq 1$ : zero has root zero at every order (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ), while a negative base has no square root. Strict positivity $a > 0$ is needed to allow every rational exponent, including negative ones, because the root must then have an inverse. The supplementary zero-base clause of Rational powers $a^{r}$ of a positive base instead covers positive rational exponents.

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

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

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