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

Given: The base $a = -8$ and the rational $r = 1/3$; the formula under test is $a^{m/n} = \big(a^{1/n}\big)^{m}$, 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 $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$).

Numerals denote canonical naturals. For a natural $k$ the symbol $k$ inside $\mathbb{R}$ means $\iota(k) = k \cdot 1_{\mathbb{R}}$, where $\iota$ is the canonical order-preserving field embedding; $\iota(k) > 0$ for $k \ge 1$, and $\iota$ 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 $2^{3} = \iota(2)\iota(2)\iota(2) = \iota(8) = 8$ (Integer powers $a^m$). This is where the numerals of this item enter $\mathbb{R}$; the order on $\mathbb{Q}$ (Order on the rationals) is not what is being used when we write $-8 < 0$ in $\mathbb{R}$.

The same rational has many representatives: $1/3 = 2/6$ in $\mathbb{Q}$, 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.

No real has sixth power $-8$: for every $x \in \mathbb{R}$, $x^{6} = \big(x^{3}\big)^{2}$ (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 $x^{6} \ge 0$, whereas $-8 < 0$ in $\mathbb{R}$ by [A1].

Exactly one real has cube $-8$, namely $-2$: if $x^{3} = -8 < 0$ then $x < 0$, since $x \ge 0$ would give $x^{3} \ge 0$; 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 $a^{1/n} \ge 0$ 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: $(-a)b = -(ab)$ and $(-a)(-b) = ab$).