Explicit rational and real power regimes

For $a>0$, integer $m$ and integer $n\ge 1$, define the rational regime by

$$a^{m\mathrm{/}n}:=(a^{1\mathrm{/}n}{)}^m,$$

where $a^{1\mathrm{/}n}$ is the unique positive root of $s^n=a$ in Existence and uniqueness of $n$-th roots: a unique $a^{1\mathrm{/}n}\ge 0$ with $(a^{1\mathrm{/}n}{)}^n=a$, and integer powers include reciprocals for negative $m$. This is Rational powers $a^r$ of a positive base; Rational powers do not depend on the representative proves independence of the fraction representative and Laws of rational exponents gives the rational exponent laws. For the rational-exponent regime at base zero, retain the convention of Rational powers $a^r$ of a positive base and Integer powers $a^m$: $0^0=1$, $0^r=0$ for rational $r>0$, and $0^r$ is undefined for rational $r<0$. This records the integer zero-exponent convention together with the positive-rational supplementary clause; it does not extend the real-exponent convention.

The separate real-exponent regime is Real powers for positive bases, with the zero-base positive-exponent convention: for $a>0$ and $x\in \mathbb{R}$, $a^x=\exp (x\log a)$, with the laws of The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents. For real exponents at base zero, set $0^x=0$ only when $x>0$; $0^0$ and $0^x$ for $x<0$ are undefined, exactly as in Real powers for positive bases, with the zero-base positive-exponent convention. The rational and real exponent operations are separate definitions; at zero base and exponent zero, state which regime is intended: the rational/integer convention gives $1$, while the real-power convention leaves the expression undefined. No arbitrary real power is assigned to a negative base. The real-power laws are supplied by that theorem, not inferred from the rational root formula. Use the real-number structure of The real numbers.

In particular $s(q)=q^{1\mathrm{/}2}$ for $q>0$ is the unique positive solution of $s(q{)}^2=q$. Its smoothness and derivative are a separate well-definedness result, not an assumption about fractional notation.