Real powers from suprema of rational powers, with the reciprocal convention below base one

For $a>1$ and $x\in\mathbb R$, set $S_a(x):=\{a^q:q\in\mathbb Q,\ q<x\},\qquad a^{[x]}:=\sup S_a(x).$ The set is nonempty because a rational lies below $x$ (The rationals embed densely in the reals). It is bounded above: choose a natural $m>x$ (Every complete ordered field is Archimedean), so every $q<x$ satisfies $q<m$ and $a^q\le a^m$ (Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$). The supremum therefore exists in $\mathbb R$ by the least-upper-bound property of a complete ordered field (Complete ordered field (least-upper-bound property), Upper bound, least upper bound, and strict upper bound), and it is strictly positive, because it is at least the element $a^q$ of $S_a(x)$ for any rational $q<x$ and every rational power of a positive base is positive (Rational powers $a^r$ of a positive base).

For $0<a<1$, define $a^{[x]}:=1/\bigl((a^{-1})^{[x]}\bigr)$; for $a=1$, define $1^{[x]}:=1$. The notation $a^{[x]}$ distinguishes this rational-supremum construction from the exponential construction until their agreement is proved.

The direct supremum formula is intentionally restricted to $a>1$. When $0<a<1$, the set $\{a^q:q<x\}$ is unbounded above as $q$ tends to negative infinity.