KindiakMath

Defining the Rational Exponential

What does a^x mean? If a = 0, then 0^x = 0 no matter what x we choose (except perhaps x = 0, but that’s a debate for another day). Suppose a > 0. Let’s work with the easiest case: x being a positive natural number.

Definition 1. For any natural number n \geq 1, we define a^n inductively as follows:

a^1 := a,\quad a^{n+1} := a \cdot a^n, \quad n \in \mathbb N_0.

Here, 0 \in \mathbb N_0.

This definition captures the idea of exponentiation as repeated multiplication. Not too shabby. This immediately gives us three fundamental formulas that will motivate our various definitions later on.

Theorem 1. Let a be any real number. For natural numbers m \geq 1, n \geq 1,

a^{m+n} = a^m a^n, \quad (a^m)^n = a^{mn}.

Proof. Either use repeated multiplication, or more rigorously, mathematical induction.

We want to define a^x reasonably so that these rules are satisfied. We will abbreviate our exploration as follows.

Definition 2. Let a > 0 be a real number and K be a set. Define the predicates \phi, \psi on K^2 by

\begin{aligned} \phi(m,n) := (a^{m+n} = a^ma^n), \quad \psi(m,n) := ((a^m)^n = a^{mn}). \end{aligned}

We are now going to extend our definition of a^x, so that \phi \wedge \psi on \mathbb Q^2.

The most natural extension is to consider a^0. Whatever our definition of a^0 is, it should be consistent with

\phi(m,n) : \quad a^{m+n} = a^m a^n.

Theorem 2. Let a > 0 be a real number. If \phi(m,n) on \mathbb N_0^2, then a^0 = 1.

Proof. By assumption, since \phi(1,0),

\displaystyle a^1 = a^{1+0} = a^1 a^0 \quad \Rightarrow \quad a^0 = \frac{a^1}{a^1} = 1.

Definition 3. For any real number a > 0, a^0 := 1. Consequently, a^n > 0 for any n \in \mathbb N_0.

One can verify that \phi \wedge \psi on \mathbb N_0^2.

Now, recall the the predicate

\psi(m,n):\quad (a^m)^n = a^{mn}.

If \psi on \mathbb Q_{\geq 0}^2, then \psi(m/n, n) implies

\displaystyle (a^{\frac mn})^n = a^{\frac mn \cdot n} = a^m \quad \iff \quad a^{\frac mn} = \sqrt[n]{a^m}.

This gives us the following observation.

Theorem 3. Defining a^{\frac mn} := \sqrt[n]{a^m}, \phi \wedge \psi on \mathbb Q_{\geq 0}^2.

These definitions allow us to define a^x for rational x \geq 0. What about x \leq 0? If \phi(m, n) on \mathbb Q, then by considering \phi(-n,n),

\displaystyle 1 = a^0 = a^{-n+n} = a^{-n}a^n \quad \Rightarrow \quad a^{-n} = \frac{1}{a^n}.

Theorem 4. Defining a^{-n} := 1/a^n, \phi \wedge \psi on \mathbb Q^2.

Surprisingly, that is the straightforward part of the discussion. The real challenge begins now: what is a^x for real x, so that \phi \wedge \psi on \mathbb R^2? We need more tools using limits to establish this result.

—Joel Kindiak, 15 Dec 24, 0146H

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