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I have been trying to rigorously define real exponentiation. Online there doesn't seem to be ANY definition of real exponentiation that covers every case of base and exponent.

In school and on non-analytical media and textbooks, exponents are never defined past integers. $2^\frac{1}{2}$ will 100% always be introduced as $\sqrt{2}$ although the $n$th root function is never defined outside perfect squares etc.

In most real analysis media, there is the definition $b^n=e^{n\ln(b)}$, or in a non-circular notation, $b^n=\exp(n\ln(b)$. This is great but it doesn't handle bases $b\leq 0$.

In this post with the same name I tried, and failed to construct exponentiation over $\mathbb{R}$. This is my latest attempt. I want to know if it is flawed, and if so can you correct me.

$$b^n=\begin{cases} \exp(x\ln(b)) & 0<b\\ 0 & b=0\land n\neq 0\\ 1 & b=0\land n=0\\ \exp(x\ln(-b)) & b<0\land E(n)\\ -\exp(x\ln(-b)) & b<0\land O(n)\\ \text{undefined} & b<0\land \neg E(n)\land\neg O(n)\\ \end{cases}$$

Where $E(n)$ means $n$ is a rational number with an even numerator and an odd denominator:

$$E(n)\iff\exists m,k\in\mathbb{Z}(n=\frac{2m}{2k+1})$$

And $O(n)$ means $n$ is a rational number with an odd numerator and an odd denominator:

$$O(n)\iff\exists m,k\in\mathbb{Z}(n=\frac{2m+1}{2k+1})$$

The first case is the usual construction $b^n=e^{x\ln(b)}$, defining positive bases.

The second and third cases handle $0$ and are as controversial as zero itself.

Difficulty: Negative Bases

The third case covers expressions like $(-3)^2$ because $2=\frac{2(1)}{2(0)+1}$, as well as $(-27)^\frac{2}{3}$. This case breaks down to:

$$b^n=\exp(x\ln(-b))=(-b)^n$$

So, for example: $$(-3)^2=\exp(2\ln(--3))=\exp(2\ln(3))=3^2$$

Certain "even" rational exponents are also covered here.

$$(-27)^\frac{2}{3}=27^\frac{2}{3}=9$$

The fourth case preserves the sign of the base and covers numbers with "odd" rational exponents, like $(-27)^\frac{1}{3}$

$$(-27)^\frac{1}{3}=-\exp(\frac{1}{3}\ln(--27))=-\exp(\frac{1}{3}\ln(27))=-(27^\frac{1}{3})=-3$$

The fifth case is the reason for complex numbers.

Again, I want to know if my reasoning is flawed and if it is can you correct the mistake?

Isaac Sechslingloff
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  • No I am not defining $\exp$, I am defining $b^n$. – Isaac Sechslingloff Sep 06 '24 at 18:25
  • Which definition are you using for real numbers? One approach is to show that $a^b$ is well-defined if we treat $b$ as the limit of a rational sequence. – CyclotomicField Sep 06 '24 at 18:32
  • The equivalence class of cauchy sequences definition. – Isaac Sechslingloff Sep 06 '24 at 18:41
  • So I would use the rational sequence approach and stick with the positive numbers since negative bases won't be well-defined for irrational exponents. Show that if a cauchy sequence converges to $b$ then $a^b$ converges and likewise for $a$. – CyclotomicField Sep 06 '24 at 18:45
  • Aside from switching between $x$ and $n$ your logic seems sound.

    Maybe rethink why you are trying to do this. The notation $a^b$ is overloaded, it has different meanings depending on the context. For example $0^0$ you can more or less define as you want.

     – David Birkenmayer Sep 06 '24 at 19:41
  • Is exponentiation with negative base that important? I thought it only appears in calculus trap question...And $\ln$ of negative number does make sense if you don't choose the principal branch – Lxm Sep 07 '24 at 20:19

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