How is the definition for exponentiation extended to rationals and reals?

Question

If you have defined multiplication, division, addition and minus on all the real numbers, then you can define positive exponents by

$b^1=b$ and $b^{n+1}=b^n\cdot b$.

From this definition it is possible to prove the following identities for positive integers m and n:

$$b^{m+n}=b^m \cdot b^n$$

and

$$b^{m-n} = \frac{b^m}{b^n}$$

so to extend the definition of exponentiation to keep it consistent with the proved theorems $b^0 = 1$ and $b^{-n} = \frac{1}{b^n}$ are defined $n$ positive integer.

How the definition is extended to rational exponents and real numbers?

Answer 1

For rational exponents, they are extended by nth-roots, using $(\sqrt[n]{x})^n=x \rightarrow x^\frac{1}{n} = \sqrt[n]{x}$.

For real exponents, we need to use the definition of the reals. They are defined as the limit of rational series.¹

¹_{Real numbers are defined as an equivalence relation among the convergent series of the rational numbers, in which two series belong together if their difference converges to zero.}

Answer 2

When rigorously laying out foundations of all of analysis, the usual method is to use the definition

$$ a^b = \exp(b \log(a)) $$

While you could try and define exponentiation directly, it is somewhat awkward. Since you have to define $\exp$ and $\log$ anyways, it's most convenient to take advantage of that work, and the fact this identity is so simple.

This definition also has an advantage when you want to generalize beyond the case of positive $a$ and real $b$, since it reduces all of the complicated issues of multi-valuedness and what a complex exponent would even mean to the study of $\exp$ and $\log$ which are comparatively very straightforward.

Answer 3

$$b^{m/n}$$ is defined as the solution of the equation

$$x^n=b^m,$$

and

$$b^r$$ were $r$ is irrational as

$$\lim_{m/n\to r}b^{m/n}.$$