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

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?

• en.wikipedia.org/wiki/Exponentiation#Rational_exponents – Aretino May 21 '18 at 14:30
• The integer definition gives $b^{nm} = (b^n)^m=(b^m)^n$ for integers $n,m$, so perhaps we want a definition that allows $b^{p\frac{1}{q}} = (b^{1/q})^{p}=(b^p)^{1/q}$ (at least for $b>0$ and integers $p,q$, $q\neq 0$). – Michael May 21 '18 at 14:40

## 3 Answers

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.

• Why should $\sqrt[n]{x}$ exist? – egreg May 21 '18 at 15:32
• @egreg $\sqrt[n]{x}$ is defined as the real number, whose n-th power is $x$. There is no guarantee that it exists and there is no guarantee that it is unique, these are detailed in different theorems. The things are here simple only inside the world of the positive reals with positive integer $n$, but it is enough in the topic of this question. – peterh May 21 '18 at 15:37

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.

• This is a circular definition, as it involves the exponential of real numbers, not counting the logarithm which is not defined at this stage. – Yves Daoust May 21 '18 at 14:59
• @YvesDaoust This is not at all circular. The logarithm can be defined independently of exponentiation by $\log x=\int_1^x\frac{1}{t}\,dt$; it is an increasing differentiable function, so its inverse $\exp$ is differentiable and increasing. – egreg May 21 '18 at 15:30
• @YvesDaoust: ... or defining $\exp$ by power series, or by differential equation, or by the limit of $(1+\frac{x}{n})^n$ over integer $n$.... There are lots of paths to get there, and most of them don't involve exponentiation at all. In fact, the only one that does that comes to mind is is the one that defines $e$ as the limit of $(1+\frac{1}{n})^n$ and then $\exp(x)$ as $e^x$. – Hurkyl May 21 '18 at 15:37
• @Hurkyl: sure. All of this is missing in the answer. The OP being at a stage where only integer powers are defined should have asked what $\exp$ and $\log$ stand for. – Yves Daoust May 21 '18 at 15:39

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

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

and

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

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

• How do you prove that $x^n=b^m$ has a solution to begin with? – egreg May 21 '18 at 15:31
• @egreg: $x^n$ is a continuous function. – Yves Daoust May 21 '18 at 15:32
• You're doing calculus, aren't you? So now you have to prove that $\lim_{m/n\to r}b^{m/n}$ exists. Note that you have to prove that for every sequence of rationals $(q_n)$ converging to $r$ the limit exists and all such limits are equal. A way out is to prove existence and uniqueness of the extension of a uniformly continuous function on a dense subset of a closed and bounded interval. Not for the faint of heart: using the integral definition of the logarithm is way easier. – egreg May 21 '18 at 15:35
• Thanks, so is this saying that rational exponents require a new definition, as you have defined, eventually extending to the real numbers? Also, how has it happened that the theerems derived from the first definition now fit for all real numbers? Thanks – James Doherty May 21 '18 at 18:14
• @JamesDoherty: the extensions to rationals and reals were chosen in such a way that the properties continue to hold, this didn't happen by chance. – Yves Daoust May 21 '18 at 21:10