How do I prove exponent rules for all real numbers?

Question

It is easy to prove exponent rules for only the positive integers. For example, a^m.a^n = (a.a.a.a...)m times . (a.a.a.a.a....)n times so we just add the exponents m and n to get a^m+n. But how do I apply the same logic on rational or irrational or negative exponents ? How do I prove that these laws of exponents work for all real numbers NOT only just the positive integers ? for example (a.b)^n = a^n.b^n or any other law of exponents.

Answer 1

The short answer, as already mentioned in a comment, is to begin by defining the exponential function by $$\exp(x) = 1 + x + {x^2\over 2!} + {x^3\over 3!} + {x^4 \over 4!} + \cdots$$ for any real (or even complex) $x$. Then $\log(x)$ is defined to be the inverse of $\exp(x)$, and exponentiation is defined by $$ a^b := \exp(b \log(a)).$$ The rules of exponents can then be proved by reasoning with the power series of the exponential function.

Contrary to what some commenters have suggested, I do not think that it is easy to find a readable, self-contained account of all the details required to flesh out the above sketch. I would suggest starting with the Prologue to Walter Rudin's book Real and Complex Analysis. This is rather terse but it does at least sketch the main steps.

Another approach, which avoids the exponential function, is sketched in Exercise 6 of Chapter 1 of Rudin's book Principles of Mathematical Analysis. This approach first carefully defines what $a^n$ (and therefore $a^{1/n}$) means for any real $a$ and any positive integer $n$. Then it defines $a^b$ to be the least upper bound of all numbers $a^t$ where $t$ is a rational number less than $b$. The advantage of this approach is that you can see more explicitly how to build up everything step by step from the axioms for the real numbers. But it's the definition in terms of the exponential function that is ultimately more useful in practice.

Answer 2

Claim: Given $n\in N, x,c\in R, c>0, x^n=c$ has at least 1 solution over the reals.

Let $A=\{x\in R| x^n\le c\}$

$\forall n, 0^n=0<c \implies 0\in A$.

If $c=1$, $x=1$ and we have our solution.

Let $x,y \in R, 0<x<y. $

$x^2<xy.$

$xy<y^2$

So $x^2<y^2$.

More generally $x<y\implies x^n<y^n$.

If $c<1$ and $x<c$, $x^n<x<c<1$ so $A$ is bounded above by 1.

If $1<c, c<x, c<c^n<x^n, so \ x\notin A.$ So $x\le c$. So $A$ is bounded above by $c$.

So either $c=1$ or A is bounded above by 1 or $c$. By the Least Upper Bound property of the reals a non-empty subset that is bounded above has a least upper bound. The least upper bound of A is the root we seek. And it thus makes sense to let $x=c^{1/n}$. From there we have $m\in N, x^m=c^{m/n}$ by the usual exponent rules. We now have a definition for raising to a rational power.

By the Archimidean Property $\forall r\in R, n \in N, \exists m/n \in Q| m/n<r<(m+1)/n$.

$n$ can be made arbitrarily small and whatever value we assign to $a^r$ it's difference from $a^{m/n}$ can be made arbitrarily small. This allows assigning it a value of the limit of a base raised to a sequence of rational numbers.