The exponential function and powers with real exponents

The power with a real exponent

In the lesson on powers with rational exponents, $a^x$ was defined for $a > 0$ and rational $x$, and it was proved that the definition does not depend on the fraction used to write the exponent, that the laws of exponents hold and that, if $a > 1$, the power increases with the exponent: if $r < s$ are rational, $a^r < a^s$. On the other hand, an expression such as $2^{\sqrt{2}}$ is left without meaning, because $\sqrt{2}$ is not a fraction and no root or integer power defines it.

Now, $\sqrt{2}$ lies above the rationals $1$, $\frac{7}{5}$, $\frac{141}{100}$, $\frac{1414}{1000}$, which approach it from below, and the powers $2^1$, $2^{\frac{7}{5}}$, $2^{\frac{141}{100}}$, $2^{\frac{1414}{1000}}$ form an increasing sequence, whose values are approximately 2, 2.639, 2.657 and 2.665. None of them exceeds $2^{\frac{3}{2}} = \sqrt{8}$, because all those exponents are less than $\frac{3}{2}$. If this is so, the natural course is to define $2^{\sqrt{2}}$ as the least number that is not exceeded by any power $2^r$ with $r$ rational and $r \leq \sqrt{2}$, that is, as a supremum; its existence is guaranteed by the completeness of the reals, presented in the lesson on number systems.

The definition by means of suprema

Let $a > 1$ and let $x$ be a real number. We call $A_x$ the set of the powers $a^r$ with $r$ rational and $r \leq x$. The power with a real exponent $a^x$ is the supremum of $A_x$. If $0 < a < 1$, one defines $a^x = \left(\frac{1}{a}\right)^{-x}$, where the base $\frac{1}{a}$ is greater than $1$; and one defines $1^x = 1$ for every real $x$.

The definition contains two claims that must be proved: that the supremum exists and that, if $x$ is rational, the new number $a^x$ is the same as the one already known. Without the second, the symbol $a^x$ would have two meanings for rational exponents.

The definition for bases less than $1$ also extends the earlier one: if $0 < a < 1$ and $r$ is rational, by the laws of rational exponents $\left(\frac{1}{a}\right)^{-r} = \frac{1}{\left(\frac{1}{a}\right)^r} = a^r$. Moreover, the definition directly yields a rule that will be used often: for every base $a > 0$ and every real $x$, the power of the reciprocal base is the power with the opposite exponent. Indeed, if $a > 1$, the base $\frac{1}{a}$ is less than $1$ and, by definition, $\left(\frac{1}{a}\right)^x = a^{-x}$; if $0 < a < 1$, the base $\frac{1}{a}$ is greater than $1$, and the definition of $a^{-x}$ is precisely $\left(\frac{1}{a}\right)^{x}$.

Monotonicity

The first use of the definition is to place $a^x$ among the rational powers: with a base greater than $1$, every rational exponent less than $x$ gives a power less than $a^x$, and every greater rational exponent, a greater power. From this follows monotonicity for arbitrary real exponents.

If $a > 1$, monotonicity can be read in both directions: for arbitrary reals $u$ and $v$, $u < v$ if and only if $a^u < a^v$, and $u = v$ if and only if $a^u = a^v$. Indeed, if we had $u \geq v$, we would have $a^u \geq a^v$. With a base between $0$ and $1$, the equivalence holds with the inequality reversed. This is how powers with the same base are compared, by comparing their exponents.