migongoniwt

2022-06-23

How to define an irrational to the power an irrational?

aletantas1x

First, show there's an $n$th root by considering the l.u.b. of numbers whose $n$th powers are no more the target number.
Then, show that the Then, show that the nth root is unique, by noticing that nth power is one-to-one function.th root is unique, by noticing that Then, show that the nth root is unique, by noticing that nth power is one-to-one function.th power is one-to-one function.
Define, temporarily,
${x}^{\frac{k}{l}}=\left({x}^{\frac{1}{l}}{\right)}^{k}$
Then, notice that for any $x$, and integer exponents, ${x}^{ab}=\left({x}^{a}{\right)}^{b}=\left({x}^{b}{\right)}^{a}$, because this kind of exponentiation is defined by multiplication.
Therefore if $r$ is rational, with two different representations $r=\frac{p}{q}=\frac{m}{n}$, then
$\begin{array}{rcl}\left({x}^{r}{\right)}^{qn}& =& \left({x}^{\frac{p}{q}}{\right)}^{qn}=\left({x}^{\frac{1}{q}}{\right)}^{qpn}=\left[\left({x}^{\frac{1}{q}}{\right)}^{q}{\right]}^{pn}={x}^{pn}\end{array}$
But also
$\begin{array}{rcl}\left({x}^{r}{\right)}^{qn}& =& \left({x}^{\frac{m}{n}}{\right)}^{nq}=\left({x}^{\frac{1}{n}}{\right)}^{nmq}=\left[\left({x}^{\frac{1}{n}}{\right)}^{n}{\right]}^{mq}={x}^{mq}\end{array}$
Now, $np=mq$, since the fractions are equal, so both ways of expressing $r$ as a fraction give the same value for of ${x}^{r}$ (again because raising to integer exponents is one-to-one). Since there's a well-defined rational exponent, all you have to do is define:
${a}^{x}=sup\left\{{a}^{r}:r\in \mathbb{Q},r\le x\right\}$

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