migongoniwt

2022-06-23

How to define an irrational to the power an irrational?

aletantas1x

Beginner2022-06-24Added 22 answers

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,

${x}^{\frac{k}{l}}=({x}^{\frac{1}{l}}{)}^{k}$

Then, notice that for any $x$, and integer exponents, ${x}^{ab}=({x}^{a}{)}^{b}=({x}^{b}{)}^{a}$, 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

$\begin{array}{rcl}({x}^{r}{)}^{qn}& =& ({x}^{\frac{p}{q}}{)}^{qn}=({x}^{\frac{1}{q}}{)}^{qpn}={\textstyle [}({x}^{\frac{1}{q}}{)}^{q}{{\textstyle ]}}^{pn}={x}^{pn}\end{array}$

But also

$\begin{array}{rcl}({x}^{r}{)}^{qn}& =& ({x}^{\frac{m}{n}}{)}^{nq}=({x}^{\frac{1}{n}}{)}^{nmq}={\textstyle [}({x}^{\frac{1}{n}}{)}^{n}{{\textstyle ]}}^{mq}={x}^{mq}\end{array}$

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:

${a}^{x}=sup\{{a}^{r}:r\in \mathbb{Q},r\le x\}$

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,

${x}^{\frac{k}{l}}=({x}^{\frac{1}{l}}{)}^{k}$

Then, notice that for any $x$, and integer exponents, ${x}^{ab}=({x}^{a}{)}^{b}=({x}^{b}{)}^{a}$, 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

$\begin{array}{rcl}({x}^{r}{)}^{qn}& =& ({x}^{\frac{p}{q}}{)}^{qn}=({x}^{\frac{1}{q}}{)}^{qpn}={\textstyle [}({x}^{\frac{1}{q}}{)}^{q}{{\textstyle ]}}^{pn}={x}^{pn}\end{array}$

But also

$\begin{array}{rcl}({x}^{r}{)}^{qn}& =& ({x}^{\frac{m}{n}}{)}^{nq}=({x}^{\frac{1}{n}}{)}^{nmq}={\textstyle [}({x}^{\frac{1}{n}}{)}^{n}{{\textstyle ]}}^{mq}={x}^{mq}\end{array}$

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:

${a}^{x}=sup\{{a}^{r}:r\in \mathbb{Q},r\le x\}$