migongoniwt

Answered question

2022-06-23

How to define an irrational to the power an irrational?

Answer & Explanation

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,
${x}^{\frac{k}{l}}=\left({x}^{\frac{1}{l}}{\right)}^{k}$
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:
${a}^{x}=sup\left\{{a}^{r}:r\in \mathbb{Q},r\le x\right\}$

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?