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let $a\in \mathbb{R},x\in \mathbb{R}\backslash \mathbb{Q}$ .

now : how is defined $a^x$ ?(without using Logarithm)


i know that : if : $a>1$ then: $$a^x:=\sup\{a^r:r\in \mathbb{Q},r<x\}=\inf\{a^r:r\in \mathbb{Q},r>x\}$$

and if $b\in(0,1)$. then $\dfrac{1}{b}>1$ and :

$$b^x:=(\dfrac{1}{b})^{-x}$$.


my question :

Can be defined $a^x$ such that : $a<0 \in \mathbb{R} ,x$ be Irrational number .

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  • $\begingroup$ (+1) Interesting and seems to work. Maybe you should add that $a<0$ in the title. $\endgroup$ – Yanko Jun 10 '17 at 21:41
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    $\begingroup$ No, if $a<0$ then $a^x$ is not defined among the reals. As a complex number, it can take though infinitely many (but nonreal) values, all having absolute value $|a|^x$ and angles $(2k+1)\pi\cdot x$ for $k\in\Bbb Z$. $\endgroup$ – Berci Jun 10 '17 at 21:43
  • $\begingroup$ Why is it only defined for odd numbers? $\endgroup$ – Yanko Jun 10 '17 at 22:11
  • $\begingroup$ Don't need the two definitions for a>1, b < 1. Just do a^x = lim q->x a^q. Doesn't matter if a <,=,or > 1. Can not define for a < 0 because limits don't exist. $\endgroup$ – fleablood Jun 10 '17 at 22:38
  • $\begingroup$ @fleablood But that means you expect the limit to exist, which is a too strong assumption for the generalization! $\endgroup$ – yo' Jun 12 '17 at 22:48

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