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Exponential function with $a>0$, $x\rightarrow a^{x}$ for all $x\in \mathbb{R}$It is a construction starting from natural to real, but...

What happens when $0>a\in \mathbb{R}$ and $x\in \mathbb{R}$, my question is what happens with these cases $(1.)$ $(-1)^{1/3},(-1)^{2/6}$ are the same?, $(-8)^{1/3},(-8)^{2/6}$ are the same?

I think so, but I don't know how to justify, since it is not fulfilled $(a^{x})^{y}=a^{xy}$ when $a<0$, $x,y\in \mathbb{R}$ (2.Why?) I saw a solution here but I don't know where that sign comes from '' - '' $$ −1 = (−1)^3 = (−1)^{6 / 2} \color{red}= -\sqrt{(-1)^6} = -\sqrt{1} = -1$$ I'm not sure if this sign results from 'modifying' the theorem, if so, how would it be?

If $b$ is a positive real number and $n$ is a positive integer, then there is exactly one positive real solution to $x^{n} = b$

I hope your help, thanks in advance. Well, i want to understand about $a^{x}$ ,$a<0$ which properties meet or don't meet, for example taking $(1+1/x)^{x}$ tends to $e$ when $x\rightarrow \infty$, what happen if 'take a photo' in $x=-1/4$,there are many questions, if you could give me any notes or text other than wikipedia, thanks.

user475475
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    You have to go to complex numbers to define $a^{x}$ for $a <0$. There will be non-trivial problems there. You will get multiple values for $a^{x}$ and you will have to wait till you study Complex Analysis to understand how $a^{x}$ behaves. The rules applicable to the case $a>0 $ will not be applicable here. – Kavi Rama Murthy Nov 24 '19 at 23:14
  • Yes, but what I want is to understand what is 'wrong', without using complex analysis, 1. is true, right? – user475475 Nov 24 '19 at 23:27

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If you want a function defined on a subset of the real numbers, then $a^x$ is very poorly behaved when $a<0$. The properties of exponents may not apply, and you can make false proofs such as $$1=\sqrt 1=\sqrt{(-1)^2}=\left(\sqrt{-1}\right)^2=-1.$$

In particular, the property $\sqrt{ab}=\sqrt a\sqrt b$ does not hold for all real numbers $a,b$.

It is generally true that you can write $a^{p/q}$ for $a<0$ as long as $q$ is odd. However, as you noticed, this statement is sort of nonsensical since $(-1)^{1/3}$ makes sense but $(-1)^{2/6}$ does not. As a small victory, we can always require that fractions are fully simplified so that at least the existence of $a^{p/q}$ is well-defined. However, no matter what, $a^x$ will not be a very nice function because it turns out that every real number is arbitrarily close to a rational number with reduced form $p/q$, with $q$ even. In other words, $a^x$ still can not be defined on any interval of the real numbers.

There are two solutions: for single numbers, we can use context to make sense of $(-8)^{1/3}$, for example. We also write $a^{1/3}=\sqrt[3]a$ to help avoid ambiguity. More generally, we need to move into the complex realm in order to understand $a^x$ for $a<0$, or really $a^z$ for any complex numbers $a,z$. (Except, perhaps $0^0$, which we still haven't really decided on.)

pancini
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  • And how would that be written in complexes or what would be its solutions there? I read that a bit, but could you explain to me, please, thank you. – user475475 Nov 25 '19 at 20:43