Rational exponents: prove some states

Question

In some rational exponent expressions the solution isn't a real number why?

Example (explain what I mean):

$$\begin{align} \Big(-x\Big)^{1/n}=\left\{\text{is not a real number}\right\} \end{align}$$

Such that $x$ is any positive integer number and $n$ is an even integer nonzero number.

Then the expression above can't be solved in real numbers. How to prove that?

Answer 1

There are a kind of numbers which called complex numbers.as you know you can write ${a}^{1\over b} $ in this form : $ \sqrt[b]{ a}$ now if $a<0 $ a Complex number will make.for example $ i = \sqrt[]{ -1}$ is a complex number.the complex number usually appeared in the polynomials and sometimes in linear algebra.

Answer 2

Since for real numbers $a$ and $b$, we want to define $a^b$ by $$ a^b := \exp(b \log a), $$ we see that we must restrict $a$ to be a positive real number, since $\log : \mathbb{R}_{>0} \to \mathbb{R}$ is undefined on $\mathbb{R}_{\leq 0}$.

Note that this definition of $a^b$ for $a >0$ contains the following restricted definition : $$ a^n := \underbrace{a \cdot a \cdot \cdots \cdot a}_{n \text{ times}} \quad \forall a >0, \forall n \in \mathbb{N}, $$ as well as this one : $$ a^{\frac{1}{n}} := \sqrt[n]{a}, \forall n \in \mathbb{N}. $$ The benefits from defining powers in terms of the inverse functions $\exp$ and $\log$ are that we can extend the last two definitions to irrational exponents and that $\exp$ and $\log$ can be rigorously defined in terms of power series and integrals, for example.

However, if $a < 0$, we can't use the definition in terms of $\exp$ and $\log$, but we can still try to mimic the restricted definitions given above. For example, we naturally define : $$ a^n := \underbrace{a \cdot a \cdot \cdots \cdot a}_{n \text{ times}} \quad \forall a \leq 0, \forall n \in \mathbb{N}, $$ which is indeed well-defined. However, when it comes to rational exponents, we can't make the general definition : $$ a^{\frac{1}{n}} := \sqrt[n]{a} \quad \forall a \leq 0, \forall n \in \mathbb{N}, $$ since it is well known that for particular choices of $a$ and $n$ there would be no element $a^{\frac{1}{n}}$ in the set of real numbers that would satisfy $$ (a^{\frac{1}{n}})^n = a. $$ For example, take $a = -2$ and $n = 2$. This contrasts with the case where $a$ is restricted to $\mathbb{R}_{> 0}$. In the latter case, when we actually construct the set of real numbers, it is possible to show that there will always be such a real number $a^{\frac{1}{n}}$.

For particular choices of $a \leq 0$ and $n$, it might happen that such a real number $a^{\frac{1}{n}}$ exists : for example, take $a = -8$ and $n = 3$ and you find that $-2 \in \mathbb{R}$ can be taken.

To conclude, if we go back to the case $a = -2$ and $n = 2$, we find that the difficulty lies in the fact that the relation on $\mathbb{R}$ $$ x^2 = -2 $$ is empty. Complex numbers have been introduced to provide solutions to such equations.