# Rational exponents: prove some states

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?

• As stated your supposition is false. For example if $a=6$ and $n=2$ the solution is real. You need to modify your assumptions. After that just use the fact that $i=\sqrt{-1}$ along with the rule of exponents and it's not hard to prove. – Wintermute Oct 3 '13 at 19:21
• @mtiano Ohh sorry I had a mistake! – Mohammad Fakhrey Oct 3 '13 at 19:23

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.

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.

For future reference, here is a simpler proof that does not rely on exponentiation by real numbers, which is the basis of one of the previous answers using the exponential and lograthim functions.

Preliminaries: Order Axioms of the Real Numbers---there is a non-empty subset $P\subset\mathbb{R}$ called the set of strictly positive real numbers that satisfies the following axioms:

• (O1) If $a,b\in P$, then $a+b\in P$.
• (O2) If $a,b\in P$, then $ab\in P$.
• (O3-"Trichotomy Property") If $a\in\mathbb{R}$, then exactly one of the following holds: $$a\in P,\ \ \ \ a=0,\ \ \ -a\in P.$$ Notationally, the Trichotomy Property is equivalent to: $$a>0,\ \ \ a=0,\ \ \ -a>0.$$

Theorem Statement: Let $a\in\mathbb{R}$. If $a<0$ and $n\in\mathbb{N}$ is even, then there exists no $x\in\mathbb{R}$ such that $x^n=a$.

Proof: By definition, if $a<0$, then $-a>0$ (*). The proof is by contradiction. Assume there exists $x\in\mathbb{R}$ such that $x^n=a$. From (O3), either: (i) $x>0$, (ii) $x=0$, or (iii) $x<0$.

• Case (i): If $x>0$, then by inductive application of (O2), $a=x^n\in P$ (or $a>0$). This contradicts (*) under (O3).
• Case (ii): If $x=0$, then $a=x^n=0$, which also contradicts (*) under (O3).
• Case (iii): If $x<0$, then $-x>0$. Therefore, $$a=x^n=1\cdot x^n=(-1)^nx^n=(-x)^n.$$ The step where $1=(-1)^n$ is justified since $n$ is even. The last step is a basic rule of integer exponentiation. Like case (i), inductive application of (O2) on $(-x)>0$ shows that $a=(-x)^n>0$, which again contradicts (*).

Therefore, since assuming the existence of $x\in\mathbb{R}$ leads to contradictions, it must be the case that there is no such $x$.