It is clear to write $(-1)^{\frac{1}{3}}=-1$
What about $$(-1)^{\frac{2}{6}} $$ ? Is it $$(-1)^{\frac{2}{6}}=(-1^2)^{\frac{1}{6}}=1\\or ?\\(-1)^{\frac{2}{6}}=(-1)^{\frac{1}{3}}$$ with respect to $$\mathbb{Q}=\left\{\frac{m}{n}|m,n \in \mathbb{Z} ,n\neq0 ,(m,n)=1\right\}$$ $$\frac{2}{6} \text {is not in} \mathbb{Q}$$ I am sorry to ask this question, but I get stuck on this ...and no result .
Thanks in advance .
On
When we are talking about $n$th root of a number, we should keep in mind that there are more than one answers (in Complex field). For example, $(-1)^{1/3}$ can be $-1$, $\frac{1+\sqrt{3}i}{2}$, or $\frac{1-\sqrt{3}i}{2}$. These roots are computed using $x^3+1=(x+1)(x^2-x+1)=0$. Similarly, you can compute $(-1)^{2/6}$ by finding the roots of $x^6-1=(x^3-1)(x^3+1)$. Three of the roots will match the ones I mentioned above, while others are extraneous.
On
A precise definition of rationals exponents is the key here. And on a first level it is best to keep complex numbers out of discussion. Then we have the following theorem:
Theorem: If $a$ is a positive real number and $n$ is a positive integer then there is a unique positive real number $b$ such that $b^{n} =a$. This number $b$ is called the $n$-th root of $a$ and denoted by $a^{1/n}$ or $\sqrt[n] {a} $.
Further using this theorem we can prove that if $n $ is an odd positive integer and $a<0$ then there is a unique $b<0$ such that $b^{n} =a$ and like before $b$ is called the $n$-th root of $a$ and denoted by $a^{1/n}$ or $\sqrt[n] {a} $. Next it is easily seen that if $n$ is even positive integer and $a<0$ then there is no real number $b$ such that $b^{n} =a$ and hence the symbol $a^{1/n}$ or $\sqrt[n] {a} $ is not defined in the real number system if $a<0$ and $n$ is an even positive integer.
Next let $a\neq 0$ be a real number and $r\neq 0$ be a rational number. Then we can express $r$ as $r=m/n$ where $m$ is an integer and $n$ is a positive integer and $(m, n) =1$. The expression $a^{r} $ is defined to be $(a^{1/n})^{m}$ whenever $a^{1/n} $ is defined.
Together with the above definitions we also have by definition $a^{0}=1$ if $a\neq 0$ and $0^{r}=0$ if $r>0$. Based on these definitions it is easy to see that $(-1)^{2/6}=(-1)^{1/3}=-1$.
From the above it should be clear that expressions like $0^{0},0^{-1},(-1)^{1/2}$ are not defined in the real number system.
Above definitions are typically covered when a student is introduced to the topic of surds and usual properties of exponents hold under these definitions. Apart from the theorem mentioned at the beginning the definition is purely algebraic and proofs of the properties of rational exponents are given using the theorem mentioned above combined with suitable algebraic manipulation.
Since $\frac{2}{6} = \frac{1}{3}$, one must conclude that $$ (-1)^{\frac{2}{6}} = (-1)^{\frac{1}{3}} = -1.$$ In this case, the notation is unambiguous. That being said, there is (perhaps) something interesting going on here. What you have highlighted is that $a^{bc}$ is not always the same thing as $(a^b)^c$, and have given an example where this potential identity breaks down.
Do you think that you can determine sufficient conditions for $(a^b)^c$ to be equal to $a^{bc}$?