Following by Exponent rules, I got really confused. I notices the following:
$(-1)^1 = 1$
so
$(-1)^{2\cdot0.5} = -1$
Now,I can get two different answer:
$((-1)^{0.5})^2 = (\sqrt{-1} )^ 2 = i^2 = -1$
or
$((-1)^2)^{0.5} = (1)^{-0.5} = \sqrt1 = 1$
What am I missing here?
Thank you very much in advance.
NG
The problem is that the rule $a^{bc}=(a^b)^c$ is true for $a>0$ and real $b$,$c$ but became wrong for complex numbers. That's why you got a contradiction. There are a few fake proofs using this fact. For example for any $t\in\mathbb R$ $$ e^{ti}=e^{2\pi i\frac{t}{2\pi}}=\left(e^{2\pi i}\right)^{\frac{t}{2\pi}}=1^{\frac{t}{2\pi}}=1 $$
To refer to $b^k; k \not \in \mathbb Z$ we need a definition.
The standard definition for $b^{\frac 1n}$ is IF $b\in \mathbb R$ and $b > 0$ then $b^{\frac 1n}$ is the unique positive real number, $c$ so that $c^n = b$. It is important to note there are $n-1$ complex, non-positive other such numbers.
And as long as $b > 0$ then we can say $(b^{\frac 1n})^n)=(b^n)^{\frac 1n}=b$ (and in general $(b^n)^m = (b^m)^n$) as all terms are unique and positive.
We can't do the same $b <0$ as $(b^2)^{\frac 12} \ne b$. By definition, $(b^2)^{\frac 12}$ is positive and $(b^2)^{\frac 12} = -b$. And likewise, in real numbers $b^{\frac 12}$ simply does not exist so $(b^{\frac 12})^2$ is ... nonsense.
Now if extend our definition to complex numbers and allow for $i^2 = -1$ we no longer have a well-defined definition definition for $b^{\frac 12}$. There are two $c$ so that $c^2 = b$ and if neither of them are real which do we choose.
We accept this by noting $b^{\frac 1n}$ is multi-valued and that the "rule" $(x^a)^b = (x^a)^b = x^{ab}$ is no longer a rule when it comes to complex numbers and negative bases. Howeve $|(x^a)^b|=|(x^b)^a|$ and if $n\in \mathbb R$ $|x^n| = |x|^n$.
There's a famous false proof that $1= 1^{\frac 12} = (-1*-1)^{\frac 12} \overset != (-1)^{\frac 12}(-1)^{\frac 12} = i*i = -1$. The $\overset !=$ is not valid.
