Fractional exponents proof

Question

The online proofs I've seen for fractional exponents goes as follows:

If $x^{a} \cdot x^{b}= x^{(a+b)}$, then $x^{0.5} \cdot x^{0.5}= x$. This is the same as $\sqrt{x}\cdot \sqrt{x}$. Therefore $x^{0.5}= \sqrt{x}$

But, since $-1\cdot-1=1~$ and $1\cdot1=1$, with $-1$ is not equal to $1$, how can we say that $\sqrt{x}=x^{0.5}$ just because it follows an existing trend.

Appreciate some help

Answer 1

Because imaginary numbers defined as such for the solution $x^2+1=0$ which gives the answer $+i,-i$ so it can be written as $(x+i)(x-i)=x^2-i^2$ so you can deduce that $i^2=-1$. so if you don't want to make mistakes. You should know about rectangular form(a+j.b) and polar representation($r.e^{j\theta}$) of complex number and conservation of them. $r=\sqrt({a^2+b^2}), \theta=tan^{-1}\frac {-b} a$ and for converting back you use euler's identity $r.e^{j\theta}=cos(\theta)+j.sin(\theta)$. The rectangular form is useful in addition and subtraction. Polar form is useful in multiplication and division.

Answer 2

You're right that that would not be a satisfactory argument.

Just because $a^2=b^2$ doesn't mean $a=b$.

Instead we have the more general: $x^{\frac ab}:=\sqrt[b]x^a=(\sqrt[b] x)^a$, which can be taken as the definition.