Why expression under root has to be positive?

Question

I have function defined like this : f(x,y) = $\sqrt[127,5]{\frac{x^²+y^²-4y}{4x-x^2-y^2}}$

I thouth that domain is $4x-x^2-y^2 \neq 0$ but when I looked on wolfram, the domain is everything under the root has to be $\gt0$.

Why mine result is wrong ? I thougth of this number $127,5$ like $\frac{10}{1275}$ and that would be translated into $\frac{2}{255}$.

From my point of view, that should be equal to $\sqrt[255]{(\frac{x^2+y^2-4y}{4x-x^2-y^2})^2}$. If I am not wrong, 255 is not even. Even, if that number is even, square should take care of negative result in expression.

Did I took something for granted that I shouldn't or I forgot something ?

Answer 1

Actually, you have to take the function as it is, without any manipulation, even if they're correct.

Think about the following example: $$ f(x)=\frac{x^2-1}{x-1}, $$ which has the domain $\mathbb{R}-{\{1\}}$. Nevertheless, f can be rewritten as $$ f(x)=\frac{(x+1)(x-1)}{x-1}=x+1, $$ but this is not actually true. Because it changes the domain. The same happened in your case.

Answer 2

The "bit under the root" has a name, in case you want to search further - it is called a $radicand$.

The fundamental reason (from my undergraduate days, things may have changed) is that $y=\sqrt{x}$ is defined to be the positive solution to $x=y^{2}$ but this is by convention.

Later, roots such as $8^{\frac{1}{3}}$ are sometimes defined (in complex space) to be principal roots which are not always real.

So, ultimately it is a matter of convention, but conventions need to be followed.