Why isn't $x^\frac{-1}{2}$ equal to $-\sqrt{x}$

Question

A fraction multiplied by $-1$ can be written in different ways:

$\frac{-a}{b}$

$\frac{a}{-b}$

$-\frac{a}{b}$

So $x^\frac{-1}{2}$ can also be written as $x^\frac{1}{-2}$ then why can't we take the whole $-2$ down and turn it into $-\sqrt{\bullet}$ and say:

$x^\frac{1}{-2}$ = $-\sqrt{x}$

Answer 1

$x^{-y}$ is $\frac1{x^y}$, not $- x^y$.

Added: $\sqrt[-2]{x}$ is the positive real number $u$ such that $u^{-2}=x$, so it's $u=\frac1{\sqrt x}$.

Answer 2

Continuing on from my other answer -

Property 1:

The quantity $x^{-a}$ is one such that $x^a x^{-a}=1$. This gives us $$x^{-a}=\frac{1}{x^a}$$ Property 2:

The quantity $x^{1/a}$ is one such that ${\left(x^{1/a}\right)}^a=x$. If we define the operation $\sqrt[a]{}$ as the inverse of taking something to the exponent $a$, we can write explicitly that $x^{1/a}=\sqrt[a]{x}$.

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With this in mind, let's compare the expressions $x^{-(1/2)}$ and $x^{1/(-2)}$. Using the first property mentioned, we can write $$x^{-(1/2)}=\frac{1}{x^{1/2}}=\frac{1}{\sqrt{x}}$$ On the other hand, taking a look at our other expression and using the second property, $$\left(x^{1/(-2)}\right)^{-2}=x$$ Which, by our first property means $$\frac{1}{\left(x^{1/(-2)}\right)^{2}}=x$$ So taking the square root of both sides, $$\frac{1}{x^{1/(-2)}}=\sqrt{x}$$ Meaning $$x^{1/(-2)}=\frac{1}{\sqrt{x}}$$ Thankfully we reach the desired conclusion that $$x^{-(1/2)}=x^{1/(-2)}=x^{\frac{-1}{2}}=x^{\frac{1}{-2}}.$$

Answer 3

In the last part you put:

$x^\frac{1}{-2}$ = $-\sqrt{x}$

With it you are meaning that

$x^\frac{1}{-2}$ = $-{x}^\frac{1}{2}$

but that is not true. In any case, $x^\frac{1}{-2}$ would mean $\sqrt[-2]{x}$, which is not the same as $-\sqrt[2]{x}$.

Note that is is only the index of the radical (the $2$) which is being multiplied by $(-1)$, not the entire expression $\sqrt[2]{x}$.

If you think the $(-1)$ that multiplies the index should exit multiplying all the radical, why not do the same with any other number? For example, $2$ is the same as $2*1$, so $\sqrt[2]{x}$ would be equal to $2*\sqrt[1]{x}$, which is equal to $2*x$. In the same way, $3$ equals $3*1$, so $\sqrt[3]{x}$ would be $3*\sqrt[1]{x} = 3*x$ and therefore taking the $n$-root of a number $x$ would consist on multiplying $n$ by $x$.

Absurd, isn't it?