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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:

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

Manar
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    The rules of algebra tell you what you can do to an expression to produce the same result. There are no rules to tell you what you cannot do because you may produce a different result. We have a case of the latter. $4^{-1/2}=\frac{1}{2}$ while $-\sqrt{4}=-2$, in other words, you are getting a different result, and this is enough to conclude that the "rule" you are trying to establish here is not valid. –  Dec 25 '20 at 15:18
  • If $a$ is positive and $b$ is real, then $a^b$ is positive. In particular, $4^{-1/2}$ is positive, and therefore not equal to $-2$. – GEdgar Dec 25 '20 at 15:39

3 Answers3

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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 $x^{1/a}=\sqrt[a]{x}$.

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 $$x^{-(1/2)}=x^{1/(-2)}=x^{\frac{-1}{2}}=x^{\frac{1}{-2}}.$$

K.defaoite
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$x^{-y}$ 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 $u=\frac1{\sqrt x}$.

  • Yeah I know that but let's assume that the $1$ in the nominator wasn't the one multiplied by $-1$ but $2$ in the dominator is. – Manar Dec 25 '20 at 15:16
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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?

Ronald Becerra
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