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I know this is a very basic question, but could someone please mathematically explain, why this is true:

$\sqrt{x} \cdot \frac{1}{x} = \frac{1}{\sqrt{x}}$

Wolfram|Alpha can confirm this.

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2 Answers

up vote 4 down vote accepted

Let us suppose that $x$ is positive. If it is not, our expression is not defined. Note that essentially by definition, $$x=\sqrt{x}\sqrt{x}.$$
It follows that $$\frac{\sqrt{x}}{x}=\frac{\sqrt{x}}{\sqrt{x}\sqrt{x}}=\frac{1}{\sqrt{x}}.$$ (For the last step, we divided top and bottom by $\sqrt{x}$.)

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Ah... I see. Basically $x = \sqrt{x} \cdot \sqrt{x} = {\sqrt{x}}^{2} = x$. Now it makes sense how the cancellation occurs. Thank you! – spryno724 Feb 2 '12 at 4:24
You don't really need to assume $x > 0$, do you? This is true for any nonzero complex number $x$ and any branch of $\sqrt{x}$. For that matter, it's true in any ring for any element that has both an inverse and a square root. – Robert Israel Feb 2 '12 at 4:34
@Robert Israel: True, the $x>0$ stuff could be omitted. But I wanted to remind the OP that, at the level (s)he is working, there are restrictions. – André Nicolas Feb 2 '12 at 4:54
up vote 1 down vote

$$\sqrt{x}\cdot\frac{1}{x}= \color{Red}{\sqrt{x}}\cdot\frac{1}{\color{Red}{\sqrt{x}}\cdot\sqrt{x}}=\frac{1}{\sqrt{x}}.$$

A more general method: $x^{1/2}\cdot x^{-1}=x^{1/2-1}=x^{-1/2}=(x^{1/2})^{-1}.$

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