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