# Thread: How are these two items equal?

1. ## How are these two items equal?

Can someone show me how

$\displaystyle Sqrt(x)/x$

is equal to $\displaystyle 1/Sqrt(x)$?

I know I can flip the denominator to:
$\displaystyle 1/X * Sqrt(x)$

But I still can't understand how it gets to $\displaystyle 1/Sqrt(x)$.

What am I missing

2. $\displaystyle \displaystyle \frac{\sqrt{x}}{x} = \frac{x^{1\over 2}}{x} \times \frac{x^{-1\over 2}}{x^{-1\over 2}}= \frac{1}{x^{1\over 2}}= \frac{1}{\sqrt{x}}$

3. Alternatively, dividing each member of the fraction by $\displaystyle \sqrt{x}$ (Pickslides multiplied by its inverse) :

$\displaystyle \displaystyle \frac{\left ( \frac{\sqrt{x}}{\sqrt{x}} \right )}{\left ( \frac{x}{\sqrt{x}} \right )} = \frac{1}{\sqrt{x}}$

4. Yet another way:
$\displaystyle \sqrt{x}$ is defined to be the number such that $\displaystyle (\sqrt{x})^2= (\sqrt{x})(\sqrt{x})= x$ so the denominator can be written as $\displaystyle (\sqrt{x})(\sqrt{x})$

$\displaystyle \frac{\sqrt{x}}{x}= \frac{\sqrt{x}}{(\sqrt{x})(\sqrt{x})}= \frac{1}{\sqrt{x}}$

5. Yet another way!

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

6. HallsofIvy's way is how I'd normally show it,
however, it does depend on whether one is working with surds or indices,
or various simplification techniques.

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

If they are equal, we will have RHS=LHS after cross-multiplying...

as in $\displaystyle \frac{1}{2}=\frac{2}{4}=\frac{3}{6}...\Rightarrow\ 1(4)=2(2),\ 1(6)=3(2),\ 2(6)=4(3)....$

hence

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

since

$\displaystyle \sqrt{x}\ \sqrt{x}=x(1)$