# Domain of sqrt( ln{x}-ln[x])?

May 1, 2018

The Null Set $\phi .$

#### Explanation:

I presume, $\left\{x\right\} = x - \left[x\right] , x \in \mathbb{R}$.

Let us consider $\mathbb{R}$ as

$\mathbb{R} = \ldots \cup \left[- 2 , - 1\right) \cup \left[- 1 , 0\right) \cup \left[0 , 1\right) \cup \left[1 , 2\right) \cup \left[2 , 3\right) \cup \ldots$.

$\text{Now } \forall x \in \left[- 2 , 1\right) , \left[x\right] = - 2.$

$\therefore \ln \left[x\right] \text{ is undefined}$.

$\therefore \ln \left\{x\right\} - \ln \left[x\right] \text{ becomes meaningless, and hence, so does}$

f(x)=sqrt(ln{x}-ln[x]".

$\text{Similar is the case when } x \in \left[- 1 , 0\right) \mathmr{and} x \in \left[0 , 1\right)$.

Let us examine $f \text{ for } x \in \left[1 , 2\right) .$

"If, "x=1, "then, "[x]=1, &, {x}=0.

$\therefore \ln \left\{x\right\} , \text{ undefined, & as such, } f \left(x\right) = \sqrt{\ln \left\{x\right\} - \ln \left[x\right]}$

meaningless.

$x \in \left(1 , 2\right) \Rightarrow \left[x\right] = 1 , \left\{x\right\} = x - \left[x\right] = x - 1 \therefore 0 < \left\{x\right\} < 1$.

$\therefore \ln \left\{x\right\} < 0. \therefore \ln \left\{x\right\} - \ln \left[x\right] < 0.$

$\therefore f \left(x\right) = \sqrt{\ln \left\{x\right\} - \ln \left[x\right]}$ is undefined.

$\text{For "x in [2,3), f(2)" is again meaningless, & if, x in } \left(2 , 3\right) ,$

$\left[x\right] = 2 , 0 < \left\{x\right\} < 1 , \therefore \ln \left\{x\right\} < 0.$

$\therefore \ln \left\{x\right\} - \ln \left[x\right] < 0. \therefore f \left(x\right) = \sqrt{\ln \left\{x\right\} - \ln \left[x\right]} \text{ is undefined}$.

Continuing in this fashion, we conclude that, the domain of $f$

is the Null Set, $\phi$.