what is the domain of $\frac{1}{\sqrt{x-[x]}}$ where [x] denotes the greatest integer function and find the range .
My approach :
since [x] greatest integer function is discontinuous on all integral value , therefore the domain of this function will be $R^+ -\{Z\}$ where Z is integer and $R^+$ is all real positive numbers. but answer is $R -\{Z\}$ how all real numbers are possible here. please suggest thanks….
Best Answer
Let $f(x)=\dfrac 1{\sqrt{\{x\}}}$ where $\{x\}$ designates the fractional part, it belongs to $[0,1)$.
This comes from the definition of integer part (LHS below): $$\lfloor x\rfloor\le x<\lfloor x\rfloor+1\implies 0\le\{x\}=x-\lfloor x\rfloor<1$$
In particular $\{x\}\ge 0$, even for negative numbers, so $\sqrt{\{x\}}$ is defined everywhere.
First of all notice that your function is $1$-periodic so you can study it on $[0,1]$.
Indeed $\{x+1\}=x+1-\lfloor x+1\rfloor=x+1-(\lfloor x\rfloor+1)=x-\lfloor x\rfloor=\{x\}$ so $f$ is $1$-periodic as well.
As you noticed $\{x\}=0$ whenever $x$ is an integer, so the local domain is $(0,1)$ and the global domain extended by periodicity is $\mathbb R\setminus\mathbb Z$.
For the range since on $(0,1)$ we have $0<\{x\}<1$ then we get $f(x)>1$ and the range is thus $(1,+\infty)$.