# what is the domain of $\frac{1}{\sqrt{x-[x]}}$ where [x] denotes the greatest integer function and find the range .

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....

• The domain doesn't have anything to do with discontinuities per se. Sep 1, 2019 at 13:55
• If $x=-\frac52$, what in your opinion is the value of $x-[x]$? Sep 1, 2019 at 13:55
• -2.5 -(-3) = .5 , ok got it.. thanks.. domain is all Real numbers what about range of this function Sep 1, 2019 at 14:02
• Since $[x] \leq x$, $x-[x] \geq 0$, so the expression can define a function wherever the denominator is not zero (i.e. whenever $[x] \neq x$).
– Joe
Sep 1, 2019 at 14:02
• To find the range, think about what values the denominator can equal.
– Joe
Sep 1, 2019 at 14:09

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 $$(1,+\infty)$$.

Since $$x\geq [x]$$ then $$x-[x]\geq 0$$ the equality when $$x$$ is integer. The greatest integer function maps integers into integers and non integers to the closet least integer.