Find the domain and range of the function : $f(x) = \sqrt{\lfloor x\rfloor -x}$

Question

Find the domain and range of the function : $f(x) = \sqrt{\lfloor x \rfloor -x}$ where $\lfloor x \rfloor $ is the floor function (greatest integer function)

This is how I did it : $f(x) = \sqrt{\lfloor x \rfloor -x}$
For $f(x)$ to be defined, $\lfloor x \rfloor-x$ must be greater than or equal to $0$
So, $\lfloor x \rfloor-x \geq 0$
So, $x \leq \lfloor x \rfloor$
$x$ is never less than $\lfloor x \rfloor$ but it can be equal to it if $x \in \Bbb Z$

So, shouldn't the $domain$ of $f$ be $\Bbb Z$?

In my textbook, the $domain$ of $f$ is given to be $\varnothing$ but I think that it is defined for all $x \in \Bbb Z$ and for all $x \in \Bbb Z$, the value of $f(x)$ is $0$, so in my opinion, the $range$ of $f$ should be {0} which is also given to be $\varnothing$ in my textbook

I have one more question, if for any real function $f$, $Domain(f) = \varnothing$, then is $Range(f) = \varnothing$ always true as well? I think so...

Thanks

Answer 1

You’re correct. The domain would be $\mathbb Z$, and for every integer input the output will be $0$.

If the domain is $\varnothing$, then the range is $\varnothing$ as well because the function cannot map to any value.

Answer 2

$\bullet$ Any real number $x$ can be written as $\lfloor{x}\rfloor+h$ with $h$ a real number in $[0; 1)$.

$\bullet$ Since $\sqrt {X}$ is only defined for input $X\geq 0$, the function $f(x)= \sqrt{\lfloor{x}\rfloor - x}$ requires $\lfloor{x}\rfloor - x \geq 0$.

Now $\lfloor{x}\rfloor - x \geq 0$

$\iff \lfloor{x}\rfloor - (\lfloor{x}\rfloor+h)\geq 0$

$\iff- h \geq 0$

$\iff - h = 0$ OR $ -h \gt 0$

$\iff - h = 0$

$ \iff h = 0$

Note : the second disjunct is an impossibility : since $h\in [0; 1)$ , $-h\in (-1; 0]$, and hence, cannot be strictly greater than $0$. Logic allows to eliminate an impossible disjunct : $P\lor False \equiv P$.

So, the function only takes inputs of the form $x= \lfloor{x}\rfloor + 0= \lfloor{x}\rfloor$.

The domain of function $f$ is the range of the floor function, namely $\mathbb Z$.

$\bullet$ For any integer input $x, \lfloor{x}\rfloor =x $, so , substituting ,

$f(x)= \sqrt {\lfloor{x}\rfloor - x} = \sqrt { x-x} = \sqrt 0=0$.

So , the range of $f$ is {0}.

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As to the question : what is the range of a function if the domain is $\emptyset$?

A function $f$ from $A$ to $B$ is, before all, a relation from $A$ to $B$ , hence a subset of the cartesian product $A\times B$ ( = set of all ordered pairs $(x,y)$ with $x\in A$ and $y\in B$).

So, if a function $f$ is a relation from $\emptyset$ to any set $S$, then it is a subset of $\emptyset\times S = \emptyset$.

Note that , since any subset of the emptyset is empty, $f = \emptyset$.

The range of $f$ is ( by definition) the set of all $y_{\in S}$ such that $y=f(x)$ for some $x$ in domain$(f)$.

Since domain$(f)$ is empty , range$(f)$ is empty.