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

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Answer 1 · 2018-05-01T12:33:55

The Null Set $phi.$

I presume, ${x}=x-[x], x in RR$ .

Let us consider $RR$ as

$RR=...uu[-2,-1)uu[-1,0)uu[0,1)uu[1,2)uu[2,3)uu...$ .

$"Now "AA x in [-2,1), [x]=-2.$

$:. ln[x]" is undefined"$ .

$:. ln{x}-ln[x]" becomes meaningless, and hence, so does"$

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

$"Similar is the case when "x in [-1,0) and x in [0,1)$ .

Let us examine $f" for "x in [1,2).$

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

$:. ln{x}," undefined, & as such, "f(x)=sqrt(ln{x}-ln[x])$

meaningless.

$x in (1,2)rArr[x]=1, {x}=x-[x]=x-1 :. 0 lt {x} lt 1$ .

$:. ln{x} lt 0. :. ln{x}-ln[x] lt 0.$

$:. f(x)=sqrt(ln{x}-ln[x])$ is undefined.

$"For "x in [2,3), f(2)" is again meaningless, & if, x in "(2,3),$

$[x]=2, 0 lt {x} lt 1, :. ln{x} lt 0.$

$:. ln{x}-ln[x] lt 0. :. f(x)=sqrt(ln{x}-ln[x])" is undefined"$ .

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

is the Null Set, $phi$ .