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#.