Again Floor Function

Algebra Level 2

Find the domain of $x$ satisfying

$\Big\lfloor x + \big\lfloor x + \lfloor x \rfloor \big\rfloor \Big\rfloor + 6= 0.$

Notation : $\lfloor \cdot \rfloor$ denotes the floor function .

-5 < x \leq -4

-2 < x \leq -1

-2 \leq x < -1

-2\leq x \leq -1

1 solution

Zee Ell
Oct 8, 2016

$x = \lfloor x \rfloor + \{ x \}$

where {x} is the fractional part of x, 0 ≤ {x} < 1 .

Then it is easy to see, that:

$\lfloor x + \lfloor x + \lfloor x \rfloor \rfloor \rfloor = \lfloor \lfloor x \rfloor + \{ x \} + \lfloor \lfloor x \rfloor + \{ x \} + \lfloor x \rfloor \rfloor \rfloor =$

$= \lfloor \lfloor x \rfloor + \{ x \} + \lfloor 2 \lfloor x \rfloor + \{ x \} \rfloor \rfloor = \lfloor \lfloor x \rfloor + \{ x \} + 2 \lfloor x \rfloor \rfloor = 3 \lfloor x \rfloor$

Therefore, our equation is equivalent to:

$3 \lfloor x \rfloor + 6 = 0$

$3 \lfloor x \rfloor = - 6$

$\lfloor x \rfloor = -2$

Hence, our answer should be:

$\boxed { -2 ≤ x < -1 }$

Again Floor Function

1 solution

0 pending reports