Calculation of $x$ in $x \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = 88$ How can I calculate real values of $x$ in $x \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = 88$, where $\lfloor x\rfloor$ is the floor function?
My attempt:
Let $\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = k\in \mathbb{Z}$. Then
$$
k \leq x\lfloor x\lfloor x\rfloor\rfloor<k+1
$$
and our equation becomes
$$
x\cdot k = 88 \implies x=\frac{88}{k}
$$
For $x>0$, simple guessing shows that $3.1<x<3.2$. But how can we account for $x<0$?
 A: Through guess and check 
$x=\dfrac{22}{7}$
We know apriori, $x\approx 3.16 \Longrightarrow \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = 28$, so setting $k=28$ yields $x=22/7$
You can verify this yourself:
$$
\tfrac{22}{7}\lfloor\tfrac{22}{7}\lfloor \tfrac{22}{7}\lfloor \tfrac{22}{7}\rfloor\rfloor\rfloor =88 
$$
A: Hint
Let $x=\frac p q\in[k,k+1)$ irreductible and positive so $88\times \frac q p\in\mathbb N$ hence 
$$p\in\{1,2,4,8,11,22,44,88\}$$
and we have
$$k^4\leq x \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = 88\leq (k+1)^4$$
so $k=3$ and then what's the possible values of $x$? Repeat the same reasoning for $x$ negative.
A: If $0\le x<3$, then the expression is $<3^4<88$.
So in the positive case, $k=\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\ge 27$  and also $k=\frac{88}x\le \frac{88}3$, i.e. $k\in\{27,28,29\}$. Trying $x=\frac{88}{28}=\frac{22}{7}$ we verify that we have found a solution, whereas $\frac{88}{27}$ and $\frac{88}{29}$ don't work. (Since $x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor$ is strictly decreasing (in the domain $x\ge1$) we might not even fully test other values once we have found or otherwise guessed a solution).
To deal with negative $x$, we might substitue $x$ with $-y$ and investigate
$$y\lceil y\lceil y \lceil y\rceil\rceil\rceil=88$$
with $y>0$ instead. 
Once again,  for $0<y\le 3$, the left hand side is $\le 81<88$, hence $y>3$, but then the left hand side is $\ge 64 y>192>88$.
A: There isn't a reason to neglect negative values, but note that if $x=-3-\epsilon$ with $\epsilon$ small ($0\lt \epsilon \lt \frac 1{37}$) you have successively $$\lfloor x \rfloor=-4$$ $$\lfloor -4(-3-\epsilon) \rfloor=12$$ $$\lfloor 12(-3-\epsilon) \rfloor=-37$$$$-37(-3-\epsilon)\gt111$$ And if $x=-3$ the value is $81$, so you can't attain the value $88$ with a negative value of $x$.
A: If there is a rational solution for $x$, we know that its numerator must divide $88$ evenly.  This leaves us with $1$, $2$, $4$, $8$, $11$, $22$, $44$, and $88$ as possible choices of numerator for $x$.
It turns out that $x=\frac{22}{7}$.  You can check all other divisors of $88$ to verify that none of them work as a numerator besides $22$.
