# A nice equation with the floor function

1. Jun 18, 2016

### Gp7417

• Moved from a technical forum, so homework template missing
Hi all!
I should solve the following equation involving the ceil or floor function:
$$x \cdot \lceil x \cdot \lceil x \rceil \rceil = 82$$
I have found just one solution 41/9, but I have no idea about how finding the other solutions.
Thanks you in advance!

2. Jun 18, 2016

### jbriggs444

What convinces you that there are other solutions?

3. Jun 18, 2016

### Staff: Mentor

Can the expression stay the same or decrease if you increase x?
Can the expression stay the same or increase if you decrease x?

4. Jun 18, 2016

### Gp7417

The problem says: 'Find all possibile solutions'. I find this solution numerically.
The expression must be the same.

5. Jun 18, 2016

### Staff: Mentor

Right, so can this be possible with a different x?
That does not tell you how many solutions there are.

6. Jun 18, 2016

### Gp7417

I propose this solution, check it out!
We must find solution on the positive real axis.
Let be $x= n + \delta$, where $n$ is the integer part and $\delta$ is the fractional part with $0 \leq \delta < 1$.
Given that $x \lfloor x \lfloor x \rfloor \rfloor=82$ and $x \lfloor x \lfloor x \rfloor \rfloor \leq x^3$ , we get $x \geq 82^{\frac{1}{3}} \approx 4.344$. Therefore $n \geq 4$.
If 5 is not solution the equation, then every number greater than 5 are not solutions. Therefore $n=4$.
With the previous statements, the equation becames:
$(4+\delta)\cdot \lfloor 16+4\delta \rfloor=82$
This equation is equivalent to the following equations:
$0 \leq \delta < 1/4$ and $(4+\delta)\cdot(16+0)=82$
$1/4 \leq \delta < 2/4$ and $(4+\delta)\cdot(16+1)=82$
$2/4 \leq \delta < 3/4$ and $(4+\delta)\cdot(16+2)=82$
$3/4 \leq \delta < 4/4$ and $(4+\delta)\cdot(16+3)=82$
Only the third equation admits a solution that is $\delta=5/9$.

Last edited: Jun 18, 2016
7. Jun 18, 2016

### Staff: Mentor

What I don't understand here is why there is a solution at all?
Obviously x has to be between 4 and 5, excluding 4 and 5. So the inner ceiling is 5 and the outer at least 21. But four times twenty-one is already 84.

8. Jun 18, 2016

### Staff: Mentor

You write the ceil function and then derive things for the floor function? For the floor function that works.
See the floor/ceil mixup.

9. Jun 18, 2016

### Staff: Mentor

10. Jun 18, 2016

### Gp7417

With $\lceil x \rceil$ I want to denote the floor function of $x$.

11. Jun 18, 2016

### SammyS

Staff Emeritus
That's how to denote the ceiling function.

Use lfloor and rfloor for the floor function. $\ \lfloor x \rfloor$

12. Jun 18, 2016

### Ray Vickson

No. The function $f(x) = x \lfloor x \lfloor x \rfloor \rfloor$ is strictly increasing in $x$, sometimes linearly and sometimes through a jump discontinuity.
To see this, let $x = n + r$ where $n \geq 0$ is an integer and $0 \leq r < 1$. Then the inner floor function is $\lfloor n + r \rfloor = n$, so
$$f(n+r) = (n+r) \lfloor (n+r)n \rfloor$$
The function $g(n,r) = \lfloor (n+r)n \rfloor$ is positve and nondecreasing in $n$ and $r$, so $f(n+r) = (n+r) g(n,r)$ is strictly increasing.

That means that $f(x) = 82$ has at most one root.