How to solve $ \left \lfloor x^2 - x - 2 \right \rfloor = \left \lfloor x \right \rfloor $

3k Views Asked by At

I need some help to solve the next equation: $$ \left \lfloor x^2 - x - 2 \right \rfloor = \left \lfloor x \right \rfloor $$

Where $ \left \lfloor \cdot \right \rfloor $ is the floor function.

What I've tried: $$ x^2 - x - 2 - x < 1 $$ $$ x^2 - x - 2 \leq x < x^2 - x - 1 $$

When I try to solve this system, I don't get the right solution. Is it right what I'm doing? How can I do to solve this?

1

There are 1 best solutions below

0
On

Your system of equations isn't quite right (for example, $x$ may be the smaller of the two).

Here's one algebraic way to go about this:

To solve $\lfloor f(x) \rfloor = \lfloor x \rfloor$ write $x = n + \epsilon$ where $0\leq \epsilon <1$ and $n$ is an integer so that $\lfloor x\rfloor = n$.

Then the condition $\lfloor f(x)\rfloor = n$ is $n \leq f(n+\epsilon) < n+1$ (and the above conditions on $n$ and $\epsilon$).