# Solve for equations involving floor function [duplicate]

Solve for real $$x$$: $$\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2x \rfloor} = x - \lfloor x \rfloor + \frac{1}{3}$$

Hello! I hope everybody is doing well. I was not able to solve the above problem. And this problem becomes even more difficult with the fact that $$x$$ is not necessarily a positive real.

Here is what I did:

Let $$x=p+r, 0 \leq r <1$$ Now we divide into cases whether $$r$$ is less than $$0.5$$ or not and I reach here(1st part): $$0 \frac{3}{2p} - \frac13 < \frac12$$. Now I could have done that $$\frac95<2\le [x]\le4<\frac92$$ but since $$x$$ is not necessarily positive, I guess this is not always true.

Any help would be appreciated. Thanks.

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• Are you going to sign off your earlier question before asking another? – S. Dolan Oct 18 at 14:06
• Oh, I am sorry I almost forgot that you can mark an answer correct too. – Vasu090 Oct 18 at 14:10

Your method looks fine. So, for the first case:-

$$\frac{3}{2p}=r+\frac{1}{3}$$

and therefore $$3r+1=\frac{9}{2p}.$$

Now $$0\le r <\frac{1}{2}$$ and so $$p$$ is positive. The possibilities for $$p$$ are $$2,3,4$$.

These give $$x$$ as $$2\frac{5}{12}, 3\frac{1}{6}, 4\frac{1}{24}.$$

Now, if $$r\ge \frac{1}{2}$$, then $$\frac{1}{p}+\frac{1}{2p+1}=r+\frac{1}{3}.$$

Again $$p$$ must be positive. For $$p=1$$, the LHS is too large. For $$p\ge2$$, the LHS is too small.

Therefore there are no further solutions.

I would approach the problem in this way.

Let's put $${1 \over {\left\lfloor x \right\rfloor }} + {1 \over {\left\lfloor {2x} \right\rfloor }} = x - \left\lfloor x \right\rfloor + {1 \over 3} = \left\{ x \right\} + {1 \over 3} = t$$

Since the fractional part of $$x$$ has values in the range $$[0, \, 1)$$ we have the the RHS is $$0 \le \left\{ x \right\} = t - {1 \over 3} < 1\quad \Rightarrow \quad {1 \over 3} \le t < {4 \over 3}$$

At the same time, we know that $$x - 1 < \left\lfloor x \right\rfloor \le x$$ thus for the LHS we get this set of inequalities $$\left\{ \matrix{ {1 \over 3} \le {1 \over {\left\lfloor x \right\rfloor }} + {1 \over {\left\lfloor {2x} \right\rfloor }} = t < {4 \over 3} \hfill \cr {1 \over x} + {1 \over {2x}} \le {1 \over {\left\lfloor x \right\rfloor }} + {1 \over {\left\lfloor {2x} \right\rfloor }} < {1 \over {x - 1}} + {1 \over {2x - 1}} \hfill \cr} \right.$$

Translating this into inequalities for $$x$$, we shall have that $$\left\{ \matrix{ {1 \over 3} < {1 \over {x - 1}} + {1 \over {2x - 1}} \hfill \cr {1 \over x} + {1 \over {2x}} < {4 \over 3} \hfill \cr} \right.$$ Solving the quadrics give $$\left\{ \matrix{ x \in \left( {1/2,\;3 - \sqrt {22} /2} \right) \cup \left( {1,\;3 + \sqrt {22} /2} \right) \hfill \cr x \in \left( { - \infty ,\;0} \right) \cup \left( {9/8,\;\infty } \right) \hfill \cr} \right.\quad \Rightarrow \quad x \in \left( {9/8,\;3 + \sqrt {22} /2} \right) \approx \left( {1.1,\;5.34} \right)$$

So we have just to find the possible solutions for $$\left\{ x \right\} = {1 \over {\left\lfloor x \right\rfloor }} + {1 \over {\left\lfloor {2x} \right\rfloor }} - {1 \over 3} \quad \left| {\;x \in \left[ {1,\;3/2} \right)\,\; \cup \;\left[ {3/2,\;2} \right)\; \cup \; \cdots \,\; \cup \;\left[ {5,\;11/2} \right)} \right.$$