Find range of $x$ satisfying $$\left \lfloor \frac{3}{x} \right \rfloor +\left \lfloor \frac{4}{x} \right \rfloor=5$$ Where $\lfloor\cdot\rfloor$ is the floor function

My try:

As far as domain of LHS is concerned we have $x \ne 0$ and since RHS is positive, we have $x \gt 0$

Now since LHS is sum of two positive integers, let us suppose:

$$\left \lfloor \frac{3}{x} \right \rfloor=m$$ and

$$\left \lfloor \frac{4}{x} \right \rfloor=5-m$$

Thus we have:

$$ m \le \frac{3}{x} \lt m+1$$ $$5-m \le \frac{4}{x} \lt 6-m$$

Adding both we get:

$$5 \le \frac{7}{x} \lt 7$$ $\implies$

$$1 \lt x \le \frac{7}{5}$$

Hence $$x \in (1, 1.4]$$

But answer in book is given as $$x \in (1,\frac{4}{3})$$

What went wrong?

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    $\begingroup$ Why the downvote?!!! The OP has asked a really good question and has shown us their effort very clearly, which is way more than what most questioners on MSE do. $\endgroup$ – stressed out Mar 6 '19 at 12:11
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    $\begingroup$ You have showed all solutions must satisfy $x\in(1,1.4]$. However you have not showed every $x$ in that set is a solution- in fact the solution set is a proper subset. $\endgroup$ – Macavity Mar 6 '19 at 13:05

Since $\frac{4}{x}>\frac{3}{x}$, we have three cases:

  1. $\left [ \frac{3}{x} \right ]=0$ and $\left [ \frac{4}{x} \right ]=5.$ Easy to show that it's impossible.

  2. $\left [ \frac{3}{x} \right ]=1$ and $\left [ \frac{4}{x} \right ]=4,$ which is impossible again and

  3. $\left [ \frac{3}{x} \right ]=2$ and $\left [ \frac{4}{x} \right ]=3,$ which gives the answer: $\left(1,\frac{4}{3}\right]$.

Can you end it now?

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  • $\begingroup$ yes i got my mistake thanks $\endgroup$ – Umesh shankar Mar 6 '19 at 11:49
  • $\begingroup$ @Umesh shankar Substitute this in the equation. You'll obtain: $2+2=5$. I think it's not true. $\endgroup$ – Michael Rozenberg Mar 6 '19 at 11:49
  • $\begingroup$ yes you are right $\endgroup$ – Umesh shankar Mar 6 '19 at 11:50
  • $\begingroup$ But how can i arrive at your answer from my method? $\endgroup$ – Umesh shankar Mar 6 '19 at 11:50
  • $\begingroup$ @Umesh shankar I think your method can not give an answer.See my hint. It works. $\endgroup$ – Michael Rozenberg Mar 6 '19 at 11:52

The inequality after "Adding both we get:" is true, but it is not the whole story. You have lost information here, which means that not every solution to this inequality is a solution to both of the constituent inequalities.

Here is a simpler example: suppose we seek the range of solutions to $$0<x<2$$ and $$1<x<3$$ Obviously the answer is $1<x<2$; but by your method, adding both gives $$1<2x<5$$ which has a wider range of solutions. Your addition has lost the information that $x<2$ and $1<x$.

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If you want to know what went wrong, then the problem starts when you sum two inequalities because that's an irreversible step. For example

$$1<x<2 \text{ and } 2<y<3 \implies 3<x+y<5$$

is a correct step. However, it's not reversible in the sense that if we're given that $3<x+y<5$, we can't claim that $1<x<2$ and $2<y<3$. It is indeed possible to have $2<x<4$ and $y=1$. Right?

Your method becomes one-sided after you sum the inequalities. Because of that, what you will find gives you the set of possible solutions. So, it might very well be bigger than the set of actual solutions. In other words, what you have is a "necessary" condition on what solutions are allowed, you do not have a "sufficient" condition.

A method that gives you exactly the solutions of the system should have if and only if steps. Or if not, at the end after you have found your candidate solutions, you must be able to overrule candidate solutions that do not work.

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Your method is fine, but you dropped the inequalities prematurely. Indeed, by comparing the extreme members, we must fulfill

$$\max\left(\dfrac m3,\dfrac{5-m}4\right)<\min\left(\dfrac{m+1}3,\dfrac{6-m}4\right)$$

which is only possible with $m=2$.


$$\max\left(\dfrac 23,\frac34\right)\le\frac1x<\min\left(1,1\right)$$ constrains $x$ to be smaller than $\dfrac43$.

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  • $\begingroup$ But the OP's method gives that $1.4$ is a solution, which is incorrect. Also, doesn't $m$ in two inequalities get canceled? So, why does it matter what has the OP supposed $m$ to be? $\endgroup$ – stressed out Mar 6 '19 at 12:58
  • $\begingroup$ @stressedout: I am explaining was is missing in the OP's solution. $m$ is canceled out but cannot take any value and the conditions in which it appears cannot be dropped. $\endgroup$ – Yves Daoust Mar 6 '19 at 13:35
  • $\begingroup$ My comment was based on your previous answer. Your new answer is complete and it's clear now but it is slightly different from what the OP has done. $\endgroup$ – stressed out Mar 6 '19 at 13:39
  • $\begingroup$ @stressedout: of course it is different, as it is a fix ! I am telling the OP where he went wrong, in fact dropping the initial equalities after adding them. $\endgroup$ – Yves Daoust Mar 6 '19 at 13:41

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