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Is anyone able to help me with the following equation concerned the floor function $\lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2=0$

I don't know how to deal with the floor terms properly.

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The solution of,

$$(1) \quad x^2-3\cdot x+2=0$$

Is, $x=1$ or $x=2$.

To find solutions to,

$$(2) \quad [x^2]-[3\cdot x]+2=0$$

We must solve six systems of equations. However, we'll solve eight for instructive purposes. The first is,

$$[x^2]=0$$ $$[3 \cdot x]=2$$

$$(a) \quad {2 \over 3} \le x \lt 1$$

The second is,

$$[x^2]=1$$ $$[3 \cdot x]=3$$

Which is solved similarly.

$$(b) \quad 1 \le x \lt {4 \over 3}$$

The third is,

$$[x^2]=2$$ $$[3 \cdot x]=4$$

Slightly trickier, we get,

$$(c) \quad \sqrt{2} \le x \lt {5 \over 3}$$

The fourth is,

$$[x^2]=3$$ $$[3 \cdot x]=5$$

$$(d) \quad \sqrt{3} \le x \lt 2$$

The fifth is,

$$[x^2]=4$$ $$[3 \cdot x]=6$$

$$(e) \quad 2 \le x \lt \sqrt{5}$$

The sixth is,

$$[x^2]=5$$ $$[3 \cdot x]=7$$

$$(f) \quad {7 \over 3} \le x \lt \sqrt{6}$$

The seventh is,

$$[x^2]=6$$ $$[3 \cdot x]=8$$

$$(g) \quad {8 \over 3} \le x \lt \sqrt{7}$$

Something goes wrong with this one. We must discard this solution because it is inconsistent. We're basically done.

The eighth is,

$$[x^2]=7$$ $$[3 \cdot x]=9$$

Which has an empty solution. We're done. We've covered all the cases. Higher numbers result in no solution. Lower numbers result in imaginary solutions.

Putting these together, it's clear that,

$${2 \over 3} \le x \lt {4 \over 3}$$ or $$\sqrt{2} \le x \lt {5 \over 3}$$ or $$\sqrt{3} \le x \lt \sqrt{5}$$ or $${7 \over 3} \le x \lt \sqrt{6}$$

Solves, $(2)$.

Why does this work? The main insight is that the floor function keeps things at integer values for suitable $x$. If we find these bounds in each of the floor terms, we can use this to solve the equation. We also saw that there are only a finite number of equations to solve, as only a few have intersecting solution sets. The ones that do, make up the solution to $(2)$.

See wolfram for the full solution and graph.

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    $\begingroup$ There are solutions for some $x<1$. $\endgroup$ – Marconius Sep 26 '15 at 14:53
  • $\begingroup$ @Marconius thanks :) updated the answer. $\endgroup$ – Zach466920 Sep 26 '15 at 15:18
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From the definition of the floor function we have

[correction here] $$\begin{align} x^2-1 &< &\lfloor x^2 \rfloor &\le x^2 \\ 3x-1 &< &\lfloor 3x \rfloor &\le 3x \end{align}$$

and so $$\lfloor x^2 \rfloor - \lfloor 3x \rfloor + 2 > x^2-3x-1+2=x^2-3x+1$$

So all solutions to the floor equation must lie between the two roots of $x^2-3x+1=0$. That is

$$\frac{3-\sqrt5}{2}\le x \le \frac{3+\sqrt5}{2}$$

Now trying various cases:

$$\begin{align} \lfloor 3x \rfloor=2,&\lfloor x^2 \rfloor=0&\implies x\in\left[\frac{2}{3},1\right)\cap[0,1)=\left[\frac{2}{3},1\right) \\[1em] \lfloor 3x \rfloor=3,&\lfloor x^2 \rfloor=1&\implies x\in\left[1,\frac{4}{3}\right)\cap[1,\sqrt2)=\left[1,\frac{4}{3}\right) \\[1em] \lfloor 3x \rfloor=4,&\lfloor x^2 \rfloor=2&\implies x\in\left[\frac{4}{3},\frac{5}{3}\right)\cap[\sqrt2,\sqrt3)=\left[\sqrt2,\frac{5}{3}\right) \\[1em] \lfloor 3x \rfloor=5,&\lfloor x^2 \rfloor=3&\implies x\in\left[\frac{5}{3},2\right)\cap[\sqrt3,2)=[\sqrt3,2) \\[1em] \lfloor 3x \rfloor=6,&\lfloor x^2 \rfloor=4&\implies x\in\left[2,\frac{7}{3}\right)\cap[2,\sqrt5)=[2,\sqrt5) \\[1em] \lfloor 3x \rfloor=7,&\lfloor x^2 \rfloor=5&\implies x\in\left[\frac{7}{3},\frac{8}{3}\right)\cap[\sqrt5,\sqrt6)=\left[\frac{7}{3},\sqrt6\right) \\[1em] \end{align}$$

So the full solution is:

$$x\in\left[\frac{2}{3},\frac{4}{3}\right)\cup\left[\sqrt2,\frac{5}{3}\right)\cup\left[\sqrt3,\sqrt5\right)\cup\left[\frac{7}{3},\sqrt6\right)$$

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  • $\begingroup$ I don't understand your starting point. I thin that it should be $x^2-1\le\lfloor x^2\rfloor< x^2$ and$3x-1\le\lfloor 3x\rfloor< 3x$ $\endgroup$ – Emilio Novati Sep 26 '15 at 16:00
  • $\begingroup$ Shouldn't it be $x^2-1<\lfloor x^2\rfloor\leqslant x^2$ since $\lfloor y\rfloor\leqslant y<\lfloor y\rfloor+1$? $\endgroup$ – CIJ Sep 26 '15 at 16:19
  • $\begingroup$ Thanks, you are both correct. It has been fixed. The analysis that follows the two floor inequalities is not affected. $\endgroup$ – Marconius Sep 26 '15 at 18:47
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$$\lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2=0$$

\begin{matrix} x-1 &\lt &\lfloor x\rfloor &\lt &x+1\\ x^2-1 &\lt &\lfloor x^2\rfloor &\lt &x^2+1\\ \\ 3x-1 &\lt &\lfloor 3x\rfloor &\lt &3x+1\\ -3x-1 &\lt &-\lfloor 3x\rfloor &\lt &-3x+1\\ \\ x^2-3x-2 &\lt &\lfloor x^2 \rfloor-\lfloor 3x\rfloor &\lt &x^2-3x+2\\ x^2-3x-2 &\lt &-2 &\lt &x^2-3x+2\\ -2 &\lt &-x^2+3x-2 &\lt &2\\ -2 &\lt &x^2-3x+2 &\lt &2\\ 0 &\lt &x &\lt &3 \end{matrix}

The places where $\lfloor x^2\rfloor$ changes value are $\{1, \sqrt 2, \sqrt 3, 2, \sqrt 5, \sqrt 6, \sqrt 7, \sqrt 8, 3 \}$

The places where $\lfloor 3x\rfloor$ changes value are $\{\frac 13, \frac 23, 1, \frac 43, \frac 53, 2, \frac 73, \frac 83, 3 \}$

The sorted list of the union of these two sets is $$\left\{\frac 13, \frac 23, 1, \frac 43, \sqrt 2, \frac 53,\sqrt 3 , 2, \sqrt 5, \frac 73, \sqrt 6, \sqrt 7, \frac 83, \sqrt 8, 3 \right\}$$

\begin{matrix} \text{breakpoint} & \lfloor x^2\rfloor &\lfloor 3x\rfloor & \lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2\\ \frac 13 & 0 & 1 & 1 &\\ \frac 23 & 0 & 2 & 0 & \checkmark\\ 1 & 1 & 3 & 0 & \checkmark\\ \frac 43 & 1 & 4 & -1 &\\ \sqrt 2 & 2 & 4 & 0 & \checkmark\\ \frac 53 & 2 & 5 & -1 &\\ \sqrt 3 & 3 & 5 & 0 & \checkmark\\ 2 & 4 & 6 & 0 & \checkmark\\ \sqrt 5 & 5 & 6 & 1 &\\ \frac 73 & 5 & 7 & 0 &\checkmark\\ \sqrt 6 & 6 & 7 & 1 &\\ \sqrt 7 & 7 & 7 & 2 &\\ \frac 83 & 7 & 8 & 1 &\\ \sqrt 8 & 8 & 8 & 2 &\\ 3 & 9 & 9 & 2 &\\ \end{matrix}

Hence the solution set is

$$ x \in \left[ 2/3, 4/3 \right) \cup \left[ \sqrt 2, 5/3 \right) \cup \left[ \sqrt 3, \sqrt 5 \right) \cup \left[ 7/3, \sqrt 6 \right) $$

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  • $\begingroup$ I'm sorry but this solution is incorrect as, $$[1.4^2]-[3\cdot1.4]+2=-1$$ $\endgroup$ – Zach466920 Sep 26 '15 at 16:39
  • $\begingroup$ @zach I see that. I'm working on it. $\endgroup$ – steven gregory Sep 26 '15 at 18:06
  • $\begingroup$ @Zach466920 I have rewriiten my answer. It seems to be a weaker version of what Marconius did. $\endgroup$ – steven gregory Sep 27 '15 at 0:46
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For any $x$ that provide a solution for your problem, $x$ must satisfy these two inequalities

$$ x^2-3x+2 \geq 0\\ x^2-3x+2<1 $$

From the first equation we see that $(x-2)(x-1)\geq 0$ so we have narrowed the solution space to $\mathbb{R}\setminus ]1,2[$.

The second equation gives us $x^2-3x+1<0$ which gives us the solution space $]\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}[$, approximated to $]0.382,2.618[$

Knowing that both inequalities must be satisfied we get an intersection of these two solution spaces, namely $$ x\in ]\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}[ \quad \setminus \quad ]1,2[\quad = \quad]\frac{3-\sqrt{5}}{2},1] \cup[2, \frac{3+\sqrt{5}}{2}[ $$

Which can be approximated to

$$ x \in ]0.382,1]\cup [2,2.618[ $$

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