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I am trying to solve the following question involving floor/greatest integer functions.

$3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$ with the notations $\lfloor x \rfloor$ denoting the greatest integer less than or equal to $x$ and $\{x\}$ to mean the fractional part of $x$.

I used the following property for floor functions.

$n\leq x$ if and only if $n \leq \lfloor x \rfloor$ where $n\in \mathbb{Z}$

Let $p=\lfloor x^{2} \rfloor$, then

$p\leq \lfloor x^{2} \rfloor < p+1$

$\rightarrow p \leq x^{2} < p+1$

$\rightarrow \sqrt{p} \leq x < \sqrt{p+1}$ , since $\sqrt{p} \in \mathbb{Z}$

$\rightarrow \sqrt{p} \leq \lfloor x \rfloor < \sqrt{p+1}$ We then have $\sqrt{p} = \lfloor x \rfloor$

Since $\{x\}=x-\lfloor x \rfloor,$

$3\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2\{x\}= 3\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2(x-\lfloor x \rfloor)= 5\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2x=0$

Substituting $p$, $\sqrt{p}$ for $\lfloor x^{2} \rfloor$ and $\lfloor x \rfloor$ respectively, and also letting $x= \sqrt{p}, $ we get $p = 3\sqrt{p}$ solving for $p$ gives $p=0, 9$, and hence $x=0, 3$

The problem is that according to the solution for the problem, $x$ also equals to $\frac{3}{2}$ for $\{x\}=\frac{1}{2}$ since $2\{x\}\in \mathbb{Z}$. However, by definition for $\{x\}$, $0 \leq \{x\} < 1$, then $0 \leq 2\{x\} < 2$. How can $\{x\}=\frac{1}{2}$ and how do I use this to obtain $x=\frac{3}{2}$. I am not sure what I am missing. IF I made any mistakes in my reasoning. Can someone point it out to me please. Thank you in advance.

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  • $\begingroup$ It's not true that $\sqrt{p}\in\mathbb{Z}$ Say $x=1.5$ Then $p=\lfloor 2.25\rfloor = 2,$ and $\sqrt{p}=\sqrt{2}$ $\endgroup$
    – saulspatz
    May 10, 2020 at 0:35
  • $\begingroup$ @saulspatz thank you for pointing that out. $\endgroup$
    – Seth Mai
    May 10, 2020 at 0:45

2 Answers 2

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Let $x = n + r$ where $n = [x]$ and $r = \{x\}$.

Then we have $3n - [n^2 + 2nr + r^2]=2r$

$3n - n^2 - [2nr + r^2] = 2r$

and.... oh, hey, the LHS is an integer the RHS being $2\{x\}$ means $\{x\} = 0$ or $0.5$.

Two options $x$ is an integer and $x = [x] = n$ and $r=\{x\} = 0$ and we have

$3n-n^2=0$ and $n^2 = 3n$ and $n= 0$ or $n = 3$.

So $x = 0$ and $x=3$ are two solutions.

(Check: $x=0\implies 3[x] - [x^2] = 3*0 - 0 = 0 = \{0\}$. Check. And $x = 3\implies 3[x]-[x^2] = 3[3]- [3^2] = 3*3-9 = 0=\{3\}$. Check.

And if $x = n + \frac 12$ and $r = \frac 12$ then

$3n - n^2 - [2n\frac 12 + \frac 14] = 2\frac 12$

$3n - n^2 - [n + \frac 14] = 1$

$3n -n^2 - n = 1$

$n^2 - 2n + 1 =0$ so $(n-1)^2 = 0$ and $n = 1$.

$x = 1+\frac 12 = 1\frac 12$.

(Check: If $x = 1.5$ then $3[x] - [x^2] = 3[1.5] - [1.5^2] = 3*1 - [2.25]=3-2=1 = 2*\frac 12 = 2\{1.5\}$. Check.)

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Write $\{x\}=x-\lfloor x\rfloor$. Then we have $$ 5\lfloor x\rfloor - \lfloor x^2\rfloor = 2x $$Since the LHS is an integer, the RHS must be as well. There are two cases: $x$ is an integer, or $x$ is a half-integer.

  • $x$ an integer. Drop the brackets: $$ 5x-x^2=2x;\qquad x=0,3 $$
  • $x$ is a half-integer. Write $x=y+1/2$. Then $x^2 = y^2+y+1/4$, and again we can drop the brackets: $$ 5y-(y^2+y)=2y+1; \qquad y=1, x=3/2 $$
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  • $\begingroup$ may I ask how you arrive at $x$ is a half integer. I mean can't $x$ be anything else in between $0$ and $1$? $\endgroup$
    – Seth Mai
    May 10, 2020 at 0:46
  • $\begingroup$ After we substitute $\{x\}=x-\lfloor x\rfloor$, both sides are integers. Then if $2x$ is an integer, $x$ is either an integer or a half-integer. $\endgroup$
    – Integrand
    May 10, 2020 at 0:47
  • $\begingroup$ I think my issue is the following: from $0 \leq \{x\} < 1$, we get $0 \leq 2\{x\} < 2$. So how do I determine where else $\{x\}$ could be. $\endgroup$
    – Seth Mai
    May 10, 2020 at 0:56
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    $\begingroup$ Clearly the LHS is an integer. Then $2\{x\}$ is an integer as well. This means either $2\{x\}=0$, i.e. $x$ is an integer, or $2\{x\}=1$, i.e. $x$ is a half-integer. $\endgroup$
    – Integrand
    May 10, 2020 at 1:00
  • $\begingroup$ I think I see it now, Since $3\lfloor x \rfloor - \lfloor x^{2} \rfloor\in \mathbb{Z}$ and $0 \leq \{x\} < 1$ then, $\frac{3\lfloor x \rfloor - \lfloor x^{2} \rfloor }{2} = \{x\}$ implies that $\frac{3\lfloor x \rfloor - \lfloor x^{2} \rfloor }{2} \leq \{x\}<1$ which forces $\{x\}=\frac{1}{2}$ $\endgroup$
    – Seth Mai
    May 10, 2020 at 1:05

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