Find the solutions of $3\{x\} = 2x + [x]$ where $\{.\}$ denotes the fractional part and $[.]$ denotes the greatest integer function.

Question

Find the solutions of $3\{x\} = 2x + [x]$ where $\{.\}$ denotes the fractional part and $[.]$ denotes the greatest integer function.

I tried to approach the problem as follows:

$3\{x\} = 3[x] + 2 \{x\}$ (as $x = [x] + \{x\}$)

$\{x\}=3[x]$

now we know that LHS belongs to $0$ to $1$ ($0$ included $1$ excluded) hence $3[x]$ belongs to $[0,1)$ hence $[x]$ belongs to $[0,1/3)$ therefore $[x] = 0$ and hence $x$ belongs to $[0,1)$ but at $x=0.5$ and in fact, all the values except $0$ don't satisfy the given equation. why is this contradiction arising?

Answer 1

$$3\{x\} = 3[x] + 2 \{x\}\\\Rightarrow\{x\}=3[x]\\\Rightarrow[x]+\{x\}=4[x]\\\Rightarrow x=4[x]$$

There is no other number than $0$ which is four times of it's greatest integer value. Therefore, $x=0$

Answer 2

Let's say $x= a+b$ where a is integer and b is rational part, $0\le b<1$

$$3b = 2a + 2b + a$$ $$b = 3a$$

But b is between 0 and 1. So:

$$0\le 3a < 1$$ $$0\le a < \frac{1}{3}$$

$a$ was integer. Only one integer satisfies the inequality is 0. If $a=0$, it follows $b=0$ and $x=0$.

Answer 3

$3\{x\}=2x+[x]$, let $x=n+q, n\in I, 0\le q<1$. Then $3q=2n+2q+n \implies 0\le 3n=q < 1 \implies n=0, q=0$. $x=0+0=0$ is the only solution.

Answer 4

$$\mathrm{3}\left\{{x}\right\}=\mathrm{2}{x}+\left[{x}\right] \\ $$ $$\mathrm{3}\left\{{x}\right\}=\mathrm{2}\left(\left\{{x}\right\}+\left[{x}\right]\right)+\left[{x}\right] \\ $$ $$\mathrm{3}\left[{x}\right]=\left\{{x}\right\}\Rightarrow\left[{x}\right]=\mathrm{0} \\ $$ $$\mathrm{3}\left({x}−\left[{x}\right]\right)=\mathrm{2}{x}+\left[{x}\right] \\ $$ $${x}=\mathrm{5}\left[{x}\right]=\mathrm{0} \\ $$