Solving $4\lfloor x \rfloor=x+\{x\}$. What is wrong in my solution? Can anyone tell?

Question

$\lfloor x \rfloor$ = Floor Function, and $\{x\}$ denotes fractional part function

Solve for $x$ $$4\lfloor x \rfloor= x + \{x\}$$

$ x = \lfloor x \rfloor + \{x\}$

$\implies x - \{x\} = \lfloor x \rfloor$

$\implies 4x - 4\{x\} = x + \{x\}$

$\implies 3x = 5\{x\}$

Note that ,

$\implies 0 \leq \{x\}<1$

$\implies 0 \leq 5\{x\} < 5 \iff 0 \leq 3x <5 \iff\boxed{ 0 \leq x < \dfrac{5}{3}}$

But the answer is $ x = 0$

Answer 1

You have correctly deduced that $\color{blue}{0\leq x<\frac 53}$. To get the solution, the given equation $4\lfloor x\rfloor=x+\{x\}$ must also be satisfied.

Note that $x=1$ does not satisfy the given equation. If $x>1$, then $\lfloor x\rfloor=1$ and hence $4=x+\{x\}<\frac 53+1=8/3$ (by the blue colored inequality) , which is not possible.

Hence it follows that $0\leq x<1$.

Since, $\lfloor x\rfloor=0$, it follows that $\color{red}{x+\{x\}=0}$ (due to the given equation).

The identity $x=\lfloor x\rfloor+\{x\}$ gives $x=\{x\}$, which along-with the red-colored equation gives: $x=0$.

Answer 2

How did you get from $$4\lfloor x \rfloor = x + \{x\}$$ to $$x=\lfloor x\rfloor + \{x\}?$$

The two equations are not equivalent!

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But OK, ignoring the very awkward way the solution is written, there is nothing really wrong with your solution, in the sense that you correctly prove that if $4\lfloor x \rfloor = x + \{x\}$, then $0\leq x<\frac53$. However, your mistake is then saying that every $x$ satisfying the equation $0\leq x < \frac53$ is also a solution to the original equation, and that is simply not true. For example, $x=\frac12$ does not satisfy the equation.

Answer 3

$0 \leq x <\frac 53 $ is correct. This implies that $[x]=0$ or $[x]+1$. In the first case we get $x+\{x\}=0$ which implies $x=0$. Suppose $[x]=1$. Then we have $1 \leq x <\frac 53 $ from which we get $4<\frac 5 3 +1$ which is wrong. Hence we must have $x=0$.

Answer 4

We have : $$4 \lfloor x \rfloor = x + \{x\} = \lfloor x \rfloor + 2 \{x\}$$ then : $$2 \{x\} = 3 \lfloor x \rfloor \in \mathbb{Z}$$ We deduce that : $$\{x\} = 0 \text{ or } \{x\} = \dfrac{1}{2}$$

1. If $\{x\} = 0$ then $3 \lfloor x \rfloor = 2 \{x\} = 0$. It means that $\lfloor x \rfloor = 0$ then $x = \lfloor x \rfloor + \{x\} = 0$.
2. If $\{x\} = \dfrac{1}{2}$ then $3 \lfloor x \rfloor = 2 \{x\} = 1$. It means that $\lfloor x \rfloor = \dfrac{1}{3}$ wich is impossible because $\lfloor x \rfloor \in \mathbb{Z}$.

The unique solution is $x = 0$.

Answer 5

As mentioned elsewhere, your conclusion is correct, but inconclusive. It might be easier to see what is going on if we write everything in terms of $\lfloor x\rfloor$ and $\{x\}$:

$$ \begin{align} 4\lfloor x\rfloor &= x+\{x\}\tag{1a}\\ &=\lfloor x\rfloor+2\{x\}\tag{1b}\\ \lfloor x\rfloor &=\tfrac23\{x\}\tag{1c}\\ 0&\le\lfloor x\rfloor\lt\tfrac23\tag{1d}\\ \lfloor x\rfloor&=0\tag{1e}\\ \{x\}&=0\tag{1f}\\ x&=0\tag{1g} \end{align} $$ Explanation:
$\text{(1a)}$: copied equation
$\text{(1b)}$: $x=\lfloor x\rfloor+\{x\}$
$\text{(1c)}$: solve $\text{(1a)}$ and $\text{(1b)}$ for $\lfloor x\rfloor$
$\text{(1d)}$: $\text{(1c)}$ and $0\le\{x\}\lt1$
$\text{(1e)}$: $\lfloor x\rfloor$ is an integer
$\text{(1f)}$: combine $\text{(1c)}$ and $\text{(1e)}$
$\text{(1g)}$: $x=\lfloor x\rfloor+\{x\}$