Does $\lfloor x \rfloor.\{x\}=1$ has negative solution?

Question

If we solve this equation $$\lfloor x \rfloor.\{x\}=1$$ $$x=\lfloor x \rfloor+\{x\}=n+p\\ 0\leq p<1 , n\in \mathbb{Z} \to $$we can rewrite $$\lfloor x \rfloor.\{x\}=1\\n(x-n)=1 \\\to x=n+\frac{1}{n}$$ It is obvious for $$n=2,3,4,... \text{works} \\\text{for example }n=2 \to x=2.5 \to \lfloor 2.5 \rfloor.\{2.5\}=2.\frac12=1 \checkmark $$ also it is obvious $n\neq 1,\neq -1$ but if if we plug for example $n=-2 $ does not work $$n=-2 \to x=-2+\frac{1}{-2}=-2.5 \\\to \lfloor -2.5 \rfloor.\{-2.5\}=-3.\frac{+1}{2}\neq 1$$ now my question is :Why answer does not work for negatives ?

Thanks in advance .

Answer 1

Answer does not work for negatives because for $x<0$

$\{x\}=x-\lfloor x\rfloor -1$

the equation translates in

$\lfloor x \rfloor.\{x\}=1\\(n-1)(x-n+1-1)=1 \\\to x=n+\dfrac{1}{n-1}$

Hope this helps

Answer 2

Remark: Only notice that $0 \leq \{x\}$.
[Or equivalently $ - \{x\} \leq 0$ . ]

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The answer is No;
suppose on contrary that $\lfloor x \rfloor + \{x\} < 0$;
so by the above remark, we can conclude that: $\lfloor x \rfloor < 0$;
therefor we must have: $$\lfloor x \rfloor . \{x\} \leq 0 \Longrightarrow 1 \leq 0; $$

which is an obvious contradiction.

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In your conclusion you have been defined: $n:=\lfloor x \rfloor$; and then $\{ x \}:=\dfrac{1}{n}$.

But the bug in your conclusion is:

you did not notice that:

$$0 \leq \dfrac{1}{n} < 1 \Longleftrightarrow n \in (1,\infty).$$

Answer 3

Well, let's try it.

For $x < 0$

Let $-n \le x < -n + 1 \le 0$ for some $n \ge 1$

Then $[x] = -n$.

So if $[x]*\{x\}= 1$ then $\{x\} = - \frac 1n$.

So $x = [x] + \{x\} = -n +(-\frac 1n) = -n -\frac 1n $.

But $-n - \frac 1n < -n \le x$.

... or ... one could have noticed right away that $0 \le \{x\} < 1$ so $\{x\}$ is never negative so $\{x\}= -\frac 1n$ is impossible.

... or ... $[x] \le x < 0$ while $0 \le \{x\} < 1$ so $[x]*\{x\} \le 0 < 1$.