Solve the equation $\{x\}+\frac{1}{\{x\}}=2+\left\{\frac{1}{[x]}\right\}$.

Question

the question

Solve the equation $\{x\}+\frac{1}{\{x\}}=2+\left\{\frac{1}{[x]}\right\}$, where $[a]$ and $\{ a\}$ represent the whole and fractional parts of the real number $a$ respectively.

The idea

I tried getting $\{\frac{1}{[x]}\}$ to a more acceptable form so I tried these cases:

case 1 : if $[x]=0,1,-1$ then the equation would look like this:

$\{x\}+\frac{1}{\{x\}}=2$ which means that $\{x\}=1$ which isn't possible.

case 2 : if $[x]$ isn't $0,1,-1$ then $\frac{1}{[x]} <1 \implies \left\{ \frac{1}{[x]}\right\}=\frac{1}{[x]}$

From here we simplified the equality to:

$$\{x\}+\frac{1}{\{x\}}=2+\frac{1}{[x]} \implies \frac{{{(\{x\}-1})}^2}{\{x\}}=\frac{1}{[x]}$$

From here I tried using the fact that $\{x\}+ [x]=x$ or get into some inequality but got to nothing useful. Hope one of you can help me! Thank you!

Answer 1

Let

$$ x = a + b \quad$$

Where $a \in \mathbb{Z}, b \in \mathbb{R} \quad 0 \leq b \lt 1$, We have integer part $a$ and fractional part $b$

Now we get

$$b + \frac{1}{b} = 2 + \{\frac{1}{a}\}$$

$$ \implies a,b \neq 0 \quad$$

Case 1

When $a=\{-1, 1\}$, we have $\{\frac{1}{a}\} = 0$, which implies $b=1$ but we have $0 \lt b \lt 1$, hence no solutions exists when $a=\{-1, 1\}$

Case 2

When $a \geq 2$, we have $\{\frac{1}{a}\} = \frac{1}{a}$

$$b + \frac{1}{b} = 2 + \frac{1}{a}$$

$$ ab^2 + a = 2ab + b $$

$$ ab^2 + b(-2a-1) + a = 0$$

Now this a quadratic in terms of $b$

$$ b = \frac{2a+1 \pm \sqrt{4a + 1}}{2a} $$

Now we use the inequality by which $b$ was defined

$$ 0 \lt 1 + \frac{1 \pm \sqrt{4a + 1}}{2a} \lt 1 $$

Solving which yields

$$0 \lt 4a^2$$

Hence there exists a value of $b$ s.t. $0 \lt b \lt 1$ for all $a \geq 2$

Case 3

When $a \leq -2$, we have $\{\frac{1}{a}\} = \frac{1}{a} - \lfloor \frac{1}{a} \rfloor$

As $a \leq -2$, we get $\lfloor \frac{1}{a} \rfloor = -1$

Therefore $\{\frac{1}{a}\} = \frac{1}{a} + 1$

Now we have

$$b + \frac{1}{b} = 2 + \frac{1}{a} + 1$$

$$ab^2 + a = 3ab + b$$

$$ab^2 + b(-3a - 1) + a = 0$$

Solving the quadratic again we get

$$ b = \frac{3a+1 \pm \sqrt{5a^2 + 6a + 1}}{2a} $$

Using the inequality again we get

$$ 0 \lt \frac{3a+1 \pm \sqrt{5a^2 + 6a + 1}}{2a} \lt 1 $$

Solving the inequality you get

$$5a^2 + 5a + 2 > 0$$

Which is true for all values of $a$, hence there exists $b$ s.t. $0 \lt b \lt 1$ for all $a \leq -2$

Answer

There exists a solution for all $x \leq -2$ and $2 \leq x$

Method

A common technique to solve problems involving questions about whole part and fractional part of a number is to write the number in the form $x=a+b$ and then proceed to solve some diophantine equation as done above

Answer 2

I think if you turn to $x=n+p , \ 0\le p <1, n=[x]$ it show, there is no negeative solution $$ \frac{{{(\{x\}-1})}^2}{\{x\}}=\frac{1}{[x]}\\ (p-1)^2=\frac np $$so if $x<0 \to [x]=n=-1,-2,-3,...$and $$(p-1)^2=\frac {0 \le p<1 }{(-) } \ <0 \to no-solution$$ and for $x>0$ we have $$1\le x<2 \to n=1 \to (p-1)^2=\frac { p }{1 } \to p^2-3p+1=0 \to p\sim 1.38 \to x=n+p=1.382\\ 2\le x <3 \to n=2 \to (p-1)^2=\frac p2 \to x=\frac 52\\ 3\le x <4 \to n=3 \to (p-1)^2=\frac p3 \to x\sim 3.56\\ \vdots$$

Answer 3

Simplifying yields $$\frac{(\{x\} - 1)^2}{\{x\}} = \left \{ \frac 1{\lfloor x \rfloor } \right\}$$

Note that we can not discard negative solutions since the fractional part is defined as lying between $0$ and $1$, so the RHS is positive.

We divide this into cases:

1. Obviously, $0\le x<1$ gives a zero denominator inside the fractional part in RHS, so it's impossible.
2. For $\{x\} = 0$, the LHS denominator gives zero, which is again impossible. Hence, no solution is an integer.
3. For $1<x<2$, RHS is zero, which is impossible.
4. For $x>2$, we can ignore the fractional part of the RHS. Then, we have a quadratic in $\{x\}$ (note that $\{x\}$ and $\lfloor x \rfloor$ are independent of each other): $$f(\{x\})=\lfloor x \rfloor\{x\}^2 - \{x\}(2\lfloor x \rfloor+1) + \lfloor x \rfloor = 0$$ Since the RHS is a continuous function in $\{x\}$, and $f(1) = -1$ and $f(0) = \lfloor x \rfloor>0$, there exists a solution in $(0,1)$, which can be found by the quadratic formula. Good. (The other root is greater than 1, as found by Vieta's formulae).
5. For $-1<x<0$, the RHS is zero, again impossible.
6. For $x<-1$, $$\left \{ \frac 1{\lfloor x \rfloor } \right\} = \frac 1{\lfloor x \rfloor } + 1$$ Again, we get a quadratic in $\{x\}$: $$g(\{x\}) = \lfloor x \rfloor\{x\}^2 - \{x\}(3\lfloor x \rfloor+1) + \lfloor x \rfloor = 0$$ By analogous reasoning, $g(0) = \lfloor x \rfloor <0$ and $g(1) = -\lfloor x \rfloor - 1 >0$, hence there exists a solution in $(0,1)$. (By Vieta's formulae, the other solution is greater than 1).

In conclusion, $\exists$ exactly one solution for $\lfloor x \rfloor$ if and only if $|\lfloor x \rfloor| \ge 2 \iff \lfloor x \rfloor \ge 2 \text{ or } \lfloor x \rfloor \le -2$.

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For example, some of the solutions found in case 4 (by Quadratic formula) are

• For $\lfloor x \rfloor = 2$, $x = 2.5$.
• For $\lfloor x \rfloor = 3$, $x \approx 3.566$
  and so on.

Some of the solutions found in case 6 (by Quadratic Formula) are:

• For $\lfloor x \rfloor = -2$, $x = -1.5$
• For $\lfloor x \rfloor = -3$, $x \approx -2.549$
  and so on.