Solution-verification || Solve the equation $\{ \frac {3x + 5}{x + 2} \}+ \big \lfloor \frac{3x + 2}{x + 1} \big \rfloor = \frac{25}{9}$

Question

Question:

Solve the equation

$$\left\{ \frac {3x + 5}{x + 2} \right\}+ \left\lfloor \frac{3x + 2}{x + 1} \right\rfloor = \frac{25}{9}$$

where $\lfloor a \rfloor$, $\{a\}$ represents the greatest integer $≤a$, respectively fractional part of $a$.

My idea:

Ok, so first I wrote $\frac{3x + 5}{x + 2}= 2+ \frac{x+1}{x+2}$ and $ \frac{3x + 2}{x + 1} = 2+ \frac{x}{x+1}$. From here I got $2$ cases:

1. If $x$ is positive: then we got $x<x+1$ and $x+1<x+2$, so:

$$\begin{align}&\left\{\frac{3x + 5}{x + 2} \right\}= \left\{ 2+ \frac{x+1}{x+2} \right\}= \frac{x+1}{x+2}\\&\left\lfloor \frac{3x + 2}{x + 1} \right\rfloor= \left\lfloor 2+ \frac{x}{x+1} \right\rfloor =2\end{align}$$

So we get that: $$\frac{x+1}{x+2} +2 = \frac{25}{9}\:\Longrightarrow\:x= \frac{5}{2}$$

2. If $x$ is negative: similary we get $$\left\{ \frac{3x + 5}{x + 2} \right\}= \left\{ 3 - \frac{1}{x+2} \right\}= \frac{x+1}{x+2}$$

We know that $x$ is negative which makes $ 2>\frac{x}{x+1}>1$ and: $$\left\lfloor \frac{3x + 2}{x + 1} \right\rfloor= \left\lfloor 2+ \frac{x}{x+1} \right\rfloor =3$$ Thus: $$3+ \frac{x+1}{x+2}= \frac{25}{9}\:\longrightarrow\:x=- \frac{13}{11}$$

Can you tell me if my idea and reasoning are corect? Im not sure of the case were $x$ is negative. Hope one of you can help me! Thank you!

Answer 1

I think mistake is in your cases eg.if $x=-1.9999$ according to your 2nd case you get fractional part equal to $1000$

Simplifying as you have

$$ \{ \frac {x + 1}{x + 2} \}+ \big \lfloor \frac{x}{x + 1} \big \rfloor = \frac{7}{9} $$

Trivially $\big \lfloor \frac{x}{x + 1} \big \rfloor=0$ since RHS<$1$ and fractional part is always positive

$$ 0\le{x\over x+1}\Rightarrow x\in(-\infty,-1)\cup[0,\infty)\\ \ \\ {x\over x+1}\lt1\Rightarrow 0\lt{1\over x+1}\Rightarrow x\in(-1,\infty) $$

The common region being $x\in[0,\infty)$

Now we have for some $n\in\mathbb{N}$ $$ {x+1\over x+2}=n+{7\over9}\\ \ \\ x={5+18n\over 2-9n} $$

We have from the first case than $x\ge0$ which gives $n\in[-{5\over18},{2\over9})$ the only integer in this range being $n=0$ corresponding to $x={5\over2}$

Answer 2

There is no negative value that satisfies the equation.

Claim: $\lfloor x\rfloor + \{y\} = \frac{a}{b}$ has exactly one solution for $\lfloor x \rfloor$ and $\{y\}$ where $a>b$, $a,b \in \mathbb Z^+$.

Note that $\{y\}$ by definition is $\in [0,1)$. This means $\lfloor x\rfloor \in \mathbb Z^+$ since it is already $\in \mathbb Z$ by definition. Now we can write that $a = \lfloor{x}\rfloor b + \{y\}b$, where the first two individual terms are positive integers and $0\le b\{y\}\in \mathbb Z<b$. By Euclid's division algorithm, the pair $\lfloor{x}\rfloor$ and $\{y\}b$ always exists and is unique -- it is the only possible positive integer pair that acts as quotient and remainder respectively on dividing $a$ by $b$.

With this claim, it follows we can write $\frac{25}{9} = 2 + \frac{7}{9}$ as the only representation as an integer + proper fraction.

Now $\lfloor \frac{3x+2}{x+1}\rfloor=\lfloor 3 - \frac{1}{x+1}\rfloor = 2$. But $3-\frac{1}{x+1}\geq 3$ for $x \le -1$, while $3-\frac{1}{x+1}<2$ for $x \in (-1,0)$. So, the floor part can't equal $2$ for any negative $x$, hence showing there is no negative solution. Your proof for the positive part is fine.