Question:
Determine the positive rational number $x$ with the property that $x=3^{[\frac{3x-1}{2}-[\frac{3x}{2}]]-1}$, where $[a ]$ represents the whole part of the real number a.
My idea
Can you please tell me if my idea is right or not?
We know that $x=[x]+\{x\}$, which means that $[x]=x-\{x\}$, which also means that $x=3^{[\frac{3x-1}{2}-\frac{3x}{2}+ \{y\}]-1}$ , where y=$\{\frac{3x}{2}\}$. Thus, $x=3^{[-\frac{1}{2}+y]-1}$.
$\displaystyle 0 \leq y<1 \implies -\frac{1}{2} \leq -\frac{1}{2}+y<\frac{1}{2} \implies \left[-\frac{1}{2}+y\right]= 0,-1$
From here we find that $x=\frac{1}{3},\frac{1}{9}$
Hope one of you can help me! Thank you!
Actually, your subsequent analysis, which indicates that the only candidate solutions are $~x = \dfrac{1}{3},~$ and $~x = \dfrac{1}{9}~$ is elegant. You then simply have to recognize that these are merely candidate solutions.
This implies:
No other solutions are possible.
Each of these two solutions must be manually checked.
In fact, as the comments following the posted question indicate, both candidate solutions satisfy the original problem.
However, the (elegant) approach that you took for this problem, might cause difficulties if applied to other (similar) problems. The less elegant approach that I favor, which will routinely conquer problems of this nature, is shown below.
In this answer, I will use the notation $~\lfloor a\rfloor~$ to refer to the largest integer $~\leq a,~$ and I will use the notation $~\{a\}~$ to refer to the fractional part of $~a.$
Using the excerpted analysis shown at the start of this answer as a starting point, since $~0 \leq y < 1,~$ the expression $~\displaystyle \left\lfloor ~\frac{-1}{2} + y ~\right\rfloor~$ will :
Evaluate to $~-1~$ if $~0 \leq y < (1/2).~$
Evaluate to $~0~$ if $~(1/2) \leq y < 1.~$
Express $x = P + r ~: ~P \in \Bbb{Z}, ~0 \leq r < 1.$
So, to (inelegantly) attack the problem, you must identify exactly which combinations of $~P~$ and $~r~$ will result in :
$0 \leq y < (1/2).~$
$(1/2) \leq y < 1.$
In examining $~\displaystyle \left\{ ~\frac{3x}{2} ~\right\} = \left\{ ~\frac{3P + 3r}{2} ~\right\}, ~$ the first thing to notice is that this will equal $~\displaystyle \left\{ ~\frac{P + 3r}{2} ~\right\},~$ since $~\dfrac{2P}{2}~$ is an integer.
Further, $~\displaystyle \left\{ ~\frac{P + 3r}{2} ~\right\},~$ will evaluate to either $~\displaystyle \left\{ ~\frac{0 + 3r}{2} ~\right\},~$ or $~\displaystyle \left\{ ~\frac{1 + 3r}{2} ~\right\},~$ depending on whether $~P~$ is even or odd. This implies that the two cases of $~P~$ even and $~P~$ odd must be investigated separately.
$\underline{\text{Case 1} ~P ~\text{is even}}$
With $~0 \leq r < 1,~$ you must identify which values of $~r~$ will result in
$~ y = \displaystyle \left\{ ~\frac{0 + 3r}{2} ~\right\} < (1/2),~$ and which values of $~r~$ will result in
$~ y = \displaystyle \left\{ ~\frac{0 + 3r}{2} ~\right\} \geq (1/2).$
$\displaystyle 0 \leq 3r < 3 \implies 0 \leq \frac{3r}{2} < \frac{3}{2}.~$
So, the issue is, which values of $~r~$ will result in the expression $~\left( ~\dfrac{3r}{2} ~\right)~$ being in the union of
$\left\{ ~0 \leq \left( ~\dfrac{3r}{2} ~\right) < (1/2) ~\right\} ~~~~\bigcup ~~~~\left\{ ~1 \leq \left( ~\dfrac{3r}{2} ~\right) < (3/2) ~\right\}.$
Let $~S~$ denote the set that represents the union of the following two intervals
$[0, ~1/3) ~~~\bigcup ~~~[2/3, ~1).$
Let $~T~$ denote the set that represents
$[1/3, ~2/3).$
Then, $~r \in S \implies y < 1/2~$ and $~r \in T \implies y \geq 1/2.$
So, each of the two subcases, $~r \in S,~$ and $~r \in T,~$ will be examined separately, under the prevailing assumption of this section, that $~P~$ is even.
If $~r \in S,~$ then $~y < 1/2,~$ so $~\displaystyle \left\lfloor \frac{-1}{2} + y ~\right\rfloor = -1.~$
This implies that $~x~$ must satisfy $~x = 3^{-2}~$ which will be true if and only if
$~x = \dfrac{1}{9} = P + r ~: ~P = 0, ~r = \dfrac{1}{9}.~$
Since $~0~$ is an even integer, and $~\dfrac{1}{9} \in S,~$ such a candidate value of $~x~$ is satisfactory. Therefore, one of the solutions is $~x = \dfrac{1}{9}.~$
If $~r \in T,~$ then $~y \geq 1/2,~$ so $~\displaystyle \left\lfloor \frac{-1}{2} + y ~\right\rfloor = 0.~$
This implies that $~x~$ must satisfy $~x = 3^{-1}~$ which will be true if and only if
$~x = \dfrac{1}{3} = P + r ~: ~P = 0, ~r = \dfrac{1}{3}.~$
Since (again) $~0~$ is an even integer, and $~\dfrac{1}{3} \in T,~$ such a candidate value of $~x~$ is (also) satisfactory. Therefore, one of the solutions is $~x = \dfrac{1}{3}.~$
So, to conclude this section, when $~P~$ is even, there are exactly two solutions, that are represented by the set $~\left\{ ~\dfrac{1}{9}, ~\dfrac{1}{3} ~\right\}.$
$\underline{\text{Case 2} ~P ~\text{is odd}}$
At this point, for this specific problem, one could use the previous analysis to realize that no solution will be possible when $~P~$ is odd. However, the point of this answer is not merely to solve this particular problem, but rather to provide a guide for all problems of this nature.
With $~0 \leq r < 1,~$ you must identify which values of $~r~$ will result in
$~ y = \displaystyle \left\{ ~\frac{1 + 3r}{2} ~\right\} < (1/2),~$ and which values of $~r~$ will result in
$~ y = \displaystyle \left\{ ~\frac{1 + 3r}{2} ~\right\} \geq (1/2).$
$\displaystyle 0 \leq 3r < 3 \implies \dfrac{1}{2} \leq \frac{1 + 3r}{2} < 2.~$
So, the issue is, which values of $~r~$ will result in the expression $~\left( ~\dfrac{1 + 3r}{2} ~\right)~$ being in
$\left\{ ~1 \leq \left( ~\dfrac{1 + 3r}{2} ~\right) < (3/2) ~\right\}.$
Let $~S~$ denote the set that represents
$[1/3, ~2/3).$
Let $~T~$ denote the set that represents the union of the following two intervals
$[0, ~1/3) ~~~\bigcup ~~~[2/3, ~1).$
Then, $~r \in S \implies y < 1/2~$ and $~r \in T \implies y \geq 1/2.$
So, each of the two subcases, $~r \in S,~$ and $~r \in T,~$ will be examined separately, under the prevailing assumption of this section, that $~P~$ is odd.
If $~r \in S,~$ then $~y < 1/2,~$ so $~\displaystyle \left\lfloor \frac{-1}{2} + y ~\right\rfloor = -1.~$
This implies that $~x~$ must satisfy $~x = 3^{-2}~$ which will be true if and only if
$~x = \dfrac{1}{9} = P + r ~: ~P = 0, ~r = \dfrac{1}{9}.~$
Since $~0~$ is an even integer, and since the prevailing assumption of this section is that $~P~$ must be odd, the solution of $~x = P + r = \dfrac{1}{9},~$ must be rejected, within this section.
Note that this still allows for the possibility that the solution of $~x = \dfrac{1}{9}~$ is viable in the other section, where $~P~$ is assumed even.
If $~r \in T,~$ then $~y \geq 1/2,~$ so $~\displaystyle \left\lfloor \frac{-1}{2} + y ~\right\rfloor = 0.~$
This implies that $~x~$ must satisfy $~x = 3^{-1}~$ which will be true if and only if
$~x = \dfrac{1}{3} = P + r ~: ~P = 0, ~r = \dfrac{1}{3}.~$
Since (again) $~0~$ is an even integer, this solution must also be rejected, within this section.
Therefore, when $~P~$ is odd, no solutions are possible.
$\underline{\text{Final Analysis}}$
In the elegant (and valid) approach taken in the original posting, the values of $~\{ ~1/9, ~1/3 ~\}~$ were identified as the sole candidate values possible. To complete that approach, each of these candidate values had to be manually inspected, to determine whether it satisfied the original constraint.
In the alternative (much less elegant) approach, that generalizes well for problems of this nature, $~x~$ was set to $~P + r,~$ and each individual case/sub-case was explored. In this exploration, any solution found was deemed satisfactory if and only it satisfied the assumption(s) of the case/sub-case [i.e. $~P~$ even or odd, $~r \in S~$ or $~r \in T~$].
Case 1 yielded the two solutions of $~\{ ~1/9, ~1/3 ~\},~$ and Case 2 yielded no solutions.