What Values of \(a\) Satisfy the Equation Involving Floor Functions?

[anemone]

Solve for all real $a$ of the equation below:

$\dfrac{1}{\left\lfloor{a}\right\rfloor}+\dfrac{1}{\left\lfloor{2a}\right\rfloor}=a-\left\lfloor{a}\right\rfloor+\dfrac{1}{3}$

 

[mathbalarka]

Spoiler

Note that $\{x\} = x - \lfloor x \rfloor$ never takes negative values in $\Bbb R$. As $\lfloor x \rfloor$ on the other hand is negative for negative real arguments, there is no solution for the equation in $\Bbb R^{-}$

We will split this equation into two cases. If $\{a\} < 1/2$, then it is clear that $\lfloor 2a \rfloor = 2\lfloor a\rfloor$ hence $$\frac{1}{\lfloor a \rfloor} + \frac{1}{\lfloor 2a \rfloor} = \frac1{\lfloor a \rfloor} + \frac{1}{2\lfloor a \rfloor} = \frac{3}{2} \frac1{\lfloor a \rfloor} = \{ a \} + \frac1{3} \Longrightarrow \frac{9}{2} = \lfloor a \rfloor \left ( 3 \{a\} + 1 \right ) \tag{+}$$

On the other hand, $$0 \leq \frac{3}{2} \frac1{\lfloor a \rfloor} = \{a\} - \frac13 < \frac56 \Longrightarrow \frac95 < \lfloor a \rfloor \leq \frac92$$

However, as $\lfloor a \rfloor$ is an integer, $\lfloor a \rfloor$ can be either $2$, $3$ or $4$. Using $(+)$, corresponding values of $\{ a \}$ are $5/12$, $1/6$ and $1/24$ respectively. Thus adding $\lfloor a \rfloor$ and $\{a\}$, we get the solutions $29/12$, $19/6$ and $97/24$ respectively.

For the $\{ a \} \geq 1/2$ case, note that $\lfloor 2a \rfloor = 2\lfloor a \rfloor + 1$ thus

$$\frac1{\lfloor a \rfloor} + \frac1{2\lfloor a \rfloor + 1} = \{a\} + \frac13 \geq \frac{5}{6} \Longrightarrow \frac{3\lfloor a \rfloor + 1}{2 \lfloor a \rfloor^2 + \lfloor a \rfloor} \geq \frac56 \Longrightarrow 13\lfloor a \rfloor + 6 \geq 10 \lfloor a \rfloor^2$$ Which is satisfied only for $\lfloor a \rfloor = 1$ (as $0$ is not invertible) but that's not possible, as the strict inequality holds : $$1 > \{a\} = \frac1{\lfloor a \rfloor} + \frac1{2\lfloor a \rfloor + 1} - \frac13 = \frac43 - \frac13 = 1$$

Thus we conclude that $29/12$, $19/6$ and $97/24$ are the only possible solutions.

 

[kaliprasad]

Good solution by mathbalarka
here is mine

Spoiler

The $rhs \lt \dfrac{4}{3}$

let a = b + f where b is integral part and f fractional part <1

a cannot be 1 as LHS = ${3/2}$

if f > .5 integral part of 2a > 2 * integral part of a

f cannot be > .5 as for a = 2

LHS = $\dfrac{1}{2} + \dfrac{1}{5} \lt .5+\dfrac{1}{3}$

now LHS = $\dfrac{3}{2b} \ge \dfrac{1}{3}$ or $b \le 4.5$

so b need to be checked for 2,3,and 4 to give the values for a which is same as above.

 

[anemone]

Well done, both of you! And thanks for participating! :)

 

[CellCoree]

The given equation involves the floor function, which rounds a real number down to the nearest integer. In order to solve for all real values of $a$, we must first consider the possible values of the floor function. Since the floor function rounds down, we know that $\left\lfloor{a}\right\rfloor \leq a$ and $\left\lfloor{2a}\right\rfloor \leq 2a$. This means that the left side of the equation will always be greater than or equal to the right side.

To solve for all real values of $a$, we can first consider the case where $\left\lfloor{a}\right\rfloor = a$. In this case, the equation becomes $\frac{1}{a}+\frac{1}{2a}=0$, which has no real solutions.

Next, we can consider the case where $\left\lfloor{2a}\right\rfloor = 2a$. This means that $\left\lfloor{a}\right\rfloor = a - \frac{1}{2}$, since the floor of $2a$ will always be one less than the floor of $a$. Substituting this into the original equation, we get $\frac{1}{a-\frac{1}{2}}+\frac{1}{2a}=a-\left(a-\frac{1}{2}\right)+\frac{1}{3}$. Simplifying, we get $\frac{2}{a}=\frac{2}{3}$, which has a solution of $a=3$.

Finally, we must consider the case where $\left\lfloor{a}\right\rfloor < a$ and $\left\lfloor{2a}\right\rfloor < 2a$. In this case, we can rewrite the equation as $\frac{1}{\left\lfloor{a}\right\rfloor}+\frac{1}{\left\lfloor{2a}\right\rfloor}=\left\lfloor{a}\right\rfloor+\frac{1}{3}$. Since the left side is always greater than or equal to the right side, the only possible solution is when $\left\lfloor{a}\right\rfloor = 1$ and $\left\lfloor{2a}\right\rfloor = 2$. This gives us the solution