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

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Solve for all real $a$ of the equation below:

$\dfrac{1}{\left\lfloor{a}\right\rfloor}+\dfrac{1}{\left\lfloor{2a}\right\rfloor}=a-\left\lfloor{a}\right\rfloor+\dfrac{1}{3}$
 
Mathematics news on Phys.org
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.
 
Last edited:
Good solution by mathbalarka
here is mine

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.
 
Last edited:
Well done, both of you! And thanks for participating! :)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top