Algebra – Solving Equations Involving Floor Functions

algebra-precalculusceiling-and-floor-functions

I'm looking for a solving procedure for this type of exercises.

If $[x]$ represents the floor of $x$, solve the equation:

$$\left[\frac{6x+5}8\right]=\frac{15x-7}5$$

Choose the correct answer:

a) {$\frac{4}5$}

b) {$\frac{3}4$}

c) {$\frac{7}{15}$,$\frac{4}5$}

d) {$\frac{6}{15}$}

e) {$\frac{1}{2}$,$\frac{3}4$}

f) {$\frac{1}{2}$,$\frac{4}5$}

Can someone please explain how I can solve this equation? Thank you very much!

ps. I asked a similar question, unfortunately, the same method doesn't seem to work.

Best Answer

Since $\frac{15x-7}{5}$ is required to be an integer, it is of the form $\frac{2+5j}{15}$ for $j$ an arbitrary integer. Thus, the equation to solve becomes $$ \left\lfloor\frac{6\big(\frac{2+5j}{15}\big)+5}{8}\right\rfloor = \frac{15\big(\frac{2+5j}{15}\big)-7}{5}\qquad\text{which simplifies to}\quad \left\lfloor\frac{\frac{4}{5}+2j+5}{8}\right\rfloor = j-1 $$ Now observe that the function on the right hand side increases quicker than the function on the left hand side. So we can solve the question by finding a maximal solution to the inequality $$ j-1\leq\left\lfloor\frac{\frac{4}{5}+2j+5}{8}\right\rfloor $$

A warning: But here we should be careful, because a maximal solution to this inequality is not necessarily a solution to our equality. Also, there might be more solutions than just the maximal solution $j_0$ to this inequality. But since the left hand side and the right hand side are linear in $j$, the set of solutions forms an interval in $\mathbb{Z}$. This means that to give all the solutions, it suffices to find the largest integer below $j_0$ for which the inequality is strict.

Note that for any $n\in\mathbb{Z}$ and $a\in\mathbb{R}$, the inequality $n\leq\lfloor a\rfloor$ holds if and only if $n\leq a$ holds. This helps, since now we can continue to compute $$ \begin{split} j-1 \leq \left\lfloor\frac{\frac{4}{5}+2j+5}{8}\right\rfloor & \Leftrightarrow j-1 \leq \frac{\frac{4}{5}+2j+5}{8}\\ & \Leftrightarrow 8j-8\leq \frac{4}{5}+2j+5\\ & \Leftrightarrow 10j\leq 23\\ & \Leftrightarrow j\leq 2. \end{split} $$ Therefore, we get the maximal solution for the equation by substituting $j=2$ in $x=\frac{2+5j}{15}$, i.e. $x=\frac{4}{5}$. This is indeed a solution to the equality. As we have noted before, $\frac{4}{5}$ is not necessarily the only solution. Indeed, it is easy to check that $j=1$ also gives a solution to the equation. But for $j\leq 0$ the inequality is strict, so the only solutions to the equation are $x=\frac{4}{5}$ and $x=\frac{7}{15}$.

Here is a link to the answer to another question, where the OP had to deal with the floor function. Maybe you'll find it useful.

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