# Solving an inequality with the floor operator

Assume that $$A$$, $$B$$, $$C$$ are positive real numbers and that $$I$$ is a positive integer. How could I isolate $$A$$ in the inequality $$\left \lfloor{AB/C}\right \rfloor \geq I$$ ? The best I could do:

$$$$AB/C -1 \geq I \Rightarrow \left \lfloor{AB/C}\right \rfloor \geq I$$$$

Thus, $$A\geq C(1+I)/B$$ is a sufficient condition for $$\left \lfloor{AB/C}\right \rfloor \geq I$$.

• You could simplify your problem by setting $D:=B/C$ and just ask a question about $\left \lfloor{AD}\right \rfloor \geq I$ May 20 '19 at 11:48

$$\left\lfloor\frac{AB}{C}\right\rfloor \ge I$$ $$\implies \frac{AB}{C} \ge I$$
$$\implies A \ge \frac{IC}{B}$$
• Well the floor of a number $\ge$ some integer which implies that the number must be $\ge$ that integer. I skipped the first line. May 20 '19 at 11:47
• Okay I see your point. But it doesn't seem very likely that you can pin down on a value of $A$ from just knowing the result of an inequality. For ex, and I am assuming I have used the definition of floor correctly, $AB/C \ge I$. What better can be said about $A$ then $A \ge IC/B$? May 20 '19 at 11:52
• In fact you are right. I beg your pardon for having had a doubt. The minimal value $A=\dfrac{IC}{B}$ when plugged into into relationship $\lfloor{AB/C}\rfloor \geq I$, one gets $\lfloor{I}\rfloor \geq I$ which is truebecause $I$ is an integer... which is the case. * May 20 '19 at 12:05