# Solving the Inequality $\dfrac{x-1}{\lfloor x \rfloor}\ge 0$

Find all solutions of $$\dfrac{x-1}{\lfloor x \rfloor}\ge 0$$

 I know how to solve the Inequality $\dfrac{x-1}{ x }\ge 0$ using the Wavy-Curve/Method of Intervals technique. However I don't know solve $\dfrac{x-1}{\lfloor x \rfloor}\ge 0$ because of the $\lfloor x \rfloor$ term in the denominator. I'm not sure as to how it would affect the method of solving.  Could somebody please show me how to use the Wavy Curve method in this case? Many thanks!

PS. Suppose the Inequality was $\dfrac{x-1}{ y }\ge 0$. How then could it be solved? Would the Wavy Curve method still be applicable? Edit: Lastly how wcould the inequality $\dfrac{\lfloor x \rfloor-1}{x}\ge 0$ be solved using the Wavy Curve?

• Why isn't it the trivial answer $x\in (-\infty;0) \cup [1;+\infty)$ ? – Salsifis Jun 9 '16 at 13:47
• Sir, I'm not sure. Actually I had thought that the floor function would affect the Inequality. – user342209 Jun 9 '16 at 13:50
• The domain is $[x<0]\cup[x\geq1]$. It's pretty easy to see that the inequality holds in each part, hence in the entire domain. – barak manos Jun 9 '16 at 13:52
• @ArchisWelankar: No solution????? – barak manos Jun 9 '16 at 13:52
• The floor function affects the domain of definition, but not the sign of the denominator. – Salsifis Jun 9 '16 at 13:54

Multiply both sides by $\lfloor x \rfloor$. The direction of the inequality is preserved if $\lfloor x \rfloor > 0$, and reversed if $\lfloor x \rfloor < 0$.

So if $\lfloor x \rfloor > 0$ (i.e., when $x \ge 1$), the inequality becomes $x - 1 \ge 0$. Combine these two pieces of information to get $x \in [1,\infty)$.

If $\lfloor x \rfloor < 0$ (i.e., when $x<0$), then the inequality becomes $x-1 \le 0$. The region where the conditions $x < 0$ and $x - 1 \le 0$ overlap is $x \in (-\infty, 0)$.

Thus the answer is $(-\infty, 0) \cup [1,\infty)$.

• You can use this general technique to answer the variations to your question. Multiply both sides by the denominator, and then consider two cases depending on whether the sign of the inequality was preserved or reversed. – Théophile Jun 9 '16 at 14:05

The inequality holds if and only if $\lfloor x\rfloor\neq 0$ and $x-1$ and $\lfloor x\rfloor$ have the same sign.

If $0\le x<1$ the expression is undefined; if $x<0$, numerator and denominator are negative; if $x\ge 1$ they are positive.

For $\lfloor x\rfloor>0$, $x-1\ge0$ and the ratio is non-negative.

For $\lfloor x\rfloor<0$, $x-1<0$ and the ratio is positive.

The wavy curve method isn't appropriate here as the function is discontinuous.