# Inequality floor function. [closed]

Solve $x^2-1+\lfloor x\rfloor\geq 0$ where $\lfloor \cdot \rfloor$ is greatest integer function. Any other method than graphically.

My thoughts: I replaced $x$ with $I+f$ where $I$ is the integer part and $f$ is the fractional paper. $(I+f)^2\geq 1-I$. After that I took cases. And then am stuck. Preferably I was trying to represent the expression in x assuming inequality as equality and then applying $0\leq f<1$.

## closed as off-topic by Carl Mummert, Dietrich Burde, user296602, John Hughes, Brevan EllefsenMar 23 '18 at 18:02

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Dietrich Burde, Community, John Hughes, Brevan Ellefsen
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• Guys answer please. – Stuart Kapoor Mar 23 '18 at 17:25
• @StuartKapoor Why should we? We have no obligation to. – Andrew Li Mar 23 '18 at 17:26
• When questions are marked Urgent or the like, the sense is that there is an ongoing exam or such and people are reluctant to get involved. You must have tried something. Please edit your post to show your efforts and to indicate where you are getting stuck. – lulu Mar 23 '18 at 17:29
• I replaced x with I and f where I is the integer part and f is the fractional paper . (I+f)^2>=1-I. After that I took cases. And then am stuck. Preferably I was trying to represent the expression in x assuming inequality as equality and then applying 0<=f<1 – Stuart Kapoor Mar 23 '18 at 17:32
• I have been trying this question for hours without doing it graphically. So was really eager for the answer. Otherwise it isn't a problem – Stuart Kapoor Mar 23 '18 at 17:35

Hint. Note that $x-1\leq \lfloor x\rfloor\leq x$. Hence $$x^2+x-2\leq x^2-1+\lfloor x\rfloor\leq x^2+x-1.$$ In particular, if $x^2+x-2=(x+2)(x-1)\geq 0$ then $x^2-1+\lfloor x\rfloor\geq 0$.

Can you take it from here?

• Graphical answer is 1<=x and x<=-√3 – Stuart Kapoor Mar 23 '18 at 17:52
• Have you used my hint? – Robert Z Mar 23 '18 at 17:55
• Yes from your hint am getting very close answer . -2 instead of -root3 – Stuart Kapoor Mar 23 '18 at 18:35
• Well done. Now you can refine your answer by investigating what happens in the interval $(-2,1)=(-2,-1)\cup[-1,0)\cup [0,1)$. – Robert Z Mar 23 '18 at 18:44

The reason why your post is not getting answers and is instead getting downvotes is that this seems like a homework, quiz, or test question you want us to solve. That is not the purpose of this site; we are here to help you learn. Perhaps if you edit your posts in the future to include context as well as what you have already tried you will get help faster.

That being said, I do want to help you. So lets consider this inequality. We want $x^2-1+\lfloor x\rfloor \geq 0$. Lets instead write this as $\lfloor x\rfloor \geq 1-x^2$. What can you say about the right hand side (RHS)? This is a function you can easily understand. What can you say about the left hand side (LHS)? The floor function function is a bit awkward to visualize if you are not used to it, so you should draw the graph for values of $x$ between, say, $-2$ and $2$.

Since we want the LHS to be greater than or equal to the RHS, and since the behavior of the sides should be easy to understand for most values of $x$, all you will have to do is treat a very small domain on a case by case basis.

As per your comments posted above, you should avoid (when possible) attempting to deal with integer and fractional parts separately. This can be useful in some contexts, but here I do not think it is helpful. And while we all want to be able to solve all problems with the raw power of analysis, do not look down on graphical tools.