# The inequation $\lfloor x\rfloor^2-\lfloor x\rfloor-6>0$

The question is to find the solutions to this inequation $$\lfloor x\rfloor^2 - \lfloor x\rfloor - 6 > 0$$

On factorising I got $$(\lfloor x\rfloor-3)(\lfloor x\rfloor+2)>0$$ so $\lfloor x\rfloor>3$ or $\lfloor x\rfloor>-2$ and so on

But in the text book the solution goes like this

$\lfloor x\rfloor>3$ so $x\ge 4$ and $\lfloor x\rfloor<-2$ so $x <-2$ therefore domain is...

Notice that the textbook has reversed the second inequality and it turns out to be correct as I checked it in wolfram alpha.

Can anyone please explain why the book reversed $\lfloor x\rfloor>-2$ into $\lfloor x\rfloor<-2$?

"On factorising I got $(\lfloor x\rfloor-3)(\lfloor x\rfloor +2)>0$, so $\lfloor x\rfloor>3$ or $\lfloor x\rfloor >-2$ and so on"
The bold "so" is wrong: for $a\le b\in\Bbb R$,
$$(t-a)(t-b)>0\iff t< a\vee t>b$$
You applied it incorrectly to the case $a=-2$, $b=3$
• @user453135 No, it is not the same thing. For instance, when $a<b$, all $t\in\Bbb R$ verify that condition. – user228113 Jul 26 '17 at 6:48