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There was a multiple choices saying:

Find the value of $\bigl\lfloor x\bigr\rfloor+\bigl\lfloor x^2\bigr\rfloor+\bigl\lfloor x^3\bigr\rfloor+\bigl\lfloor x^4\bigr\rfloor$ knowing that $x^2+x<0$

The right answer is $-2$.

For solving this Maths test, I solved $x^2+x<0$ which is $(-1,0)$ and then find the value of each term of the sum.

My question is: Is there any approach more formal than I did? Thank you for your time.

  • $\lfloor x\rfloor$ is floor$(x)$
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    Your approach can be done completely formally.2012-09-08
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    There is nothing casual about your approach. for example, if $-1\lt x\lt 0$ then $0\lt x^2\lt 1$ so $[x^2]=0$. The other are done the same way.2012-09-08
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    Babak, I don't know what you mean by formal. I can't imagine any approach that would be better in any significant way.2012-09-08

2 Answers 2

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As $x^2+x<0, (x-0)(x-(-1))<0$

Now the product two terms is negative, so

either ($x-0>0$ and $x-(-1)<0$) or ($x-0<0$ and $x-(-1)>0$).

If $x>0$ and $x<-1\implies -1>x>0\implies -1>0$, which is clearly impossible.

If $x<0$ and $x>-1$, $-1

$\implies -1

$\implies 0.

So, $\bigl\lfloor x\bigr\rfloor+\bigl\lfloor x^2\bigr\rfloor+\bigl\lfloor x^3\bigr\rfloor+\bigl\lfloor x^4\bigr\rfloor=-1+0-1+0=-2$

In general, if $(x-a)(x-b)<0$ where $a,

either $x and $x>b\implies a>x>b\implies a>b$, but $a(given)

or $x>a$ and $x Here $a=-1,b=0$

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    Your solution of the inequality is, I think, incorrect: the system "or", as you wrote, exists for a quadratic inequality of this kind iff the inmequality sign was exactly the opposite one. In the present case, the inequality's solution is directly $\,-1, which is a system "and" solution. All this is pretty clear if we draw the parabola $\,x^2+x=x(x+1)\,$ and locate its roots.2012-09-08
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    @DonAntonio, could you please explain "the inequality sign was exactly the opposite one"?2012-09-08
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    I mean that the inequality $\,(x-a)(x-b)>0\,\,,\,\,a has precisely the solution $\,xb\,$ , which is an "or" system, whereas the *opposite* inequality $\,(x-a)(x-b)<0\,\,,\,\,a has the solution $\,a , which is an "and" system.2012-09-08
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    @Don: you're focusing too much on form: the end result are the same. One common method to solving inequalities is to solve for the signs, then use the signs to solve for the variables. In this particular case, it means you get a spurious sign solution you have to consider and rule out.2012-09-08
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    @Hurkyl, I can't tell you. I've taught this stuff to high school kids and refreshed it to university ones, and we try to be formal and precise. The very beautiful interaction between algebra and geometry in this particular case makes it very easy both to prove and then to solve this kind of inequalities, and that's why I stressed what I did. That the solution "is the same" does not, imho, justify imprecisions of mistakes, because later we might face another cases where we can be misled.2012-09-08
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    @DonAntonio: it's not an imprecision or a mistake. One of the standard methods to solve "$AB < 0$" is to break it into two cases: "$A < 0$ and $B > 0$" and "$A > 0$ and $B < 0$", and then every solution to the original equality is either a solution to the first case or a solution to the second case. We may be able to be clever (or apply a different general tool) in this particular situation, but that doesn't make it wrong to apply this general tool instead.2012-09-08
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    Again, this might be a matter of education and/or custom: in the very particular case of a factored quadratic, as in our present case, any kid in high school around here writing both cases in general will see, at least, several points gone in that question's grading. I understand what you mean, though, and it may be that somewhere else they don't treat this case as we do here, but I've seen high school material from the USA and Spain, say, where they deal with it as we do. This is a nice convo and if you're interested in continuing it perhaps we could open a meta thread.2012-09-08
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    I'm not erasing my first comments in order for others to see what's been going on here, yet I will upvote the answer instead. +12012-09-08
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$$x^2+x<0\Longrightarrow x(x+1)<0\Longleftrightarrow -1

Thus, passing to the floor function under the above condition:

$$\sum_{n=1}^4\lfloor x^n\rfloor=-1+0-1+0=-2$$

Of course, the above is just what you did slightly more fleshed out.

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    Thanks Don for your attempt here. Thanks for your time.2012-09-08
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    Any time, @BabakSorouh.2012-09-08