# Square root of floor of square root of sum $\sqrt{\lfloor\sqrt{\left|x+y\right|}\rfloor}$

Let $x,y$ $\in\mathbb{R}$ and $\lfloor x\rfloor = 10$ and $\lfloor y\rfloor = 14$.

Need to compute: $\sqrt{\lfloor\sqrt{\left|x+y\right|}\rfloor}= \; ?$

MY IDEA:

$$10\le x < 11$$

$$14\le y < 15$$

So $$24\le x+y < 26.$$

But then i don't know what to say about:

$$\lfloor\sqrt{|x+y|}\rfloor = \ldots$$

Because i don't get what is the use here for the absolute part: what his role here?

And how can i proceed?

Thanks.

-
The absolute value is a no-op there, you can ignore it. Under the constraints, what can $\lfloor \sqrt{x+y}\rfloor$ be? – Daniel Fischer Feb 25 '14 at 22:54
i think that $4$? – feder Feb 25 '14 at 22:56
Well, that or ...? – Daniel Fischer Feb 25 '14 at 22:56
@feder Couldn't it be $5$ as well? – AnonSubmitter85 Feb 25 '14 at 22:57
that or $5$? not sure how to show that, just by intuition – feder Feb 25 '14 at 22:57

What you did so far is good. Now justify these steps: \begin{align*} 24 &\le x + y < 26 \\ 24 &\le |x + y| < 26 \\ \sqrt{24} &\le \sqrt{|x + y|} < \sqrt{26} \\ 4 &\le \left\lfloor{\sqrt{|x + y|}} \right\rfloor \le 5 \\ \sqrt{\left\lfloor{\sqrt{|x + y|}} \right\rfloor} &= 2 \text{ or } \sqrt{5} \end{align*} Indeed, as Ross Millikan states, there are two possible answers. If you expected a single answer, you must not have written the question exactly right.

-
i expected two answers so thanks alot!:) – feder Feb 25 '14 at 23:39

Since you know $x+y > 0$, you know $|x+y|=x+y$ and you can remove the absolute value bars. They were just there to confuse you. You won't get a single answer.

-
so i can tell by the two inequalities that the values of $\lfloor \sqrt{x+y}\rfloor$ could be $4$ or $5$ so the answer would be $2$ or $\sqrt5$? but how can i show algebraic the the values can be $4$ or $5$? – feder Feb 25 '14 at 23:01
Post this as a comment to his question, please. – Brian J. Fink Feb 25 '14 at 23:03
@BrianJ.Fink Why? It is common to answer questions with a hint rather than a full solution when that is deemed appropriate. – 6005 Feb 25 '14 at 23:07
Sorry I misunderstood the words "not...a single answer." – Brian J. Fink Feb 25 '14 at 23:10
@BrianJ.Fink I've made a trivial edit you should be able to remove the dv – TooTone Feb 25 '14 at 23:21