In order to solve $\lfloor \sqrt{x}\rfloor =\sqrt{\lfloor x\rfloor }$ , how to express the set defined by : $z^2 \leq x\lt z^2+1 (\space z\in Z$ )

Question

My original goal is to solve for $x$ in $\mathbb R$ :

$$\lfloor \sqrt{x}\rfloor =\sqrt{\lfloor x\rfloor }$$

(1) Since $\lfloor \sqrt{x}\rfloor \in \mathbb Z$ (by definiton), it must be the case that : $\sqrt{\lfloor x\rfloor } \in \mathbb Z$

( 2) In general, if $\sqrt A =z , \text {with}\space \space z \in \mathbb Z$, $A$ can be written as $A= z^2$. Here, $A= \lfloor x\rfloor$, so $\lfloor x\rfloor = z^2, \space z\in \mathbb Z$.

(3) In general : $ \lfloor x\rfloor = F_{\in \mathbb Z } \iff F\leq x \lt F+1$. Here, $F= z^2$ , so $\lfloor x\rfloor = z^2 \iff z^2 \leq x \lt z^2 +1$, with, as before, $z\in \mathbb Z$.

(4) Finally $x$ must be a positive or null real number , unless $\lfloor x\rfloor \ngeq 0 $, and $\sqrt {\lfloor x\rfloor}$ would not be defined.

Consequently, the solution set $S$ of the original equation is :

$$S= \{x| x\geq 0 \land z^2 \leq x \lt z^2 +1 \space \space\text {for some } \space z\in \mathbb Z\}$$

My questions are :

(1) is the expression of the solution set correct?

(2) what is this solution set equal to? Is it possible to find a more explicit definition of $S$?

I can see that $[0,1)$ is a subset of $S$, because every $x$ in this interval satisfies $ 0^2 \leq x \lt 0^2 +1$.

Same thing for $[1, 2)$ since every $x$ in this interval satisfies $ 1^2 \leq x \lt 1^2 +1$.

Same thing for $[4, 5)$, with here $z=2$.

But what if $x\in [2, 3)$?

Certainly if $x\in [2,3)$ there is a $z\in \mathbb Z$ such that $z^2 \leq x$, and the closest one is $1$ . But, no $x$ in this interval satisfies $1^2\leq x \lt 1^2 +1$.

However, Wolfram Alpha suggests that the solution set of the equation $ \lfloor \sqrt{x}\rfloor -\sqrt{\lfloor x\rfloor} = 0 $ is the positive part of the $X-$ axis, that is $\mathbb R^{+}$.

Answer 1

Edit: at first, I did not notice that your concern was How to express the solution set.

It simply is the union of intervals $S =\displaystyle \bigcup_{k = 0,1,2,\ldots} \left[ k^2, k^2+1 \right)$.

The solution I originally wrote is very similar to yours.

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A first remark is that any solution must be nonnegative, since otherwise, $\sqrt{x}$ has no sense.

Let us show that the set of solutions is $\bigcup_{k\in \{0,1,\ldots,\}} [k^2,k^2+1)$.

First, assume that $x\geqslant 0$ satisfies $\lfloor \sqrt{x} \rfloor = \sqrt{\lfloor x \rfloor}$. Let $k = \lfloor \sqrt{x} \rfloor$. Then $\lfloor x \rfloor = \sqrt{\lfloor x \rfloor}^2 = k^2$. Hence, by definition $$ k^2 \leqslant x < k^2+1. $$

Conversly, let $x \geqslant 0$ be such that its integer part is a square, that is there exists $k\in \{0,1\ldots,\}$ such that $k^2 \leqslant x < k^2+1$. Since the square-root function is strictly increasing on $\Bbb R_+$, we have $k \leqslant \sqrt{x} < \sqrt{k^2+1}$.

1. If $k=0$, there is nothing else to prove.
2. Assume that $k>0$. Recall that for all $h\geqslant 0$, we have $\sqrt{1+h} \leqslant 1+ \frac{1}{2}h$. It follows that $\sqrt{k^2+1} = k\sqrt{1+\frac{1}{k^2}} \leqslant k + \frac{1}{2k} \leqslant k + 1$. Hence, $$ k \leqslant \sqrt{x} < k+1, $$ and therefore, $ \sqrt{\lfloor x \rfloor} = k = \lfloor\sqrt{x}\rfloor$.