Source : Lycée Berthelot's higher mathematics exercices ( https://molin-mathematiques.fr/sources/exercices106.php)
Note : definitely not HW
To be solved : $\space \lfloor {\sqrt x}\rfloor = \sqrt {\lfloor x \rfloor}$
My reasoning so far:
From $\lfloor {\sqrt x}\rfloor = \sqrt {\lfloor x \rfloor}$
we can deduce
(1) that $x\geq 0$
(2) $\sqrt {\lfloor x \rfloor} \in \mathbb Z$ and is a positive integer by definition ( since a square root is always positive).
(3) since only squares of integers have integral square roots, we have $\lfloor x \rfloor = p^2$ with $p\in \mathbb Z$
(4) $p^2 \leq x \lt p^2 +1$, for some $p \in \mathbb Z$.
The conclusion I seem to reach is that the solution set is the set of
all $x$ such that $x$ is greater or equal to some number that is the square of an integer , meaning
that the solution set is the set of all positive real numbers ( with $0$).
But in fact, according to Desmos, the graphs of the LHS and of the RHS function seem never to cross oneanother, suggesting that the solution set is empty.
What do I miss?
Desmos :
Best Answer
You haven't missed anything except that when Desmos plots the graph and the two functions agree the green is plotted on top of the red. You claimed, for example, that when $1 \le x \lt 2$ both functions evaluate to $1$, which is correct. You can see the green curve is constant at $1$ over that interval and you cannot see the red curve because it is under the green curve. Again for $4 \le x \lt 5$ both functions evaluate to $2$ and again they are plotted on top of each other.