Show that $\left\lfloor \sqrt{x} \right\rfloor$ is $n$

Question

Let $n^2$ be the largest perfect square that is less than or equal to $x$, then $x$ can be defined as follows,

$$ x = n^2 + m + \epsilon, 0\lt\epsilon\lt 1, m\in\Bbb Z$$

Show that $\left\lfloor \sqrt{x} \right\rfloor$ is $n$ and show that $\left\lfloor \sqrt{\left\lfloor x \right\rfloor} \right\rfloor$ is also $n$.

Proof: let us just consider proofing why $\left\lfloor \sqrt{x} \right\rfloor$ is $n$.

Start with $\left\lfloor \sqrt{x} \right\rfloor$,

$$\sqrt{x} = \sqrt{n^2 + m + \epsilon}$$

The proof claims that since $n^2$ is the perfect largest square that is less than or equal to $x$, $\sqrt{x}$ has to be between $n$ and $n+1$,

$$\left\lfloor \sqrt{x} \right\rfloor = n$$

Could you please explain as why $\left\lfloor \sqrt{x} \right\rfloor = n$?

Answer 1

In your specification, you overlooked something:

Since $n^2 \leq x < (n+1)^2$,
$m \leq (2n)$.

This follows, since if $m \geq (2n+1)$, then $n^2 + m + \epsilon$ would be $\geq (n+1)^2$, which is a contradiction.

Then, you have that $\sqrt{x} = \sqrt{n^2 + m + \epsilon} < \sqrt{n^2 + (2n+1)} = \sqrt{(n+1)^2} = (n+1)\implies$
$n = \sqrt{n^2} \leq \sqrt{x} < (n+1) \implies$
$\left\lfloor \sqrt{x}\right\rfloor = n.$