Prove that if $x \in \mathbb{R_{\ge 0}}$ Then $\left[\sqrt{x}\right]=\left[\sqrt{\left[x\right]}\right]$

Prove that if $x \in \mathbb{R_{\ge 0}}$ Then $$\left[\sqrt{x}\right]=\left[\sqrt{\left[x\right]}\right]$$

My Try:

Case $1.$

Let $\sqrt{x}=p$, where $p \in \mathbb{Z_{\ge 0}}$

Then $$LHS=p$$ and we have $x=p^2$ So

$$RHS=\left[\sqrt{\left[x\right]}\right]=\left[\sqrt{p^2}\right]=p=LHS$$

Case $2.$

Let $\sqrt{x}=p+f$ where $p \in \mathbb{Z_{\ge 0}}$ and $0 \lt f \lt 1$

We have $$x=p^2+2pf+f^2$$

$$LHS=\left[p+f\right]=p$$

We have $$RHS=\left[\sqrt{\left[p^2+f^2+2pf\right]}\right]=\left[\sqrt{p^2+\left[f^2+2pf\right]}\right]$$

Now we have $$f^2+2pf \gt 0$$

But how to prove $$f^2+2pf \lt 1$$

Note: I am not looking for a different solution. I need help to continue with my solution

• I have not asked for new solution, i asked how to continue with my solution? – Ekaveera Kumar Sharma Jan 5 '18 at 19:05
• – Martin R Jan 5 '18 at 19:34
• You cannot prove $f^2+2pf \lt 1$ because that does not necessarily hold. Simple example: $p = 1, f=0.5$. – So your solution does not work. This answer math.stackexchange.com/a/1649208/42969 shows how you can proceed ($n$ in that answer is your $p$). – Martin R Jan 5 '18 at 20:58

The following specifically answers this part of OP's question: "I am not looking for a different solution. I need help to continue with my solution".

We have $$RHS=\left[\sqrt{\left[p^2+f^2+2pf\right]}\right]=\left[\sqrt{p^2+\left[f^2+2pf\right]}\right]$$

Now we have $$f^2+2pf \gt 0$$

Correct thus far.

But how to prove $$f^2+2pf \lt 1$$

That's not what you need to prove (and, in fact, doesn't hold true in general).

What has to be proved at this point is that:

\begin{align} \left\lfloor \sqrt{\lfloor x \rfloor} \right\rfloor = p \;\;&\iff\;\; \left\lfloor \sqrt{\left\lfloor (p+f)^2 \right\rfloor} \right\rfloor = p \\[5px] &\iff\;\; p \le \sqrt{\left\lfloor (p+f)^2 \right\rfloor} \lt p+1 \\[5px] &\iff\;\; \color{blue}{p^2 \le \left\lfloor (p+f)^2 \right\rfloor \lt (p+1)^2} \tag{1} \end{align}

But $0 \lt f \lt 1\,$, and therefore $p \lt p+f \lt p+1\,$, so:

\begin{align} p^2 \lt (p+f)^2 \lt (p+1)^2 \;\;&\implies\;\; \color{blue}{p^2} = \left\lfloor p^2 \right\rfloor \color{blue}{\le \left\lfloor (p+f)^2 \right\rfloor \lt} \left\lfloor (p+1)^2 \right\rfloor = \color{blue}{(p+1)^2} \tag{2} \end{align}

(The implication follows from the property of the greatest integer function that $\,a \lt b \implies \lfloor a \rfloor \le \lfloor b \rfloor\,$, with strict inequality $\,\lfloor a \rfloor \lt \lfloor b \rfloor\,$ if $\,b\,$ is an integer.)

The above proves $(2)\,$, which is identical with $(1)\,$, and therefore concludes the proof.

$[\sqrt x]=m$ is equivalent $m\le \sqrt x<m+1$, i.e., to $m^2\le x<(m+1)^2$. As $m^2$ is an integer, this implies $m^2\le [x]<(m+1)^2$, and hence $m\le \sqrt{[x]}<m+1$, i.e., $[\sqrt{[x]}]=m$.