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I know this is a very crude question and will receive down votes but my question still remains:

Is it true that $$\lfloor{x}\rfloor^2 = \lfloor{x^2}\rfloor$$

Or generalising it to any arbitrary power $k$: $$\lfloor{x}\rfloor^k = \lfloor{x^k}\rfloor$$

Thanks a lot!

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    Hint: try some examples. What about $x=1.5$ say? – 2017-02-11
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    @lulu wow. That gave me some insight.. Should've thought about this :/ Thanks! – 2017-02-11

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Hint : take $$x=\sqrt2+1$$ now see $$\lfloor x^2 \rfloor=\lfloor (\sqrt2+1)^2\rfloor =\\ \lfloor (2+2\sqrt2+1)\rfloor =\\3+\lfloor (2\sqrt2)\rfloor =5\\$$ vs $$(\lfloor x \rfloor)^2=\\ (\lfloor \sqrt2+1 \rfloor)^2=\\2^2=4$$ so $$\lfloor x^2 \rfloor \neq (\lfloor x \rfloor)^2$$it is better to say $$\lfloor x^2 \rfloor \geq (\lfloor x \rfloor)^2$$

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    Brilliant! Thanks for answer... I should have realised on my own. I'm tempted to say that the end result can be generalised to any power, ie, $\lfloor x^k \rfloor \geq (\lfloor x \rfloor)^k$, for any real $k$. Is it true? – 2017-02-11
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    @MihirChaturvedi :for $k=2n $ it is true ,but not always true for odd $k $ – 2017-02-11
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    Technically, they will be equal whenever x is an integer. I think the only difference is that one jumps at the integers whereas one has denser jumps at the square roots of integers. – 2017-02-11
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Well, you might want to try this by doing the following:

Let $x=n+r$ where $[x]=n$ and $0\leq\ r <1$. Then

$[x^2]=[(n+r)^2]=[n^2+2nr+r^2]=n^2+[2nr+r^2]$.

And $[x]^2=n^2$.

Equating both expressions we obtain $n^2=n^2+[2nr+r^2]$ which implies that $[2nr+r^2]=0$. If $[2nr+r^2]=0$ then $0\leq\ 2nr+r^2<1$ so $0\leq\ 2[x]r+r^2<1$.

I really hope that this helps you! Can you work it from here? :-)