Question

If [x] denotes the greatest integer of function of x,

Then estimate:

Answer 1

[\(\sqrt{2011^2+2012}\)]

Answer 2

$$2011^2<2011^2+2012=2011(2011+1+1/2011)<2012^2$$

Answer 3

so we know \(\lfloor\sqrt{2011^2+2012}\rfloor=\sqrt{2011^2}=2011\)

Answer 4

You can always start with \(n=[x]\equiv n\leq x <n+1\) if you want to be formal about it.

Answer 5

\(\sqrt\cdot\) is monotonic so \(a<b<c\) means \(\sqrt{a}<\sqrt{b}<\sqrt{c}\) hence my work above

Answer 6

Hmm...So should I always use this methodology for such problems? It isnt striking me when I see the qn.

Answer 7

Well, inequalities are tricky, but they are certainly easier to deal with than \([\dots]\) unless you have a lot of properties you can work with.

Answer 8

well we know immediately that it's bigger than \(2011^2\)... the question is whether it's bigger than \(2012^2\), etc; we want to bound it between two squares so we can take square roots of both sides

Answer 9

here we note \(2012^2-(2011^2+2012)=2012^2-2011^2-2012=(2012+2011)-2012=2011\) so we know for a fact it's snug between \(2011^2,2012^2\)

Answer 10

note \(2012^2-2011^2=(2012+2011)(2012-2011)=(2012+2011)\cdot1=2012+2011\)

Answer 11

There is also:\[
2012^2=(2011+1)^2=2011^2+2011+2011+1 = 2011^2+2012+2011
\]

Answer 12

Yeah looks rational now. Thanks for the help everyone!

Answer 13

\[\lfloor \sqrt{x^2+x+1} \rfloor =\lfloor \sqrt{2011^2+2011+1} \rfloor\]
\[x^2 \le x^2+2x+1<x^2+2x+1=(x+1)^2 \rightarrow \sqrt{x^2} \le \sqrt{x^2+x+1} <\sqrt{x^2+2x+1}\]
\[so x \le \sqrt{x^2+x+1} <x+1\rightarrow \lfloor \sqrt{x^2+x+1} \rfloor =x =2011\]

Answer 14

indeed, it's obvious from the fact that \(n^2+n=n(n+1)\) so \(n^2+n+(n+1)=n(n+1)=(n+1)(n+1)=(n+1)^2\)

Answer 15

\(n^2+n+(n+1)=n(n+1)+(n+1)=\dots\) rather

Answer 16

Thanks @oldrin.bataku for your clear explanation. I understand it very well now.