# A problem on Floor and Ceiling Functions!

Is it true or false?

$\large{\left \lceil \dfrac{1 + \sqrt{1+8n}}{2} \right \rceil - 1 = \left \lfloor \sqrt{2n} \right \rfloor}$

Here $n$ is a positive integer.

If it is false, provide me a counter-example. If it's true, please provide me a proof.

Note by Satyajit Mohanty
4 years, 10 months ago

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False. Often, they don't agree.

- 4 years, 10 months ago

I'm sorry. I forgot to mention that $n$ is a positive integer! Can you find some counter-example now?

- 4 years, 10 months ago

..

$\dfrac { 1 }{ 2 } (1+\sqrt { 1+8\cdot 4 } )-1=2.37228...$
$\sqrt { 2\cdot 4 } =2.828427...$

But because of the way ceiling and floor functions work, these two go off in opposite directions. This is just one example.

- 4 years, 10 months ago

if n = 2 it is not true

the only solutions are: 0,1,3,4,6

- 4 years, 10 months ago

It is true for $n=2$. I don't agree!

- 4 years, 10 months ago

[] means floor?

- 4 years, 10 months ago

Look. $\lceil . \rceil$ means ceiling and $\lfloor . \rfloor$ means floor.

- 4 years, 10 months ago

Ah- i see =)

- 4 years, 10 months ago

It's not true when n = 12 (calculated with Wolfram Alpha)

- 4 years, 10 months ago

It's not true when n = 12 (calculated with Wolfram Alpha)

- 4 years, 10 months ago

Hint: Consider what happens when $k^2 \leq 2n < (k+1)^2$.

Staff - 4 years, 10 months ago

Thanks :) I got it. I actually had a doubt on a solution to this problem: Peculiar Sequence of Positive Integers! as the generalized version of my solution did not match the generalized version of the other solution.

- 4 years, 10 months ago

If you see my note, I was pointing out the mistake that @Chew-Seong Cheong made in the generalization.

The true version of the statement that you are looking for, is

${\left \lceil \dfrac{1 + \sqrt{1+8n}}{2} \right \rceil = \left \lfloor \sqrt{2n} + \frac{3}{2} \right \rfloor}$

Staff - 4 years, 10 months ago