**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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestHint:Consider what happens when \( k^2 \leq 2n < (k+1)^2 \).Log in to reply

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.

Log in to reply

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

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

Log in to reply

if n = 2 it is not true

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

Log in to reply

It is true for \(n=2\). I don't agree!

Log in to reply

[] means floor?

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Comment deleted Jul 29, 2015

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

problem, please ?

Can you look at myLog in to reply

False. Often, they don't agree.

Log in to reply

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

Log in to reply

..

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

\(\sqrt { 2\cdot 4 } =2.828427...\)

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

Log in to reply