**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.

## Comments

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TopNewestHint:Consider what happens when \( k^2 \leq 2n < (k+1)^2 \). – Calvin Lin Staff · 1 year, 12 months agoLog in to reply

Peculiar Sequence of Positive Integers! as the generalized version of my solution did not match the generalized version of the other solution. – Satyajit Mohanty · 1 year, 12 months ago

Thanks :) I got it. I actually had a doubt on a solution to this problem:Log in to reply

@Chew-Seong Cheong made in the generalization.

If you see my note, I was pointing out the mistake thatThe true version of the statement that you are looking for, is

\[{\left \lceil \dfrac{1 + \sqrt{1+8n}}{2} \right \rceil = \left \lfloor \sqrt{2n} + \frac{3}{2} \right \rfloor}\] – Calvin Lin Staff · 1 year, 12 months ago

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if n = 2 it is not true

the only solutions are: 0,1,3,4,6 – Fatum Altum · 1 year, 12 months ago

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– Satyajit Mohanty · 1 year, 12 months ago

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

– Fatum Altum · 1 year, 12 months ago

[] means floor?Log in to reply

– Satyajit Mohanty · 1 year, 12 months ago

Look. \(\lceil . \rceil\) means ceiling and \(\lfloor . \rfloor\) means floor.Log in to reply

– Fatum Altum · 1 year, 12 months ago

It's not true when n = 12 (calculated with Wolfram Alpha)Log in to reply

– Fatum Altum · 1 year, 12 months ago

Ah- i see =)Log in to reply

– Fatum Altum · 1 year, 12 months ago

It's not true when n = 12 (calculated with Wolfram Alpha)Log in to reply

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– Fatum Altum · 1 year, 12 months ago

5 != 4Log in to reply

– Satyajit Mohanty · 1 year, 12 months ago

Can you give me the set of values of \(n\) for which the equation doesn't satisfy?Log in to reply

– Fatum Altum · 1 year, 12 months ago

So, the first n is 4; then I tried to see the connection, here it is : (11 , 12) then we adding 4 and get the next pair(16,17) - two pairs of n that don't satisfy; then we adding 5 and get triplet (22 , 23 , 24) again 5 and (29,30,31) - two triplets for n; adding 6 and get quartet (37.38,39,40) again 6 and (46,47,48,49) - two quartets; and so on (next 7 - 2 pairs of 5 pieces of n ) to infinityLog in to reply

– Fatum Altum · 1 year, 12 months ago

Sure, wait a minuteLog in to reply

– Satyajit Mohanty · 1 year, 12 months ago

I'm sorry. I agree.Log in to reply

problem, please ? – Fatum Altum · 1 year, 12 months ago

Can you look at myLog in to reply

False. Often, they don't agree. – Michael Mendrin · 1 year, 12 months ago

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– Satyajit Mohanty · 1 year, 12 months ago

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. – Michael Mendrin · 1 year, 12 months ago

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