Question

@ganeshie8 , a number theory question...

Answer 1

Do you need help with anything?

Answer 2

I solved a problem but I'm not sure whether I have solved it correctly or not,please check my work ;)

The question:
Suppose "n" is a natural number.prove \(n + \left[ \sqrt{n}+ \frac{ 1 }{ 2 } \right] \) is not a Square number.

Answer 3

Here's what i did (please wait it might take me about 5 min to write)

Answer 4

0

Answer 5

Suppose there is a natural number \( m \) in which \(n + \left[ \sqrt{n}+ \frac{ 1 }{ 2 } \right] = m^2 \) \( \large \color{red}{(1)} \)
as \(\left[ x \right] + \left[ y \right] \le \left[ x+y \right] \le \left[ x \right] + \left[ y \right] + 1\) so we can say:

\[\left[ \sqrt{n} \right] + \left[ \frac{ 1 }{ 2 } \right] \le \left[ \sqrt{n} + \frac{ 1 }{ 2 }\right] \le \left[ \sqrt{n} \right] + \left[ \frac{ 1 }{ 2 } \right] + 1\]

and we know that \(\left[ \frac{ 1 }{ 2 } \right] = 0\),therefore,

\[\left[ \sqrt{n} \right] \le m^2 -n\le \left[ \sqrt{n} \right] + 1\]

There is a rule that \(x < \left[ x+1 \right] \le x+1\) so,

\(\left[ \left[ \sqrt{n} \right] +1 \right] = m^2 - n\)
so,
\[\left[ \sqrt{n} \right] + n + 1 = m^2 \]

because of the equation \( \large \color{red}{(1)} \) we can write:

\[\left[ \sqrt{n} + 1 \right] = \left[ \sqrt{n} + \frac{ 1 }{ 2 } \right]\]

but there is not any natural number in which this equation exist,in this case there is not any natural number such as \(m\).

Answer 6

it became a little more than 5 min, ;)

Answer 7

@ganeshie8 , check my work,i'll be back ;)

Answer 8

Looks neat ! you should end up with
\[\left[ [\sqrt{n}] + 1 \right] = \left[ \sqrt{n} + \frac{ 1 }{ 2 } \right]\]

right ?

Answer 9

@PFEH.1999

Answer 10

yes

Answer 11

how do you know that equation is always false ?

Answer 12

i'm not sure about it,do u have any idea?

Answer 13

\[\left[ \sqrt{n} +1\right] = \left[ \sqrt{n}+ \frac{ 1 }{ 2 } \right] = \left[ \sqrt{n}+ 1 - \frac{ 1 }{ 2 } \right] = \left[ \sqrt{n}- \frac{ 1 }{ 2 } + 1 \right]\]
so,
\[\left[ \sqrt{n} \right] = \left[ \sqrt{n} - \frac{ 1 }{ 2 } \right]\]

Answer 14

@ganeshie8 ,i checked the proof you gave me! i don't know why i always try difficult ideas,it's my problem when i try to solve some problem,the idea in that proof was awesome...but you know i'm insisted,i want to complete my own proof.
i'll work on the last equation and will keep you posted ;)

Answer 15

i liked ur method more because you're using the greatest integer function properties cleverly xD i tried to finish it but i got stuck because greatest integer function requires strict inequality for lower bound : x-1 < [x] <= x im bit eager to see the finished proof xD