g.i. func

[afif mansib ch]

if N is a pos int proof the following:
\[\pi (N)=\]\[ \sum_{n=1}^{N}\left [ N/n \right ]-\left [ (N-1)/n \right ]\]

[Labib]

Could you please define $\pi(N)$ and $n$...
I really don't understand the problem...

And please write the full statement. This is an International level forum you know...

[afif mansib ch]

ok ,sorry for writing in that way. you see it has become a habit now for using facebook.
by\[\pi (N)\]
i meant the total number of divisors of N,because i couldn't find the real sign here.n is only an integer here.

[*Mahi*]

if N is a pos int proof the following:
\[\pi (N)=\]\[ \sum_{n=1}^{N}\left [ N/n \right ]-\left [ (N-1)/n \right ]\]

Let's see the two cases:
1.If $n|N$,
\[ \left \lfloor \frac N n \right \rfloor - \left \lfloor \frac N {n-1} \right \rfloor =1\]
2. Else
\[ \left \lfloor \frac N n \right \rfloor- \left \lfloor \frac N {n-1} \right \rfloor =0 \]
So the sum increases $1$ whenever a divisor of $N$ appears and thus the sum is equal to $\pi (N)$

[afif mansib ch]

i actually mentioned \[\left [ N-1/n \right ]\]

[Labib]

Are you using $[]$ as floor function??
I am still confused.
If you are, then Mahi almost solved it. He meant to write $\left \lfloor \frac{N-1}n \right \rfloor$ instead of $\left \lfloor \frac N {n-1} \right \rfloor$.
That solves the problem.
And try using the equation editor next time please. It's on the top right corner of the formatting panel.(Right above the smileys.)

[afif mansib ch]

actually i did it like mahi 1st time for both the funcs floor and ceiling individually .bt i was confused and i posted it.and i gave the reply confusing also!!!i meant to write how can it be done for[].individually?

[Labib]

If you're using ceiling, the function will not hold.

[afif mansib ch]

\[\left \lceil N/n \right \rceil-\left \lceil N-1/n \right \rceil=0\]
for two cases(not when N-1 is divisable by n).so the result won't increase to \[\pi (n)\]
i want to know wt does the func mean by [].

[Labib]

$\left [a,b\right ]=LCM \left [a,b\right ]$