I want to know how to prove that the divisors of any number are always less than that number (divided by two).
Are the Factors of N more than zero but less than (N/2) ?
Excuses me if this question is Wrong.
Edit: I want the Question to be more clear so that when someone else reads it they will understand what i asked so i am restating the question.
Question: Given A Positive Integer N are its factors More than or equal 2 but less than or equal n/2 Ignoring 1 And N itself
$N$ is clearly a factor of $N$ and is also bigger than $\frac{N}{2}$.
Otherwise, if there is a factor $d$ of $N$ greater than $\frac{N}{2}$, $\frac{N}{d}$ is less than 2.
$0*d=N $ is impossible so $0$ isn't a factor of any number but $0$.
By definition factors are integers. By convention they are positive. (Otherwise we'd go nuts making exceptions and always pointing out if $a|b$ then negative $a $ also divides $b $.
$1$ and $N $ are always factors and $N>N/2$ so you need to restate your proposition.
Proposition: if $d$ is a non-trivial factor of $N$ (non-trivial means $d\ne 1$ and $d\ne N $) then $d \le N/2$.
Proof: if $d|N$ then there exists positive integer $m $ so that $dm=N$. That means $d=N/m $.
$m\ne 1$ because $d*1\ne N$. So $m\ge 2$. So $1/m \le 1/2$ and $d=N/m \le N/2$.
I read the question as "are the divisors of N fewer than N/2"? This is obviously not true for several small numbers: e.g. 2 and 3 have 2 divisors, 4 has 3 divisors, 6 and 8 have 4 divisors, 12 has 6 divisors.
