Lemma $5.$ If $|N||,$ then

$\frac{()}{}\frac{(N)}{N}$

where equality holds only if $=N.$

Proof. Note that if $d|N||$ then $kd=N$ for some $k,$ so $k=(\frac{N}{d})|N||.$ This argument works just as well in reverse, so we have that $d|N||$ if and only if $(\frac{N}{d})|N||,$ which implies that

$(N)={\displaystyle \underset{?}{\overset{?}{}}}|)d={\displaystyle \underset{?}{\overset{?}{}}}|)\frac{N}{d}=N{\displaystyle \underset{?}{\overset{?}{}}}|)\frac{1}{d}.$

If $$ is a proper divisor of $N,$ we have

$\frac{(N)}{N}={\displaystyle \underset{?}{\overset{?}{}}}|)\frac{1}{d}>{\displaystyle \underset{?}{\overset{?}{}}}|)\frac{1}{d^{\text{'}}}=\frac{()}{}$

Otherwise, equality holds.

explain to me how to prove this lemma in detail