OPN Research - Feb 06, 2012 - Part 3

Let N = {q^k}{n^2} be an odd perfect number with Euler 
factor q^k.


The contrapositive to the result:

If sigma(q)/n < sigma(n)/q,  then sigma(q) < sigma(n).

is:

If sigma(n) < sigma(q),  then sigma(n)/q < sigma(q)/n.

But:

If sigma(n)/q < sigma(q)/n, then n < q.

Therefore:

If sigma(n) < sigma(q), then n < q.

In this case, the resulting inequality for the quantities 
{q, sigma(q), n, sigma(n)} is:

Case #1:  n < sigma(n) <= q < sigma(q).

The contrapositive of:

If sigma(n) < sigma(q), then n < q.

is:

If q < n, then sigma(q) < sigma(n).

In this case, the resulting inequality for the quantities 
{q, sigma(q), n, sigma(n)} is:


Case #2:  q < sigma(q) < n < sigma(n).


The resulting inequality in a third case that comes to mind 
for the quantities {q, sigma(q), n, sigma(n)} is:


Case #3:  n < q < sigma(q) < sigma(n).