Let N = {q^k}{n^2} be an odd perfect number with Euler
factor q^k.
Suppose that sigma(q) < n.
Then we have q + 1 < n, which implies that:
q + 2 <= n
Therefore:
q <= n - 2
Consequently:
q < n
Result #1: If sigma(q) < n, then q <= n - 2.
Now suppose that n < sigma(q).
This implies that:
n < q + 1
n - 1 < q
Therefore:
n <= q
Consequently:
n < q since gcd(q,n) = 1.
Result #2: If n < sigma(q), then n < q.
Hence, we have the following equivalence result:
Lemma: sigma(q)/n > 1 if and only if n < q.
Note that, in case sigma(q)/n > 1, then
sigma(q^k)/n >= sigma(q)/n > 1.