**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.