Let $N$ be a perfect number. Prove the following:
(a) $m \times N$ where $m > 1$ is an abundant number.
(b) $\frac{N}{d}$ where $d$ is a non-trivial divisor of $N$ is a deficient number.
We remember that $\sigma(N)=2N$ for perfect $N$.
(a) Note that we can't use the multiplicity of $\sigma$ because $N$ and $m$ are not necessarily coprime.
We start by comparing the divisors of $mN$ to those of $N$
$$ \begin{align} \sigma(m N) & = \sum_{d \mid mN} d \tag{i} \\ \\ & > m \sum_{e \mid N} e \tag{ii} \\ \\ & = m(2N) \tag{iii} \\ \\ & = 2(mN) \end{align} $$
Let's explain each step:
(i) $\sigma(nM)$ is the sum of all the divisors of $nM$.
(ii) The divisors of $N$, each multiplied by $m$, are divisors of $mN$. However this list is a proper subset of the divisors of $mN$. For example, this list does not contain the divisors of $N$ not multiplied by $m$. And so the sum of the divisors of $mN$ is strictly greater than the sum of the divisors of $N$, each multiplied by $m$.
(iii) Since $N$ is perfect, the sum of its divisors is $2N$.
Since $\sigma(mN) > 2(mN)$, then $mN$ is abundant.
(b) We start by comparing the divisors of $\frac{N}{d}$ and $N$
$$ \begin{align} d \times \sigma \bigl ( \frac{N}{d} \bigr ) & < \sigma(N) \tag{iv} \\ \\ & = 2N \tag{v} \\ \\ \sigma \bigl ( \frac{N}{d} \bigr ) & < 2 \bigl ( \frac{N}{d} \bigr ) \end{align} $$
Let's explain each step:
(iv) The divisors of $\frac{N}{d}$ are all divisors of $N$. But not all divisors of $N$ are divisors of $\frac{N}{d}$, for example $N$ itself. This means the divisors of $\frac{N}{d}$ are a proper subset of the divisors of $N$. If we multiply the divisors of $\frac{N}{d}$ by $d$, the resulting set of numbers are all divisors of $N$. However, again, not all divisors of $N$ are in that set, for example 1 is not in that set, because it was multiplied by $d>1$. So the sum of divisors of $\frac{N}{d}$ multiplied by $d$ is strictly less than the sum of divisors of $N$.
(v) Since $N$ is perfect, the sum of its divisors is $2N$.
Since $\sigma \bigl ( \frac{N}{d} \bigr ) < 2 \bigl ( \frac{N}{d} \bigr )$, then $\frac{N}{d}$ is deficient.
Note the author's solutions appear to have typos for both parts (a) and (b).