Prime Number is Deficient/Proof 2
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Theorem
Let $p$ be a prime number.
Then $p$ is deficient.
Proof
Let $p$ be a prime number.
From Divisor Sum of Prime Number:
- $\map {\sigma_1} p = p + 1$
and so:
- $\dfrac {\map {\sigma_1} p} p = \dfrac {p + 1} p = 1 + \dfrac 1 p$
As $p > 1$ it follows that $\dfrac 1 p < 1$.
Hence:
- $\dfrac {\map {\sigma_1} p} p < 2$
The result follows by definition of deficient.
$\blacksquare$