Prime Number is Deficient

Theorem

Let $p$ be a prime number.

Then $p$ is deficient.


Proof 1

A specific instance of Power of Prime is Deficient.

$\blacksquare$


Proof 2

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.

$\blacksquare$


Sources

  • 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $12$
  • 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $12$
  • 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): deficient number
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