Prime Number is Deficient
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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