# A Simpler Dense Proof Regarding the Abundancy Index

@article{Ryan2003ASD, title={A Simpler Dense Proof Regarding the Abundancy Index}, author={Richard F. Ryan}, journal={Mathematics Magazine}, year={2003}, volume={76}, pages={299 - 301} }

(B) If I (a) = r/s is in lowest terms, then s divides a. This follows since sa (a) = ra and gcd(r, s) = 1. (C) If I (a) = r/s is in lowest terms then r > ar(s). This follows from properties (B) and (A) since r/s = I (a) > I (s) = a(s)/s. (The condition that r and s be relatively prime is an important one! Note that 1(2) = 6/4 even though 6 r then r/s is not an abundancy index.

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