0

Let $p^k m^2$ be an odd perfect number (OPN) with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

My question is as is in the title:

Why does an OPN seemingly "violate" basic inequality rules?

Denote the sum of divisors of the positive integer $x$ by $\sigma(x)$, the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the abundancy index of $x$ by $I(x)=\sigma(x)/x$.

Note that it is trivial to prove that $$\frac{p+1}{p} \leq I(p^k) < \frac{p}{p-1}$$ from which we obtain $$\frac{2(p-1)}{p} < I(m^2) = \frac{2}{I(p^k)} \leq \frac{2p}{p+1}.$$ This implies that $$\frac{2}{p+1} \leq \frac{D(m^2)}{m^2} < \frac{2}{p}.$$ Taking reciprocals, multiplying by $2$, and subtracting $1$, we get $$p-1 < \frac{\sigma(m^2)}{D(m^2)} \leq p,$$ where we note that $(2/p) < p-1$.

Notice then how the OPN seems to "violate" the basic inequality rule

$$(a \leq b) \land (c \leq d) \implies (ac \leq bd).$$

Note that $$I(m^2) = \frac{D(m^2)}{m^2}\cdot\frac{\sigma(m^2)}{D(m^2)},$$ and that the premise is false while the conclusion is true in the above implication.

Jose Arnaldo Bebita Dris
  • 8,471
  • To be clear, are you saying that it is strange that $\dfrac{2(p-1)}{p} < I(m^2) \leq \dfrac{2p}{p+1}\implies \dfrac{2(p-1)}{p+1}\lt I(m^2)\lt 2$ ? – mathlove Dec 03 '20 at 11:55
  • No, that was not what I meant, @mathlove. Take $$a = \frac{2}{p}, b = \frac{D(m^2)}{m^2}, c = p - 1, d = \frac{\sigma(m^2)}{D(m^2)}$$ for example. In this case, $a \leq b$ is false (so that the premise is false), while the conclusion holds. A similar phenomenon holds with $$a = \frac{D(m^2)}{m^2}, b = \frac{2}{p+1}, c = \frac{\sigma(m^2)}{D(m^2)}, d = p.$$ – Jose Arnaldo Bebita Dris Dec 03 '20 at 12:05
  • 1
    OK. I'm not sure if this helps, but since the implication says only that if $a\le b$ and $c\le d$, then $ac\le bd$, it does not say that $ac\le bd$ only if $a\le b$ and $c\le d$. – mathlove Dec 03 '20 at 12:58
  • Thank you for your last comment, @mathlove! That certainly clears things up for me. If you could just please post that as an actual answer and include additional details if needs be, then I would upvote and accept your answer. – Jose Arnaldo Bebita Dris Dec 03 '20 at 13:03

1 Answers1

1

On OP's request, I am converting my comment into an answer.

The implication says only that if $a\le b$ and $c\le d$, then $ac\le bd$ (assuming that $a,c$ are non-negative).

It does not say that $ac\le bd$ only if $a\le b$ and $c\le d$.

(It is false that if $ac\le bd$, then $a\le b$ and $c\le d$.

For $(a,b,c,d)=(3,2,2,4)$, for example, we have $ac\le bd$ and $a\gt b$.)

mathlove
  • 151,597