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I am trying to prove the following theorem:

Theorem. A number is perfect iff the sum of the reciprocal of its divisors, excluding $1$, is $1$.

Thus far, this is the proof that I have managed to sew:

Proof. Let $n$ be perfect. Then $2n=1+a_1+a_2+\cdots+a_m+n$, where each $a_j$ is a divisor of $n$. Moreover, let $1<a_j<a_{j+1}<n$. It follows that $2=\frac{1}{n}+\frac{a_1}{n}+\frac{a_2}{n}+\cdots+\frac{a_m}{n}+1\Longrightarrow$ $1=\frac{1}{n}+\frac{1}{a_m}+\frac{1}{a_{m-1}}+\cdots+\frac{1}{a_1}$, as required. Conversely, let $1=\frac{1}{n}+\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_m}$. It follows that $1=\frac{1}{n}+\frac{a_m}{n}+\frac{a_{m-1}}{n}+\cdots+\frac{a_1}{n}\Longrightarrow$ $n=1+a_1+a_2+\cdots+a_m$. Therefore, $n$ is perfect. $\square$

Nevertheless, I do not feel very confident about it. What do you guys think?

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2  
A "Corollary" means a consequence of a more general, or a more sophisticated, or a more substantial result (often these consequences are straightforward, but sometimes are rather technical). So the question is: this is a corollary to what? –  William Feb 18 '12 at 1:30
    
I believe that I may have messed up the terminology. Thank you for pointing that out. What I meant to express was that I am trying to give (and prove) an alternative 'definition' to perfect numbers. –  Josué Molina Feb 18 '12 at 1:39
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As pointed out by WNY, the term "Corollary" is not appropriate here, the proof uses only the definition of perfect number. Your solution is almost fine, just small wording tweaks needed. Start with "Let $1, a_1, a_2, \dots, a_m, n$ be the divisors of $n$, listed in increasing order." The second half is correct, but kind of confusing. Why not after the "Conversely, let $\dots$" sentence, write a sentence that adds $1$ to each side, then a sentence that multiplies both side by $n$. –  André Nicolas Feb 18 '12 at 1:42
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Apart from slight wording changes, it is fine. One could also write: As $d$ ranges over the positive divisors of $n$, $\frac{n}{d}$ also ranges over the positive divisors of $n$. Thus $\sigma(n)=\sum_{d|n} \frac{n}{d}$. It follows that $\sigma(n)=2n$ iff $2n=\sum_{d|n} \frac{n}{d}$. Now divide both sides by $n$. (Your proof does exactly the same thing. For concreteness, it is better to use your notation than the condensed notation of this comment.) –  André Nicolas Feb 18 '12 at 1:53
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1 Answer

up vote 1 down vote accepted

Let $a$ be a perfect number, My '$a$' has $n$ factors.

According to perfect number,

$a=a_1+a_2+a_3+...a_n$

$a_1=1$

$\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}+...\frac{1}{a_n}$

Notice that

$a_1.a=a$

$a_2.a_n=a$

.

.

.

$a_n.a_2=a$

You can surely take on from here.

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I should point out that if $a = a_1 + \dots + a_n$, then $a$ has $n+1$ positive divisors, the largest of which is $a$ itself. –  JavaMan Mar 9 '13 at 8:07
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