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In my lecture notes, we have been given the theorem

If $N \in \mathbb{Z}_{+} \setminus \{1\}$ is odd and perfect, and written $\prod_{i = 1}^k p_i^{n_i}$ as shown, then $k \geq 3$, that is $N$ has at least $3$ distinct prime divisors.

The proof for this has been given by using the formula

$$2 \prod_{i = 1}^{k} \left(1 - \frac{1}{p_i} \right)$$

and my lecturer put in $k = 1$ and $k = 2$ and showed they were both $> 1$ and somehow this proves it.

Firstly, how does this prove that there are atleast 3 dividers? I thought the divider had to be an integer and subbing in $k = 1,2$ gives you two fractions which are then not dividers?

Also, how can you can the particular theorem exist and be proven if there is a chance that no odd perfect numbers exist? I don't really understand that

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If they don't exist, then any universal statement about them is true. –  Git Gud Feb 5 '13 at 20:44
    
The at least three distinct prime factors thing has been asked and answered in a recent post. –  André Nicolas Feb 5 '13 at 20:58
    
@GitGud The powers that be have deleted my 'joke', alas. –  Jp McCarthy Feb 5 '13 at 23:01
    
@JpMcCarthy Unlucky :( I forgot what the joke was :( –  Git Gud Feb 5 '13 at 23:03
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If my aunt had... –  Jp McCarthy Feb 5 '13 at 23:05
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4 Answers

The theorems are of the form, "If any odd perfect numbers exist, then they must be of the form..."

If there are no odd perfect numbers, then these theorems are vacuously true (and indeed, one of the ways of eventually proving there are no odd perfect numbers would be to prove that if any existed, then some contradictory properties would have to be true about them.)

If there are any odd perfect numbers, then these theorems narrow down where we should look.

If $p_1^{a_1}, p_2^{a_2}, \ldots, p_m^{a_m}$ are the distinct prime divisors of an integer $n$, then $$2n = \prod_i (1 + p_i + \ldots + p_i^{a_i}),$$ which you can check by expanding the RHS. The terms in the product are partial geometric series, so $$2n = \prod_i \frac{p_i^{a_i+1}-1}{p_i-1},$$ and shuffling around the terms, $$2 \prod_i (1-p_i^{-1}) = \prod_i (1-p_i^{-a_i-1}).$$ Notice that the right-hand side is less than 1. Moreover, the LHS is larger the larger the $p_i$ are, so for $n$ odd and $m=2$, $$2\prod_{i=1}^2 (1-p_i^{-1}) \geq 2(1-3^{-1})(1-5^{-1})=\frac{16}{15} > 1,$$ so $m$ cannot be $2$.

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In mathematics, to claim

(A) All unicorns are mammals.

is exactly the same as claiming

(B) There is nothing that is a unicorn but not a mammal.

One way this can be true is that if there are no unicorns at all. A universal statement such as (A) is "true by default": the only way it can be false is if there is a counterexample to it. A counterexample would need to be a non-mammal unicorn, and if unicorns don't exist at all, it means there can be no counterexample to (A), and therefore (A) is true.

(A) and (B) can also be true if there do exist some unicorns but each of them happens to be a mammal. The claims don't distinguish between that case and the possibility that there are no unicorns, period.

This interpretation of statements with "all" is used everywhere in modern mathematics, but it is not the only one possible -- in classical Aristotelian logic it was assumed that the statement "all X are Y" implies "some X are Y", so there had to be least some X we can speak about somewhere.

This difference is not really a disagreement about what is true or not; it is just a matter of language, that is, which use we put the words to. In modern mathematics we choose to use the words "all such-and-such have this-or-that property" to express one meaning; the Aristotelian logicians of yore used the same words to express a different meaning.

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The theorem must be taken to mean that if any odd perfect numbers exist, then each such number has at least three distinct prime divisors.

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Mathematical statements are all of the form "If A than B" in propositional calculus, and that is assigned a truth value corresponding to that of $\neg A \lor B$ because we want to have the rule of inference modus ponens that $(A\land (A\implies B))\implies B$. In English, that says a false hypothesis proves everything is true. Technically a correct first order logic proof of a theorem T is a proof that "if ZF (or in almost all cases ZFC) set theory is consistent), then T logically follows and there is no way that T can also be proved false." Nobody says that because we all believe that the underlying logic of set theory is consistent or we could prove all statements or we would all be out of work. What your statement actually conveys is "if some day an odd perfect number is found, then the given formula must hold." As long as no one knows if an odd perfect number exists, the statement is what is often called vacuously true because if the set of odd perfect numbers is empty you have a false hypothesis, but if that set is nonempty you presumably have a proof of the formula.

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