odds of a random prime being a number
 

I think the complement to [url=http://www.mersenneforum.org/showthread.php?t=8823]this thread[/url] deserves its own thread.

I have a marvelous proof of this, but am waiting for AMS to acknowledge receipt of manuscript before making it public.

Lemma: The probability that a random prime p is a number n is
equal to the probabilty that p-1 is a number n-1.

Proof: Obvious.

Having thus reduced the problem to a much simpler one,
and allowing for infinite regress, the original problem is solved.

:wink:

Your proof is sound, but form an aesthetic viewpoint, I've never really liked proofs by induction. There's just something too brutish-force about them for my taste. But ... we all have our quirks.

Also, even glossing over the ambiguous nature of proof-by-obviousness and assuming what you say is true, your lemma only shows that the probability is the *same*, not what the probability *is*. Perhaps a corollary or a separate claim/lemma/theorem is in order.

[QUOTE=davar55;111506]Having thus reduced the problem to a much simpler one, and allowing for infinite regress, the original problem is solved.[/QUOTE]

I don't think so. For an inductive proof, such as you suggest, you need two elements. You need the inductive step, such as you have indicated, but you also need a boundary (terminal) condition. You have failed to provide this portion of your "proof".

A random Prime might turn out to be the Prime of Miss Jean Brodie or a Prime Rib Steak. I'm pretty sure neither of these is a number, so the probability in question appears to be less than 1.

Googling "Prime" gets 225 million hits, "+Prime +Number" gets 142 million hits, so my guess is that the probability of a random prime being a number is 63%.

William

I would think that with the great computational skills evident
on this forum that the following derivation would be considered excessive:

Let P[sub]p[/sub] be the probability that a random prime p is a number n.

By the lemma, P[sub]p[/sub] = P[sub]p-1[/sub] = ... .

Hence multiplying the P[sub]p[/sub] gives

(P[sub]p[/sub])[sup]n[/sup] --> 1 or 0

depending on whether there exist any primes.

Additional Lemma: There are primes!

Proof: Start counting at 1 and continue until a number is reached
whose only factors are (well, you know). This process terminates at p=2.
Hence there are primes!

Corollary: The desired probability is 1 (if there really are primes).

:wink:

[QUOTE=Wacky;111511]I don't think so. For an inductive proof, such as you suggest, you need two elements. You need the inductive step, such as you have indicated, but you also need a boundary (terminal) condition. You have failed to provide this portion of your "proof".[/QUOTE]

Aha - a hole in davar55's "proof"!

See, I told you it was not so simple after all - which is why I am carefully refraining from revealing any of the power, the glory, the subtle elegance [the proofistic Feng Shui, if you will] that is my proof until I am sure it has been received and begun the peer review process. [As in, the referee says, "let me peer at it and get back to you..."]

The truly marvelous thing about my proof is that not only it is non-inductive, it is also non-capacitative and non-resistive. A sort of room-temperature-superconducting proof, one might [humbly] say.

[QUOTE=davar55;111514]Let P[sub]p[/sub] be the probability that a random prime p is a number n.

By the lemma, P[sub]p[/sub] = P[sub]p-1[/sub] = ... .[/quote]

Ah, but that assumes that for any number, subtracting 1 also gives a number. That is quite plausible, but also requires proof, to avoid the "it's plausible, so must be true" logical-fallacy trap.

[quote]Hence multiplying the P[sub]p gives

(P[sub]p[/sub])[sup]n[/sup] --> 1 or 0 [/quote]

This assumes the sequence terminates - can you prove that it in fact always does? [And that every intermediate term is a number?]

[quote]Additional Lemma: There are primes!

Proof: Start counting at 1 and continue until a number is reached
whose only factors are (well, you know). This process terminates at p=2.
Hence there are primes![/quote]

No, you have basically just "proved" that "2 is prime, because its only prime factor is 2, which is prime." In other words, a tautology, not a proof. I'm afraid that it's back to the drawing board with you my friend, despite your valiant and praiseworthy effort.

Well, thus begins (and perhaps ends) the review process.

The necessary intermediate steps to complete my proof
might take volumes, and perhaps a lifetime to solve a problem
that has already been solved by another
(albeit the solution is not yet revealed -- we await patiently).

:wink:

(Must prove 2 is prime ...
must prove 2 is prime ...
must prove 2 is prime ...
...
Does this EVER terminate?)

[QUOTE=wblipp;111512]A random Prime might turn out to be the Prime of Miss Jean Brodie or a Prime Rib Steak. I'm pretty sure neither of these is a number[/QUOTE]

William,

I am in complete agreement with your conclusion that the probability is less than unity.

However, beware, I do not agree with your above statement. I have known a few "Misses" who certainly were "numbers", and d*mn good looking ones at that.

[QUOTE=ewmayer;111515]Aha - a hole in davar55's "proof"!

See, I told you it was not so simple after all - which is why I am carefully refraining from revealing any of the power, the glory, the subtle elegance [the proofistic Feng Shui, if you will] that is my proof until I am sure it has been received and begun the peer review process. [As in, the referee says, "let me peer at it and get back to you..."][/QUOTE]

Ah- the ambiguity of the English language strikes again. I think the following is the proper "Pier Review" for the paper in question.

Norm