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(*https://www.mersenneforum.org/showthread.php?t=8841*)

odds of a random prime being a numberI 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. |

1 Attachment(s)
[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 |

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