odds of a random prime being a number

[ewmayer]

I think the complement to this thread 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.

[davar55]

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.

[ewmayer]

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.

[Wacky]

Originally Posted by davar55

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

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".

[wblipp]

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

[davar55]

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

Let P_p be the probability that a random prime p is a number n.

By the lemma, P_p = P_p-1 = ... .

Hence multiplying the P_p gives

(P_p)ⁿ --> 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).

[ewmayer]

Originally Posted by Wacky

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".

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.

[ewmayer]

Originally Posted by davar55

Let P_p be the probability that a random prime p is a number n.

By the lemma, P_p = P_p-1 = ... .

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.

Hence multiplying the P[sub]p gives

(P_p)ⁿ --> 1 or 0

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

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!

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.

[davar55]

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).



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

[Wacky]

Originally Posted by wblipp

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

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.

[Spherical Cow]

Originally Posted by ewmayer

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..."]

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

Norm