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## probability that a natural number is prime

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• 2a. Are Big Numbers More Likely To Be Prime? Posted by: Chester Elders chesterelders28@yahoo.com chesterelders28 Date: Sun May 23, 2010 2:17 am ((PDT)) ...
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2a. Are Big Numbers More Likely To Be Prime?
Posted by: "Chester Elders" chesterelders28@... chesterelders28
Date: Sun May 23, 2010 2:17 am ((PDT))

> I have no idea whether this proof is unique to me; but, I think I can
> prove that the probability that a natural number is prime is directly
> proportional to it's size.

How do you calculate the probability that a number of a given size
be prime?

By my own intuition, I would calculate it as follows.

Probability of N being prime = (Number of positive primes =< N) / N

which by the prime number theorem,

for large N,

is approximately,

(N/log(N)) / N = 1/log(N)

which approaches 0 as N increases without bound.

> Proof:

> Let A be a natural number.

> Let S be the set of all prime numbers such that any element of S is
> less than the step function of the square root of A.

Why not say that S is the set of prime numbers < square root of A.

S = { p such that p is prime and p**2 < A }

> Define the probability that A is a prime number as the quotient
> (A-S)/A.

A is a number.
S is a set.

What do you mean by A - S?

> Since the quotient of the square root of a natural number divided by
> the number decreases in proportion to the size of that number, it
> follows that the limit of (A-S)/A approaches 1 as A gets arbitrarily
> large.

If S were a number, not a set, this would have a better chance of making
sense.

Suppose we let S be the number of primes < square root of A.

Then your ratio (A - S)/A is more related to the probability that
A is not prime, than it is to the probability that A is prime.

If we took S to be the number of primes < A, then

the ratio (A - S)/A would be the probability that a positive number
less than A is composite.

Perhaps you might want to argue that the probability that a number < A
being composite is the same as the probability that A is composite.

> This is tantamount to saying that the probability that A is prime
> increases as A increases. QED.

Did you come close to proving that the probability that A is composite
increases as A increases?

> Wouldn't this imply that the prime numbers get arbitrarily
> dense when one considers extremely large natural numbers?

:) Composite numbers do get arbitrarily dense in sets of
extremely large natural numbers.

> What really has me worried is the seemingly inescapable conclusion
> that, for sufficiently large natural numbers, wouldn't primes tend to
> become sequential? I know that's not possible because the
> distribution of prime numbers is random.

Distribution of primes is not random. The distribution is directly
determined by smaller primes.

> I assume that probability holds uniformly across the number line.

Not sue what you have in mind when you say "probability holds uniformly"

> I would appreciate some feedback.

> Sincerely,
> Chester

:) You got it.

Kermit Rose
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