- What is the product over all of the primes p of:

(p^2+1)/(p^2-1) ?

That's a constant that requires EVERY prime in order

to calculate it.

It turns out to be 5/2. Which is not irrational.

On 2/13/2013 9:15 AM, Kermit Rose wrote:

> I expected the twin prime constant to be irrational

> because I expected that any constant

> that requires EVERY prime in order to calculate it,

>

> would necessarily be irrational.

>

> Kermit Rose

>

>

>

>

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> - --- In primenumbers@yahoogroups.com, Jack Brennen wrote:
>

Nice point, Jack.

> What is the product over all of the primes p of:

>

> (p^2+1)/(p^2-1) ?

>

> That's a constant that requires EVERY prime in order

> to calculate it.

>

> It turns out to be 5/2. Which is not irrational.

print(zeta(2)^2/zeta(4));

2.5000000000000000000000000000000000000

David - Re: Is the twin prime constant irrational?

Twin prime constant

= (3/2)(1/2)(5/4)(3/4)(7/6)(5/6)(11/10)(9/10)...(p/(p-1))((p-2)/(p-1))...

I expected that it would have been easily determined whether or not

the twin prime constant was rational or irrational.

It would not be possible for the twin prime constant to be rational

because the infinite numerator is odd, and the infinite denominator is

divisible by

2 infinitely many times.

Kermit - How about this infinite product here?

(99/10)*(111/110)*(1111/1110)*(11111/11110)*...

The partial products are:

9.9

9.99

9.999

9.9999

and so on...

The product quite obviously converges to an even number (10), but all of

the numerators are odd and all of the denominators are even. Even and

odd really have no meaning when it comes to infinity and limits. As

this example shows, a series of partial products, all of which have

odd numerator and even denominator, can converge to not only a rational

number, but an even integer.

On 2/15/2013 8:53 AM, Kermit Rose wrote:

> Re: Is the twin prime constant irrational?

>

>

>

> Twin prime constant

> = (3/2)(1/2)(5/4)(3/4)(7/6)(5/6)(11/10)(9/10)...(p/(p-1))((p-2)/(p-1))...

>

>

> I expected that it would have been easily determined whether or not

> the twin prime constant was rational or irrational.

>

> It would not be possible for the twin prime constant to be rational

> because the infinite numerator is odd, and the infinite denominator is

> divisible by

> 2 infinitely many times.

>

> Kermit

>

>

>

>

>

>

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