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Old 2006-01-24, 16:56   #1
mfgoode
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Question Random numbers and proper factors

In my current reading of Maths I have come across the following problem which I am at a loss to ascertain.
Take three positive whole numbers at random.
What is the chance they have no proper factor in common?
Answer around 83%- to be precise, 0.83190737258070746868......:surprised
Can anyone elucidate?
Mally
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Old 2006-01-24, 17:11   #2
R.D. Silverman
 
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Quote:
Originally Posted by mfgoode
In my current reading of Maths I have come across the following problem which I am at a loss to ascertain.
Take three positive whole numbers at random.
What is the chance they have no proper factor in common?
Answer around 83%- to be precise, 0.83190737258070746868......:surprised
Can anyone elucidate?
Mally
Yes. I can.
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Old 2006-01-24, 17:37   #3
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I suppose the analysis will be similar to that for two randomly chosen integers. We had that one before in this forum. The result for two random integers was 1/zeta(2), the results for 3 random integers should be 1/zeta(3) which matches your constant.

Alex
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Old 2006-01-24, 17:59   #4
mfgoode
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Thumbs up Random numbers and proper factors.

Quote:
Originally Posted by akruppa
I suppose the analysis will be similar to that for two randomly chosen integers. We had that one before in this forum. The result for two random integers was 1/zeta(2), the results for 3 random integers should be 1/zeta(3) which matches your constant.

Alex
Okay Alex well and good.
Could you link this number with another well known number in electro dynamics ? Its amazing !
Mally

Last fiddled with by mfgoode on 2006-01-24 at 17:59
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Old 2006-01-24, 18:18   #5
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The quotient \zeta {(n)}/\pi^n for n even is always rational, but for odd values of n, such as in this case there is no known explicit finite expression using elementary functions and constants.

Last fiddled with by alpertron on 2006-01-24 at 18:19
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Old 2006-01-25, 15:39   #6
mfgoode
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Quote:
Originally Posted by R.D. Silverman
Yes. I can.
Kindly explain the derivation as I have requested.
Thanking you,
Mally
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Old 2006-01-25, 15:47   #7
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Quote:
Originally Posted by alpertron
The quotient \zeta {(n)}/\pi^n for n even is always rational, but for odd values of n, such as in this case there is no known explicit finite expression using elementary functions and constants.

We know that Z(2) = (pi^2)/6. Would you call this a rational number?
Similarly Z(4) = (pi^4)/90.
Mally
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Old 2006-01-25, 17:23   #8
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Quote:
Originally Posted by mfgoode

We know that Z(2) = (pi^2)/6. Would you call this a rational number?
Note that Alpertron was talking about the quotient (Z(n)/pi^n) being rational, not Z(n) itself being rational.
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Old 2006-01-25, 23:49   #9
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What does "two randomly chosen integers" mean? You can't put a uniform distribution on the entire set of integers, so there is some interpretation involved.

Specifically, you seem to be using the following heuristic: Let $n$ be an random integer and $p$ be a prime. Then $Pr(p | n) = 1/p$.

This is perfectly sensible and number theorists use this kind of reasoning all the time. I was wondering, though, how you formulate this precisely,.
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Old 2006-01-26, 00:17   #10
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Thinking about this some more... you are making some statement over the interval (0, n) and then take a limit as n ->\infty. Is there somewhere easily accessible these details are written down? I'm still grappling with whether this really corresponds to "random integers," but I'll do the philosophical musings on my own time.

Simpler question: How do I put this as a postscript to my previous post, rather than a new posting?
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Old 2006-01-26, 16:28   #11
mfgoode
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Thumbs up Random numbers and proper factors.

Quote:
Originally Posted by cheesehead
Note that Alpertron was talking about the quotient (Z(n)/pi^n) being rational, not Z(n) itself being rational.
Thank you cheesehead that makes it much clearer and precise.
These rational numbers I take it are of even no.s and no odd has been found to describe the quotient rationally.
Mally
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