Random numbers and proper factors

mfgoode · 2006-01-24, 16:56

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

R.D. Silverman · 2006-01-24, 17:11

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.

akruppa · 2006-01-24, 17:37

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

mfgoode · 2006-01-24, 17:59

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

alpertron · 2006-01-24, 18:18

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.

mfgoode · 2006-01-25, 15:39

Quote:

Originally Posted by R.D. Silverman

Yes. I can.

Kindly explain the derivation as I have requested.
Thanking you,
Mally

mfgoode · 2006-01-25, 15:47

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

cheesehead · 2006-01-25, 17:23

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.

John Renze · 2006-01-25, 23:49

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,.

John Renze · 2006-01-26, 00:17

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?

mfgoode · 2006-01-26, 16:28

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

2006-01-24, 16:56	#1
mfgoode Bronze Medalist Jan 2004 Mumbai,India 2²×3³×19 Posts	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

2006-01-24, 18:18	#5
alpertron Aug 2002 Buenos Aires, Argentina 3²×149 Posts	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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2006-01-24, 17:37	#3
akruppa "Nancy" Aug 2002 Alexandria 2,467 Posts	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

2006-01-25, 23:49	#9
John Renze Nov 2005 2⁴×3 Posts	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,.

2006-01-26, 00:17	#10
John Renze Nov 2005 60₈ Posts	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?