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June 1st, 2016, 09:37 PM  #1 
Senior Member Joined: May 2015 From: Arlington, VA Posts: 397 Thanks: 27 Math Focus: Number theory  The infinitude between random numbers
Envisage two random numbers on the unbounded real number line. Aren't the chances almost certain that the two numbers would be separated by practically an infinite distance, and that a third random number would likely lie practically an infinite distance from each of the first two, and so on?

June 1st, 2016, 11:29 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,617 Thanks: 2608 Math Focus: Mainly analysis and algebra 
No. All the distances are guaranteed to be finite. Furthermore, you seem to be implying a uniform probability distribution for selecting the numbers. This is not possible. 
June 2nd, 2016, 10:09 PM  #3 
Senior Member Joined: May 2015 From: Arlington, VA Posts: 397 Thanks: 27 Math Focus: Number theory 
Are you saying that all random numbers are finite? How do you guarantee this? I am saying that by far most unbounded random numbers are likely infinite since they populate the unbounded real number line. And doesn't the difference between two points approaching infinity also approach infinity? Here are a couple of axioms I have derived: 1. Unbounded, sequential random numbers differ infinitely across exhaustive permutations. Their finite differences are in pseudorandom sequence. As for finite distances: 2. Bounded, sequential random numbers differ unequally across exhaustive permutations. Their equal differences are in pseudorandom sequence. 
June 2nd, 2016, 11:11 PM  #4 
Senior Member Joined: Apr 2014 From: UK Posts: 898 Thanks: 329 
As soon as a random number is determined, it cannot be infinite, thus 2 random numbers cannot have an infinite difference. Infinity isn't a number, but if it were, the chance of picking it randomly would be $\displaystyle \frac{1}{\infty}$ which is arguably 0 
June 3rd, 2016, 04:14 AM  #5  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,617 Thanks: 2608 Math Focus: Mainly analysis and algebra  Quote:
Quote:
Weird Dave's point is crucial in this matter. A probability distribution must sum to 1. For a continuous domain, this means that a probability distribution function (pdf), $f(x)$, must satisfy $\int \limits_{\infty}^{\infty} f(x) \,\mathrm d x=1$. Similarly, for a discrete domain, a pdf, $f(n)$, must satisfy $\sum \limits_{n=\infty}^{\infty} f(n)=1$. A result of these requirements is that no constant function can be a pdf over an infinite domain because the integral/sum doesn't converge.  
June 3rd, 2016, 12:39 PM  #6 
Senior Member Joined: May 2015 From: Arlington, VA Posts: 397 Thanks: 27 Math Focus: Number theory 
I should have had a v8! You and weird Dave are gradually increasing my knowledge of numbers. In physics, the pdf can be the wavefunction, I believe, integrated from negative infinity to infinity, tending toward zero at those limits. 
June 3rd, 2016, 12:43 PM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,617 Thanks: 2608 Math Focus: Mainly analysis and algebra 
Yes, in quantum mechanics, the wave function is a probability distribution for the quantity. All pdfs that are nonzero at $\pm \infty$ tend to zero at those limits in order that the sum/integral converges.

June 3rd, 2016, 02:17 PM  #8 
Senior Member Joined: Dec 2007 Posts: 687 Thanks: 47 
If you mean to pick two different real numbers $a$ and $b$, say wlog $a<b$, then yes, trivially there are infinitely many real numbers in the closed interval $[a,b]$. If you mean some "infinite real number", then you need to define what you mean by this. 

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