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 June 1st, 2016, 09:37 PM #1 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 464 Thanks: 29 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? Thanks from manus
 June 1st, 2016, 11:29 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 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. Thanks from manus and Joppy
 June 2nd, 2016, 10:09 PM #3 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 464 Thanks: 29 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. Thanks from manus
 June 2nd, 2016, 11:11 PM #4 Senior Member   Joined: Apr 2014 From: UK Posts: 965 Thanks: 342 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 Thanks from Loren and manus
June 3rd, 2016, 04:14 AM   #5
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Quote:
 Originally Posted by Loren 1. Unbounded, sequential random numbers differ infinitely across exhaustive permutations. Their finite differences are in pseudorandom sequence.
This makes no sense to me. Numbers cannot be unbounded. Unbounded is an adjective that applies to sets or functions. Sequential numbers are not random numbers because there is an infinitely recurring pattern. You can't order the real numbers at all, so what numbers are we talking about? What does it mean for numbers to differ inifinitely? They are either different or they are not.The difference between any two numbers is finite, because every number is finite. I don't think you understand the difference between "random" and "psuedo-random" either. If you have a sequence of random numbers from a particular distribution, their differences are also random.

Quote:
 Originally Posted by Loren 2. Bounded, sequential random numbers differ unequally across exhaustive permutations. Their equal differences are in pseudorandom sequence.
How can numbers that "differ unequally" have "equal differences"? This apparent contradiction makes it completely unclear what you mean by either phrase. Equal differences cannot be in either a random or a psuedo-random sequence.

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: 464 Thanks: 29 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: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra Yes, in quantum mechanics, the wave function is a probability distribution for the quantity. All pdfs that are non-zero 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

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