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Is order a subset of randomness?
Q. Here's an example: Suppose that you have an MP3 playlist alphabetically ordered, but you prefer to listen to it in a random order. You turn on the shuffle but the songs still play in a perfectly alphabetical order. Unlikely? Yes, but definitely possible. Would that indicate that order is just a subset of chaos? While I agree that there's an element of subjectivity when using the terms "order" and "random", the actual rules for very well defined. An alphabetical order is quite precise and isn't a matter of one's whim. Anything which isn't in alphabetical order, isn't ordered, so chaos cannot be a subset of order.
Asked by Belzetot - Sat Aug 4 02:18:53 2007 - - 10 Answers - 6 Comments

A. It is consistency that differentiates order from chaos. As the above answer says, any sequence could be some order if we choose to define it that way... but it won't be order unless it can be repetitively consistent... repetition is an essential part of judging consistency. Which therefore means strength of probability differentiates between random and order. To put into order is to enhance probability of the desired consistency. Does that make order a subset of chaos... in my view no... it is chaos put into a bind to lose its randomness in order to tilt probability towards a defined consistency. What an awesome thought provoking question!!!
Answered by small - Sat Aug 4 03:20:41 2007

Why aren't primes thought of as the subset of numbers that can be multiplied to generate all numbers ?
Q. Isn't that easier to get your head around. (Of course, on this definition, 1 would be needed as a prime. But what harm is that ?)
Asked by Shariputra - Thu Jun 4 00:51:08 2009 - - 4 Answers - 0 Comments

How do you prove {n/2 , n E N} is the smallest subset of R that contains 1/2 and is closed under addition?
Q. How do you prove {n/2 , n E N} is the smallest subset of R that contains 1/2 and is closed under addition?
Asked by Spidey M - Mon Jan 21 18:38:00 2008 - - 0 Answers - 0 Comments

A. 1. it is closed under addition for any 2 elements of the set, let say p/2 and q/2 where p, q are natural numbers, p/2+q/2=(p+q)/2 since p, q are natural numbers, so is p+q. 2. It is the smallest one the set contains all natural numbers plus the (odd natural numbers)/2 a) if one of the form (odd natural number)/2 is missing then it can be written as 1/2 +a natural number and it is not closed under addition a) if one natural number is missing (let say m) then also is missing m/2; if m/2 is natural you divide again by 2 until m/2 is not natural; then apply a)
Answered by testmathbp - Mon Jan 21 18:52:03 2008