April 9th, 2017, 06:22 PM  #101  
Senior Member Joined: Aug 2012 Posts: 1,700 Thanks: 448  Quote:
 
April 9th, 2017, 06:29 PM  #102  
Senior Member Joined: Aug 2012 Posts: 1,700 Thanks: 448  Quote:
Quote:
Last edited by Maschke; April 9th, 2017 at 06:42 PM.  
April 9th, 2017, 10:15 PM  #103  
Senior Member Joined: Mar 2017 From: . Posts: 274 Thanks: 5 Math Focus: Number theory  Quote:
3n+1 is not even because of a probabilistic approach. It is even because of the formula itself. We very well know that if we multiply an odd number by another odd number we get an odd number and when we add an odd number and another odd number, we always get an even number. When we phrase that in terms of probability, we will say the probability of getting an even number after multiplying an odd number by an odd number is zero. The statement is very true. for any number. When it comes to the example you have used on prime numbers, it doesn't relate at all to what we are doing here and that is an assumption that is not based on any fact and it doesn't hold.  
April 9th, 2017, 10:45 PM  #104  
Senior Member Joined: Mar 2017 From: . Posts: 274 Thanks: 5 Math Focus: Number theory  Quote:
Take number 47, the guy is not interested in the value of any outcome but is interested in whether the numbers are even or odd. So he asks you 3n+1.. odd or even? to rescue yourself, if you are an accomplished mathematician, you would say even even without the need to calculate the value. What if he asks you $\frac{3n+1}{2}$ whether the number is odd or even.. you don't have the time to calculate the value.. would you be sure of your answer? what if he asks you a number like 4326.. when you divide it by two.. will you get an even number or an odd number? How would you rescue your life? by plain guess work. Is there any difference between such a scenario with a scenario where you are asked to guess heads or tail in a coin flip?  
April 9th, 2017, 11:10 PM  #105  
Senior Member Joined: Mar 2017 From: . Posts: 274 Thanks: 5 Math Focus: Number theory  Quote:
If we continue on with the chain, the results for consecutive odd numbers will be alternating even and odd numbers. this shows that if I choose any random odd number, i multiply it by 3,add 1 then divide by 2, the number is we get is as likely to be odd as it is to be even. I don't know how to better explain this point if you still don't get it but take your time. or let's take this example.. take all odd integers that are below say 1000. We have shown that for each consecutive odd numbers the outcomes will be alternating and so the even outcomes will be 500 and odd outcomes will be 500. So if I pick any random odd number below 1000, chances are, the outcome will be among the 500 odd numbers or the 500 even numbers.  
April 9th, 2017, 11:18 PM  #106 
Senior Member Joined: Mar 2017 From: . Posts: 274 Thanks: 5 Math Focus: Number theory  
April 9th, 2017, 11:37 PM  #107  
Senior Member Joined: Mar 2017 From: . Posts: 274 Thanks: 5 Math Focus: Number theory  Quote:
If you still have a problem understanding all those points then also asking someone who understands it better may be helpful because the way I explain them may still not align with the way you view the problem. This discussion would go on and on because we are seeing the problem differently yet each of us has a point. From the angle I have shown where we are not calculating the values, I have clearly shown how it is not deterministic. We will never know a cat is in which box unless we open a box. We can't know whether the next number is odd or even unless we calculate it. Also we have exhausted on the point of infinitary probability by showing that it is impossible to have any number from which the sequence infinitely rises. Such a number from which such outcomes would be obtained exists only in theory. Last edited by Mariga; April 9th, 2017 at 11:40 PM. Reason: To grammatically correct a sentence.  
April 10th, 2017, 04:06 AM  #108 
Global Moderator Joined: Dec 2006 Posts: 18,544 Thanks: 1476 
What does your method tell us about the corresponding problem where the multiplier 3 is replaced by a larger odd number?

April 10th, 2017, 04:39 AM  #109 
Senior Member Joined: Mar 2017 From: . Posts: 274 Thanks: 5 Math Focus: Number theory 
Unless we encounter a cycle, the sequence should infinitely diverge. we are either dividing by 2 or multiplying by 2.5 
April 10th, 2017, 08:27 AM  #110  
Senior Member Joined: Jun 2015 From: England Posts: 728 Thanks: 204  Quote:
You don't understand it so it must be wrong or dismissed? As far as I can see you are using what is known as the classical or Laplace definition of probability. This is known to be at best incomplete and at worst flawed. Have you heard of the Bertrand Paradox? https://en.wikipedia.org/wiki/Bertra..._(probability) I quote the key phrase from my link Quote:
The statement The probability is 0.5 is meaningless since it possesses no mechanism or outcome. The statement the probability of a fair coin toss being heads is 0.5 is meaningful, although that is no guarantee of its veracity. I think the problem with probability in this proof lies with the fact that it is an applied concept that has borrowed from construct(s) in pure maths that are not (necessarily) addressed by the application. Here are some problems. 1) Proving the Conjecture requires you to prove that there must exist a finite n such that the necessary convergence occurs. (This is what Masche means by terminating) But you are offering the convergence of an infinite sequence as part of your proof. Yet the convergence of an infinite sequence is unaffected by the prior convergence of any finite number of terms. So with my fair coin if I always start with a head the probability at that first toss will be 1, yet the infinite sequence will still converge to 0.5. Last edited by skipjack; April 10th, 2017 at 09:03 AM.  

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