My Math Forum Conjecture on "proof on proofs" for infinite cases

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 January 28th, 2012, 12:29 PM #1 Member   Joined: Aug 2011 Posts: 85 Thanks: 1 Conjecture on "proof on proofs" for infinite cases By the title I mean roughly a broad conjecture of mine on infinite sets that I arrived at after learning about many infinite cases proofs. So it states: In infinite sets if you have a phenomenon occur twice, it will occur infinite times. I would say at the minimum the scope is the natural numbers but also at the outmost maybe all complex numbers from the fact that all polynomials can be made to equal a natural number. So in primes that I am most interested one finds: Examples on proven results: - 2 and 3 are primes and there are infinitely many - the first two critical roots of Zeta function have real part 1/2 and there are infinite such roots If conjecture is true (given the scope): - there are two instances of twin primes (3,5) and (5,7) therefore there are infinitely many twin primes - generally any prime gap that shows twice will occur infinite times Of course this is lighthearted and not to be taken too seriously, I don't claim I can prove this or anything, but this a gut feeling I had. Can you give me an anti-example, of something occuring in an infinite set more than twice albeit a finite number of times?
 January 28th, 2012, 04:33 PM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Conjecture on "proof on proofs" for infinite cases Certainly; I can think of a couple offhand, but there will be many, many examples of phenomena that occur a finite (but > 2) number of times in an infinite set. 1. In the infinite group $(\mathbb{C}^\ast,\times),$ there are four elements of finite order (1,-1,i,-i) 2. Every positive integer can be represented as the sum of at most 19 fourth powers of integers. There are a finite number, however, that cannot be represented as a sum of 16 or fewer. Feel free to add more, folks - I don't really have the brainpower to conjure up any more examples, as I should be asleep.
 January 28th, 2012, 06:17 PM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Conjecture on "proof on proofs" for infinite cases There are lots of examples, but here's a cute (nontrivial) one: http://oeis.org/A071828
 January 28th, 2012, 07:18 PM #4 Senior Member   Joined: Nov 2011 Posts: 595 Thanks: 16 Re: Conjecture on "proof on proofs" for infinite cases [quote="mattpi"]Certainly; I can think of a couple offhand, but there will be many, many examples of phenomena that occur a finite (but > 2) number of times in an infinite set. 1. In the infinite group $(\mathbb{C}^\ast,\times),$ there are four elements of finite order (1,-1,i,-i) I am not sure I understand what you mean by this. What is being of finite orders? Is that a number a such that there exists n so that $a^n=a$ because then there are infinitely many of them no?
 January 28th, 2012, 07:24 PM #5 Senior Member   Joined: Oct 2008 Posts: 215 Thanks: 0 Re: Conjecture on "proof on proofs" for infinite cases The phenomena could be "the number is no more than 2". Then both 1 and 2 satisfy the phemomena but all others positive integer not
 January 28th, 2012, 07:28 PM #6 Senior Member   Joined: Nov 2011 Posts: 595 Thanks: 16 Re: Conjecture on "proof on proofs" for infinite cases Haha Duz..this is a very good one!
January 29th, 2012, 02:28 AM   #7
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Re: Conjecture on "proof on proofs" for infinite cases

Quote:
Originally Posted by Dougy
Quote:
 Originally Posted by mattpi Certainly; I can think of a couple offhand, but there will be many, many examples of phenomena that occur a finite (but > 2) number of times in an infinite set. 1. In the infinite group $(\mathbb{C}^\ast,\times),$ there are four elements of finite order (1,-1,i,-i)
I am not sure I understand what you mean by this. What is being of finite orders? Is that a number a such that there exists n so that $a^n=a$ because then there are infinitely many of them no?
Duuhh. This is why I shouldn't post after midnight. Of course you are right - I was thinking of the group $\{a+bi:\,a,b\in\mathbb{Z},\,(a,b)\neq(0,0)\}$ with multiplication.

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