1. Standard memberorfeo
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    17 Aug '06 13:05
    I have conclusive proof that no prime is divisible by 927. I checked the first dozen or so.
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    17 Aug '06 17:13
    Of the 100 first primes only one is even. That makes 1/100=1%. Right?
    Does this mean that the more primes we have, the less fraction of even primes we get?
    Am I right so far?
  3. Standard memberBowmann
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    17 Aug '06 18:232 edits
    The average distance between any two consecutive primes near a given number x can be approximated by the natural logarithm of x, and the prediction improves as x gets larger (i.e. X).
  4. Standard memberXanthosNZ
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    17 Aug '06 23:13
    Originally posted by FabianFnas
    Of the 100 first primes only one is even. That makes 1/100=1%. Right?
    Does this mean that the more primes we have, the less fraction of even primes we get?
    Am I right so far?
    I'll say it again.
    A proof that 100% of primes are odd is not the same as proof that all primes are odd.

    An example. The Riemann Hypothesis is one of the great unsolved problems in number theory. It has been proved that 100% of the zeroes fall on the critical line. However, it has not been shown that all interesting zeroes fall on the critical line and that is what is required for a proof of the Hypothesis.

    Remember if a finite number of X are Y and there are an infinite number of X then 100% of X are not Y.
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    18 Aug '06 01:00
    Originally posted by XanthosNZ
    I'll say it again.
    A proof that 100% of primes are odd is not the same as proof that all primes are odd.
    100% sounds like 'all' to me.
    If 'all' > 100% how many percent is it?
    Perhaps you mean something else.
  6. Standard memberXanthosNZ
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    18 Aug '06 01:12
    Originally posted by ThudanBlunder
    100% sounds like 'all' to me.
    If 'all' > 100% how many percent is it?
    Perhaps you mean something else.
    Nope I know exactly what I mean. Page 92 of Dr. Riemann's Zeros explains it better than I can but I'll try.

    The percentage of X that satisfy Y is defined as P(X=Y)*100 or if we define y as the set of X that satisfies Y then 100*y/X.
    Now if X is infinite and y is finite then the percentage is 0 (finite/infinite). Therefore 100% of X does not satify Y even though there could be a large number of exceptions.

    I realize it may seem contradictory and a little pedantic but in math we get that a lot. Apparently it is also possible to have an infinite number of counterexamples and still have 100% satisfied thanks to different degrees of infinite. But that's a side issue.

    I should correct something I said though, my memory failed me and apparently it hasn't been shown that 100% of Riemann Zeroes lie on the critical line. It's 98.6%. However I present a different example. In 1985 someone proved that Fermat's Last Theorem is true for 100% of n. That is not a proof of it however. It was later proved by Andrew Wiles in 1996.
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    18 Aug '06 01:36
    Originally posted by XanthosNZ
    Nope I know exactly what I mean. Page 92 of Dr. Riemann's Zeros explains it better than I can but I'll try.

    The percentage of X that satisfy Y is defined as P(X=Y)*100 or if we define y as the set of X that satisfies Y then 100*y/X.
    Now if X is infinite and y is finite then the percentage is 0 (finite/infinite). Therefore 100% of X does not satify Y e ...[text shortened]... 100% of n. That is not a proof of it however. It was later proved by Andrew Wiles in 1996.
    Perhaps you mean that proving something is true for an infinite number of values is not necessarily the same as proving something is true for all values (which to me means 100% of values by definition).
  8. Standard memberXanthosNZ
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    18 Aug '06 01:45
    Originally posted by ThudanBlunder
    Perhaps you mean that proving something is true for an infinite number of values is not necessarily the same as proving something is true for all values (which to me means 100% of values by definition).
    Yes it is possible to prove something is true for an infinite number of values but that is once again different.
    The paper on Fermat's Last Theorem that I mentioned is:

    Andrew Granville,
    The set of exponents for which Fermat's Last Theorem is true, has density one Comptes Rendus de l'Acad\'emie des Sciences du Canada 7 (1985), 55-60.

    The set having density 1 is the same as saying that the set comprises 100% of the total.
  9. Standard memberroyalchicken
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    18 Aug '06 01:50
    Xanthos is correct; almost all of the other posts in this thread are word salad. Please stop.
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    18 Aug '06 02:353 edits
    Originally posted by royalchicken
    Xanthos is correct; Please stop.
    Nobody was saying he is wrong. I was merely asking him to clarify what he meant.
    If you have nothing worthwhile to contribute except proofs by assertion then perhaps you should lead by example and butt out.

    I still believe that the non-technical term 100% means 'all'; the rest I leave to measure theory.
  11. Standard memberXanthosNZ
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    18 Aug '06 03:37
    Originally posted by ThudanBlunder
    I still believe that the non-technical term 100% means 'all'; the rest I leave to measure theory.
    That would make you wrong in a mathmatical sense.
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    18 Aug '06 03:544 edits
    Originally posted by XanthosNZ
    That would make you wrong in a mathmatical sense.
    Actually, I have yet to see 100% defined 'in a mathematical sense' rather than the usual one. Perhaps you could point me in the right direction? Certainly the concepts you discuss do not require such a definition.

    http://en.wikipedia.org/wiki/Almost_everywhere
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    18 Aug '06 07:58
    If I may continue:
    Of the first thousand, million and billion primes there are progressively smaller and smaller relative amount of primes that are even.
    The formula is 1/n is smaller when n is being larger. Right?

    XanthosNZ knows what I'm aiming at. He's using a preemptive strike while he tries to prove that I am wrong. But he does it at a mathematical level we normally don't have.
    The only thing he proves, so far, is his own superiority, no offence given, he *is* superior.
    But I would hope he doesn't try too hard until I put the final question in this poser, until the twist appear.

    For the rest of us - does my reasoning make any sense?
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    18 Aug '06 09:195 edits
    Originally posted by FabianFnas
    If I may continue:
    Of the first thousand, million and billion primes there are progressively smaller and smaller relative amount of primes that are even.
    The formula is 1/n is smaller when n is being larger. Right?

    XanthosNZ knows what I'm aiming at. He's using a preemptive strike while he tries to prove that I am wrong. But he does it at a mathemat his poser, until the twist appear.

    For the rest of us - does my reasoning make any sense?
    Not every prime is odd, since 2 is not odd and prime.

    If something happens with probability 1, you say it occurs 'almost surely' or 'almost always'. This is not the same as it 'always' happening. Whether or not you prefer to think of '%100 per cent of the time' as 'always' or 'almost always' is arbitrary, a matter of convention.

    For example, suppose you toss fair coins, independently and get score -1 for a head and +1 for a tail. Then the Strong Law of Large Numbers says that, *almost surely*

    ( (total score after n tosses) / n ) tends to 0

    as n tends to infinity. Although there are some events where this does not happen (eg. tossing H, H ,H, ...forever) these events have probability 0.

    Besides this whole thread is quite silly, I thought it was a joke at first, and you were using the word 'odd' in the sense 'unusual'.

    The reason the thread is silly is that there is no uniform probability measure on an infinite countable set. So this talk of 'n-1 out of n primes being odd and letting n tend to infinity' and then concluding 'probability a prime picked at random is odd = 1 therefore %100 primes are odd' is nonsensical. There is no uniform probability P for which you can say

    P(a prime is odd) = 1.

    So even if you're using 'X happens %100 of the time' to mean

    P(X happens) = 1

    the argument doesn't work.
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    18 Aug '06 10:002 edits
    Originally posted by XanthosNZ
    Nope I know exactly what I mean. Page 92 of Dr. Riemann's Zeros explains it better than I can but I'll try.

    The percentage of X that satisfy Y is defined as P(X=Y)*100 or if we define y as the set of X that satisfies Y then 100*y/X.
    Now if X is infinite and y is finite then the percentage is 0 (finite/infinite). Therefore 100% of X does not satify Y e ...[text shortened]... .

    I realize it may seem contradictory and a little pedantic but in math we get that a lot.
    Be warned that the contents of 'popular maths' books are quite often misleading, and sometimes simply wrong.

    This is a good example. A person wishing to be precise would simply have said 'all but finitely many of the X do not satisfy Y' instead of introducing this misleading and unnecessary verbiage on probabilities (which seems to me only to be valid in the finite, uniform case, as 'something/infinity = 0' is not a valid argument) and percentages (which, as I have indicated in an earlier post, is really a matter of language use).
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