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Math Help - help me to prove this conjecture

  1. #16
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    Quote Originally Posted by CaptainBlack View Post
    As every prime p not equal to 2 is odd, p+1 is even and so not a prime.

    But 2 is prime and 2+1=3 which is also prime.

    RonL

    [I must say that I am lost, what is the conjecture - or whatever - that is under discussion here?]
    I just realized that what you said is right.. so then how would I correct my proof to proof or disproof this conjecture??
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  2. #17
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    Quote Originally Posted by Discrete View Post
    I just realized that what you said is right.. so then how would I correct my proof to proof or disproof this conjecture??
    Simply restrict your proof to a product of primes that has at least two primes in it.

    -Dan
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  3. #18
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    Quote Originally Posted by topsquark View Post
    Simply restrict your proof to a product of primes that has at least two primes in it.

    -Dan
    but then it won't satisfy for all condition if I restrict it

    BTW can I use this counterexample

    say that the number is 3 x 5 x 7 x 11 x 13

    and the first 3 prime number is 3,5,and 7... if we use multiply these three number we get 105 but then we add 1 to this number it becomes 106 which is not a prime number.. therefore the conjecture is false
    Last edited by Discrete; July 17th 2007 at 08:33 AM.
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  4. #19
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Discrete View Post
    but then it won't satisfy for all condition if I restrict it
    I don't see that to be a disagreeable issue. For instance there are many formulas that are only applicable to certain circumstances. And there are numerous proofs I've seen that use Mathematical Induction where the theorem isn't true for the first few values of n but are true for, say, n = 3 and above.

    -Dan
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  5. #20
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    Quote Originally Posted by topsquark View Post
    I don't see that to be a disagreeable issue. For instance there are many formulas that are only applicable to certain circumstances. And there are numerous proofs I've seen that use Mathematical Induction where the theorem isn't true for the first few values of n but are true for, say, n = 3 and above.

    -Dan
    is my counter example right?
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  6. #21
    Forum Admin topsquark's Avatar
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    Maybe you'd better restate what it is you are trying to prove.

    I had thought you were trying to show that
    2 \cdot 3 \cdot 5 \cdot ~ ... ~ \cdot p + 1
    is not divisible by any prime p in the multiplicative sequence.

    -Dan
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  7. #22
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    no that's not the case, I am trying to prove that p1 x p2 x p3 x pn + 1 is a prime number for every positive integer n. where p1 x p2 x pn is the first n prime number.. I am not trying to proof any divisible here
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  8. #23
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    Quote Originally Posted by Discrete View Post
    no that's not the case, I am trying to prove that p1 x p2 x p3 x pn + 1 is a prime number for every positive integer n. where p1 x p2 x pn is the first n prime number.. I am not trying to proof any divisible here
    That is false.
    2+1 = \mbox{ prime}
    2\cdot 3 + 1 =  \mbox{ prime}
    2\cdot 3\cdot 5 + 1 =  \mbox{ prime}
    2\cdot 3\cdot 5\cdot 7 + 1 =  \mbox{ prime}
    2\cdot 3\cdot 5\cdot 7\cdot 11+1 =  \mbox{ prime}
    2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13 + 1 = 59\cdot 509
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  9. #24
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    Quote Originally Posted by ThePerfectHacker View Post
    That is false.
    2+1 = \mbox{ prime}
    2\cdot 3 + 1 = \mbox{ prime}
    2\cdot 3\cdot 5 + 1 = \mbox{ prime}
    2\cdot 3\cdot 5\cdot 7 + 1 = \mbox{ prime}
    2\cdot 3\cdot 5\cdot 7\cdot 11+1 = \mbox{ prime}
    2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13 + 1 = 59\cdot 509
    Gosh I could have said that

    RonL
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  10. #25
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    I am now totally confused! Why did you mention that there is a number outside 2 x 3 x 5 x ... that if is multiplied can produce the same number. I am confused why you keep mentioning:

    2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13 + 1 = 59\cdot 509

    can you explain
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  11. #26
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    Quote Originally Posted by Discrete View Post
    I am now totally confused! Why did you mention that there is a number outside 2 x 3 x 5 x ... that if is multiplied can produce the same number. I am confused why you keep mentioning:

    2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13 + 1 = 59\cdot 509

    can you explain
    BECAUSE IT IS A COUNTER EXAMPLE TO YOUR CONJECTURE AND ONE COUNTER EXAMPLE DISPROVES A CONJECTURE.

    2, 3, 5, 7, 11, 13 are the first six primes and one more than their product is composite.

    RonL
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  12. #27
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    Quote Originally Posted by Discrete View Post
    I am now totally confused! Why did you mention that there is a number outside 2 x 3 x 5 x ... that if is multiplied can produce the same number. I am confused why you keep mentioning:

    2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13 + 1 = 59\cdot 509

    can you explain
    Because you are trying to show that for any sequence of primes, their product plus one is a prime. This is not true because
    2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1
    which your proposition states should be a prime number is not a prime number.

    -Dan
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  13. #28
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    Quote Originally Posted by topsquark View Post
    Because you are trying to show that for any sequence of primes, their product plus one is a prime. This is not true because
    2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1
    which your proposition states should be a prime number is not a prime number.

    -Dan
    In fact:

    <br />
1+\prod_{i=1}^{N}p_i \ \ \ \mbox{ is composite for }N=6, 7, 8,9,10,12<br />

    RonL
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  14. #29
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    I got it now, thanks!
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