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Math Help - Prime Number on an interval

  1. #1
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    Prime Number on an interval

    Can someone help me prove the following:

    If n is an integer and n is greater than 2, then there exists a prime p such that n is less than p which is less than n!.

    Thanks so much for any help.
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  2. #2
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    Quote Originally Posted by JimmyT View Post
    Can someone help me prove the following:

    If n is an integer and n is greater than 2, then there exists a prime p such that n is less than p which is less than n!.

    Thanks so much for any help.
    Suppose that for some number n there is no prime between n and n!.

    Now consider (n!-1), this is not divisible by any number 2<=k<=n, and so
    is not divisible by any prime <=n. Hence either (n!-1) is prime or there is
    some prime q, n<n<n!, which divides (n!-1). Either case is a contradiction
    and so there must be a prime netween n and n!.

    RonL
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  3. #3
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    Quote Originally Posted by JimmyT View Post
    Can someone help me prove the following:

    If n is an integer and n is greater than 2, then there exists a prime p such that n is less than p which is less than n!.

    Thanks so much for any help.
    Use the Betrand Postulate (proven by Chebyshev).

    Given any integer n>2 there exists a prime,
    n<p<2n.

    Then, since n!>2n we have that it too contains a prime.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    Use the Betrand Postulate (proven by Chebyshev).

    Given any integer n>2 there exists a prime,
    n<p<2n.

    Then, since n!>2n we have that it too contains a prime.
    Prove the weaker result from the stronger? Doesn't that suggest to
    you that maybe they have not covered the stronger result?

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    Doesn't that suggest to
    you that maybe they have not covered the stronger result?

    RonL
    Are you saying that there is a flaw in my flawlwss proof written by my flawless hand?

    It seems you are saying I should not use that since I did not proof it. But the result is well known.
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  6. #6
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    Quote Originally Posted by ThePerfectHacker View Post
    Are you saying that there is a flaw in my flawlwss proof written by my flawless hand?

    It seems you are saying I should not use that since I did not proof it. But the result is well known.
    No I am suggesting that you could prove the result in a manner that
    the student might understand, and could reproduce in their own words.

    RonL
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  7. #7
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    Quote Originally Posted by ThePerfectHacker View Post
    Use the Betrand Postulate (proven by Chebyshev).

    Given any integer n>2 there exists a prime,
    n<p<2n.

    Then, since n!>2n we have that it too contains a prime.
    and yet n can equal 2 and it would work
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  8. #8
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    Quote Originally Posted by Quick View Post
    and yet n can equal 2 and it would work
    Yes it does.
    I should have said n>1 but that is trivial.
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  9. #9
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    Quote Originally Posted by ThePerfectHacker View Post
    Yes it does.
    I should have said n>1 but that is trivial.
    Why would that be trivial?
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  10. #10
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    Quote Originally Posted by Quick View Post
    Why would that be trivial?
    When a mathemation uses the phrase trivial it might refer to something which is easily seen to be true. Like that x(x^2+1)=0 it is trivial that x=0, but not that x=i. Also, many theorem do not work for some integers but work for sufficiently large integers. Like this theorem over here that n=1 does not work, a mathemation would say that is trivial. When you learn more math you will see that many theorem do not work for the simplest case, usually invlovling 1 or 0. Thus, when someone says that is does not work a mathemation might respond that what he said is trivial. Those are the two meanings of the word.
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