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Fractions of all numbers

  1. Nov 1, 2009 #1
    I really didn't know how to word the title so sorry if it's a little confusing.
    And I didn't know whether to post this in number theory or not but ah well.

    The other day, I started thinking about this and I was just wondering if it had been done before or if it's even correct;

    Half of all positive integers are divisible by 2,
    A third are divisible by 3,
    A quarter by 4 etc.

    Now, of the third that are divisible by 3, a half of those numbers are also divisible by 2.
    So 1/6 of positive integers are divisible by 3 and not 2.

    I continued this trend to find;
    [tex]\frac{1}{2}[/tex] of p. ints are divisible by 2
    [tex]\frac{1}{3}\left(1-\frac{1}{2}\right)[/tex] are divisible by 3 and not 2
    [tex]\frac{1}{5}\left(1-\left(\frac{1}{2}+\frac{1}{3}\left(1-\frac{1}{2}\right)\right)\right)[/tex] are divisible by 5 and not 3 or 2
    [tex]\frac{1}{7}\left(1-\left(\frac{1}{2}+\frac{1}{3}\left(1-\frac{1}{2}\right)+\frac{1}{5}\left(1-\left(\frac{1}{2}+\frac{1}{3}\left(1-\frac{1}{2}\right)\right)\right)\right)\right)[/tex] are divisible by 7 and not 5, 3 or 2

    If I sum all these to infinity, it should equal 1 as it will then cover all positive integers.

    So could you by any chance verify the truth of this statement,

    [tex]\sum^{\infty}_{p_{0}=2}\frac{1}{p_{0}}\left(1-\sum^{p_{0}-1}_{p_{1}=2}\frac{1}{p_{1}}\left(1-\sum^{p_{1}-1}_{p_{2}=2}\frac{1}{p_{2}}\left(...\right)\right)\right) = 1[/tex]

    I know this is quite awkward looking and I don't even know if I've used correct notation, I just sort of scrawled it down as I was thinking it.
     
  2. jcsd
  3. Nov 1, 2009 #2

    CRGreathouse

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    To make it a little easier: the sum of "divisible by 2" + "not divisible by 2 but divisible by 3" + "not divisible by 2 or 3 but divisible by 5" is "divisible by 2, 3, or 5" which is "not relatively prime to 2, 3, and 5":

    [tex]f_1 = 1/2[/tex]
    [tex]f_2 = 1/2 + 1/2\cdot1/3 = 1 - 1/2\cdot2/3[/tex]
    [tex]f_3 = 1/2 + 1/2\cdot1/3 + 1/2\cdot2/3\cdot1/5 = 1 - 1/2\cdot2/3\cdot4/5[/tex]
    . . .
    [tex]f_k = 1 - \sum_{q\in\mathbb{P},q\le p_k}\left(1-\frac1p\right)[/tex]

    Now instead of lots of sums, you only have one. All you need to show is that the sum in question tends to 0 as k goes to infinity (which is true).
     
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