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Proof that 1/2 +2/3 + 3/4 does not equal a whole number

  1. Dec 1, 2004 #1
    I am deleting
    Last edited: Dec 1, 2004
  2. jcsd
  3. Dec 1, 2004 #2


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    Does this help?

    [tex]\sum_{j = 1}^{N} \frac {j}{j + 1} = \sum_{j = 1}^{N} \left( 1 - \frac {1}{j + 1}\right) = N - \sum_{j = 1}^{N} \frac {1}{j + 1}[/tex]
    Last edited: Dec 1, 2004
  4. Dec 1, 2004 #3


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    Minor quibble-Tide you lost a negative sign.

    Tide has reduced this nicely to showing any the Harmonic series never hits an integer. This might be tough to prove so I'll kick out a hint.

    Let 2^k be the largest power of 2 less than or equal to N, so 2^k<=N<2^k+1. Assume the sum is an integer. Clear the denominators carefully so every term becomes an integer, except the 1/2^k term. This gives a contradiction.

    I'll leave it to you how to clear them 'carefully', but seeing how to do this for the first few values of N might help.
  5. Dec 2, 2004 #4


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    if you clear denominators you see that n! must divide the top. It follows for example that n must divide (n-1)! which is impossible say, if n is prime. If not, and p is the largest prime number less than n, it follows again that p must divide some number between p+1 and n. I.e. n must be at least 2p. But this contradicts the fact that there must be another prime between p and 2p, hence another between p and n.
  6. Jan 4, 2005 #5
    This is kind of related, is:

    [tex]\sum_{r=1}^{n} r![/tex]

    ever a perfect square? I know this is true for n=1 and n=3, but what about in general?

    I'd like hints only, please. :smile:
  7. Jan 4, 2005 #6

    matt grime

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    I'd probably pick some useful number and look at things mod that number. for it is divisible by 3 for n>1, so it must be WHAT if it is a square?
    Last edited: Jan 4, 2005
  8. Jan 4, 2005 #7
    It must be in the form 3(n^2) or in the form 3^(2n+1)?
  9. Jan 4, 2005 #8
    For n>4, the last digit is always '3', so it will never be a perfect square.
  10. Jan 4, 2005 #9
    I swear I was just thinking the same thing!

    m! where m>4 always ends in zero. So when n>4, we have 10p+(4!+3!+2!+1!)=10p+30+3=10q+3, but I really didn't know that if a number ends in 3 it's a perfect square.
  11. Jan 4, 2005 #10

    matt grime

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    just work out the squares of the residues mod 10, which is waht i meant by considering a suitable mod thing

    and if 3 divides N and N is a square then 9 divides N is what I meant
  12. Jan 4, 2005 #11
    Wow... That's brilliant.

    Well done!
  13. Jan 4, 2005 #12
    For the sake of curiosity...

    What you're saying is:
    if N>3 and N = 0 (mod 3), then N = 0 (mod 9).

    Can we generalize this to:
    if N>3^n and N = 0 (mod 3), then N = 0 (mod 3^{n+1})?
  14. Jan 4, 2005 #13

    matt grime

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    This is realiant on N being a square, you realize, and nothing to do with N being greater than some power of three, nor for that matter is it to do with 3.

    Let N be the n'th power of q where q is an integer. Write q as the product of primes, now, what is the decomposition of N relative to q?

    For example, let n=3 so we#re considering cubes. What are some cubes? 8, 27, 216. How may times do 2 and 3 occur each decomposition? it's always a multiple of n isn't it?
  15. Jan 4, 2005 #14
    I know this might sound naive, but why is that?

    Wait, nevermind... I figured it out. :smile:

    So, to sum up, if q divides p^n then so does q^n?

    Thanks for all your help matt.
    Last edited by a moderator: Jan 4, 2005
  16. Jan 4, 2005 #15

    matt grime

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    Did you write out q as a product of primes? Raise it to the power n? What do you get?

    for instance if q=12=2^2.3, then q^3= 2^^.3^3 doesn't it?
  17. Jan 4, 2005 #16

    matt grime

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    in response to your last corrected qusetion, only if the divisor is a prime, it is one of the equivalent criteria for being prime (in the integers)
  18. Jan 4, 2005 #17
    Yeah, naturally.

    Thanks again.
  19. Jan 4, 2005 #18
    And it is not!
    As I had said, if a number ends in 3, it is not a perfect square.

    There is no n>4 which satisfies your equation.
  20. Jan 4, 2005 #19
    Yes, sorry, that's what I meant. :rofl:
  21. Jan 5, 2005 #20
    I have one more question about your proof.
    S = 33 (mod 10), but does S really need to be divisible by 3? 43 = 33 (mod 10), but 43 isn't divisible by 3.
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