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Prove m + m/1

  1. May 20, 2007 #1

    disregardthat

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    1. The problem statement, all variables and given/known data

    Prove that if m is a positive rational number, then:

    m + 1/m = integer

    Only when m = 1

    2. Relevant equations

    Don't know any

    3. The attempt at a solution

    That's the problem, I don't know where to start. I have tryed a few things, but none of them works out for me.

    I don't know much of proofs and such, the problem I have is always where shall I start? Do you have any tips?

    EDIT: The topics name is not correct, the correct question is in the topic
     
  2. jcsd
  3. May 20, 2007 #2

    radou

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    If m is a positive rational number, then m = p/q, for some natural numbers p and q. Hence, you have p/q + q/p = integer => p/q & q/p are integers. Now, what does that mean?
     
  4. May 20, 2007 #3

    D H

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    Just because the sum of two numbers is an integer does not mean the two numbers themselves are integers. This final step needs further justification.
     
  5. May 20, 2007 #4

    disregardthat

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    I was at that stage:

    a/b + b/a = integer

    But I couldn't state that this only works when m=1

    Is it ok to say: a and b is integers, if a is more that 1, we would have a decimal number in b/a and an integer in a/b, if b is more than one, we would have a decimal number in a/b and an integer in b/a.

    Since m must be positive, and that we can't divide by 0, m must be 1 to get an integer.
     
    Last edited: May 20, 2007
  6. May 20, 2007 #5

    D H

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    How would you express a/b+b/a as a fraction? What are the conditions that this fraction be an integer?

    Edited to add:

    Now put this back in the context of the original problem, which is to prove m+1/m is an integer only if m is 1.
     
  7. May 20, 2007 #6

    radou

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    a/b + b/a is not an integer if a doesn't divide b or if b doesn't divide a (or both). So, for a/b + b/a to be an integer, a must divide b, and b must divide a.
     
  8. May 20, 2007 #7

    D H

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    radou, you just gave out the answer. Not a good idea in the homework section ...
     
  9. May 20, 2007 #8

    radou

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    Hm, I'd let Jarle decide about that.
     
  10. May 20, 2007 #9

    Dick

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    This is all true, except that the sum of two nonintegers can be a integer. So it's not so clear and I don't think the answer is given away yet. You'll want to clear the fractions in a/b+b/a=1 and assume a and b are relatively prime.
     
    Last edited: May 20, 2007
  11. May 21, 2007 #10

    disregardthat

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    I don't know how to prove it, I tryed, but I can't
     
    Last edited: May 21, 2007
  12. May 21, 2007 #11

    Dick

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    I'm not sure what you mean by 'shortened'. I would write it as a^2+b^2=a*b. Is this relation possible if a and b have no common divisors?
     
  13. May 21, 2007 #12

    HallsofIvy

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    I'm a bit unclear as to whether the problem is to show that m+ 1/m is an integer or (m+1)/m is an integer only if m= 1.

    If it is the first case, I don't think you need to worry about "divisors". Since the difference of two integers is an integer, if m+ 1/m= n, an integer, then 1/m= n- m is an integer. What does that tell you?
     
  14. May 21, 2007 #13

    D H

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    Halls, it looks like you assumed that m is an integer. It is a rational.
     
  15. May 21, 2007 #14

    Dick

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    You can assume a and b have no common divisors, right? Otherwise you could just simplify the fraction. This mean one of a or b has a prime divisor that does not divide the other. Suppose p divides a but not b. How many of the terms in a^2+b^2=a*b are divisible by p? How many are not? Is there anything wrong here?
     
    Last edited: May 22, 2007
  16. May 21, 2007 #15

    D H

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    Dick, you arrived at [itex]a/b+b/a=1[/itex] in post #9 and then used that in posts #11 and #14. However, you do not know [itex]a/b+b/a=1[/itex]. The given condition is that [itex]a/b+b/a=n[/itex], where [itex]a, b, n \in \mathbb N[/itex]. So the problem becomes one of showing that [itex]a^2+b^2=nab[/itex] only has solution if [itex]a=b=1[/itex].
     
  17. May 21, 2007 #16

    Dick

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    Ooooops. Sorry! Same argument, though.
     
  18. May 21, 2007 #17
    That should read "only has solution if [itex]a=b[/itex]"
     
    Last edited: May 21, 2007
  19. May 21, 2007 #18

    radou

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    Nah, it should read "only has solution in the set of integers".

    Boy, what a thread. :tongue2:
     
  20. May 21, 2007 #19
    a and b may both be integers and yet not be a solution. a = b is crucial. a and b were chosen from the set of integers.
     
  21. May 21, 2007 #20

    radou

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    I was referring to n, but nevermind.
     
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