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Math Help - if a|b and b|a, then a=b

  1. #1
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    if a|b and b|a, then a=b

    if a|b and b|a, then a=b

    am=b and bn=a but how do I show a=b?
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  2. #2
    Moo
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    a/b=m and a/b=n, then...
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    Quote Originally Posted by Moo View Post
    a/b=m and a/b=n, then...

    Shouldn't it be \frac{b}{a}=m \ \mbox{and} \ \frac{a}{b}=n but I don't see how that helps.

    Does Latex not work?
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    A Plied Mathematician
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    am = b and bn = a implies a = (am)n = a(mn). What does mn have to be?
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    Quote Originally Posted by Ackbeet View Post
    am = b and bn = a implies a = (am)n = a(mn). What does mn have to be?
    An integer.
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    A Plied Mathematician
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    I think you can know more than that: a = a(mn) implies a (mn-1) = 0. Since a is not zero, mn - 1 = 0, which implies...
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  7. #7
    MHF Contributor undefined's Avatar
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    I'm assuming a and b are positive integers? Otherwise consider

    a = 2
    b = -2

    If a and b are positive integers, then a|b implies a \le b and b|a implies b \le a, therefore a = b.
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    Quote Originally Posted by dwsmith View Post
    if a|b and b|a, then a=b

    am=b and bn=a but how do I show a=b?

    Modifying your question: Let a and b be integers. If  a|b and  b|a , then  a=\pm b (  b or  -b ).

     a=bn and  b=am, where  n and  m are integers , then substitute for  b to get  a=(am)n or  a=a(mn) . Divide both sides by a(\neq0), and so both m and n equal 1 or both equal -1.
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    MHF Contributor undefined's Avatar
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    Hmm, I'm not sure if this will display properly, I keep getting LaTeX errors on my previews.

    Another way to get melese's result above is

    a|b \Rightarrow |a| \le |b|

    b|a \Rightarrow |b| \le |a|

    |a| = |b|

    a = \pm b

    Since this isn't rendering properly, here is an image I got using this site:

    if a|b and b|a, then a=b-divides_abs_val.png
    Last edited by undefined; June 17th 2010 at 03:06 PM.
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  10. #10
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by undefined View Post
    Hmm, I'm not sure if this will display properly, I keep getting LaTeX errors on my previews.

    Another way to get melese's result above is

    a|b \Rightarrow |a| \le |b|

    b|a \Rightarrow |b| \le |a|

    |a| = |b|

    a = \pm b

    Since this isn't rendering properly, here is an image I got using this site:

    Click image for larger version. 

Name:	divides_abs_val.png 
Views:	43 
Size:	1.2 KB 
ID:	17884
    This works, but it makes use of the total order on \mathbb{Z}, which you don't really need. Ackbeet's proof works if \mathbb{Z} is replaced by the polynomial ring \mathbb{Z}[x], for instance.
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