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Math Help - Gelfand's Algebra - Problem 42.

  1. #1
    Junior Member DIOGYK's Avatar
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    Gelfand's Algebra - Problem 42.

    Here's the problem:
    Fractions a/b and c/d are called neighbor fractions if their difference (ab-bc)/bd has numerator +-1, that is, ad-bc=+-1. prove that:
    a) in this case neither fraction can be simplified (that is, neither has any common factors in numerator and denominator);
    b) if a/b and c/d are neighboring fractions, then (a+b)/(c+d) is between them and is a neighbor fraction for both a/b and c/d;

    I successfully solved a) part of this problem, but I'm stuck on b), and I think maybe question is asked incorrectly, because I think instead of (a+b)/(c+d), there should be (a+c)/(b+d) fraction, which is middle of both a/b and c/d and is also neighbor to both. Am I correct or is question asked correctly and I am just unable to solve it. By the way, I solved b) part successfully if we replace (a+b)/(c+d) with (a+c)/(b+d).
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  2. #2
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    Re: Gelfand's Algebra - Problem 42.

    I think you are right. For example, 1/3 and 1/2 are neighbor fractions, but (1+3)/(1+2) = 4/3 is not between them.
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    Re: Gelfand's Algebra - Problem 42.

    In the theory of Farey series, the "mediant" of a/b and c/d is defined to be (a+c)/(b+d); so my guess is that was Gelfand's intent.
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  4. #4
    Junior Member DIOGYK's Avatar
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    Re: Gelfand's Algebra - Problem 42.

    Quote Originally Posted by emakarov View Post
    I think you are right. For example, 1/3 and 1/2 are neighbor fractions, but (1+3)/(1+2) = 4/3 is not between them.
    Yes, I assumed that (a+c)/(b+d) was the correct form and solved it that way. Thanks for the reply.
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