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Greatest common divisor proof

  1. Apr 20, 2017 #1
    Hi,
    I need opinion about this problem.
    ==================================================
    question :Prove:
    If(a,b)= l and if ( "(a,b)=1" mean greatest common divisor of integers and b is 1 )
    c|a (c divides a)
    and
    d|b (d divides b )
    then
    (c,d)= 1. ( "(c,d)=1" mean greatest common divisor of integers and b is 1 ) <-- this need to be proved.
    ========================================
    (Is that following a good proof ?)
    ========================================
    Then there are 2 sets A and B.
    divisors of a ∈ A <-- do this need be proved too?
    divisors of b ∈ B

    A ∩ B = 1

    since
    c ⊂ A
    d ⊂ B

    c ∩ b = 1 which is what we are looking for.
    ===========================================

    Thank you.
     
  2. jcsd
  3. Apr 20, 2017 #2

    Math_QED

    User Avatar
    Homework Helper

    You are correct when you say you can consider the sets ##A,B## containing the divisors of ##a##, resp. ##b##. You don't need to prove that. You know that such a set always exists.

    However, when you write ##A \cap B = 1##, this is bad notation. You either write ##A \cap B = \{1\}## or ##|A \cap B | = 1##. I don't know what exactly you mean by this, but either way you must explain why this is true.

    You also wrote ##c \cap d = 1## which doesn't make sense as ##c,d## are elements and not sets.
     
  4. Apr 20, 2017 #3
    Then there are 2 sets A and B.
    divisors of a ∈ A
    divisors of b ∈ B

    A ∩ B = {1} <-- this is just restating the fact that 'a' and 'b' has only gcd which is "1" I am trying to say the divisor set A and divisor set be B has only one common element with is "1"


    c ∈ A <--because 'c' divide 'a' that means its part of 'A' set of all the divisor of a
    d ∈ B < -- same reason as above

    if
    C ={ all the divisor of c }
    D ={ all the divisor of d }

    C ⊂ A because a is one of the multiples of c. is this need to proved ?
    D ⊂ B same reason as above.

    we know A ∩ B = {1}
    since C ⊂ A and D ⊂ B

    C ∩ D ={1}
    Which means the only common divisor of c and d is '1'
     
  5. Apr 20, 2017 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You can prove it, but I guess your course did that earlier already - it is one of the basic features of divisibility.
     
  6. Apr 20, 2017 #5
    So I have proved it properly ?
    Please tell.

    Thank you.
     
  7. Apr 20, 2017 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I'm not the person grading your homework. I think it is okay, but I cannot know if the person grading your homework wants to see more steps in between.

    I moved the thread to our homework section, by the way.
     
  8. Apr 20, 2017 #7
    Its not home work, I am judging my self before taking a analysis course. That will be my first ever math course.
    This problem is from the book I will be using.
    That is why I was keen to know.
    Thank you.
     
  9. Apr 20, 2017 #8

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper

    i would just show directly that if x divides both c and d, then x also divides both a and b, hence x = ±1.
     
  10. Apr 20, 2017 #9
    Curious - what's the name/author of the book?
     
  11. Apr 20, 2017 #10
    Introduction to Analytic Number Theory
    by Tom M. Apostol
    https://www.amazon.com/Introduction...sr=8-1-fkmr0&keywords=number+analysis+apostol


    BTW, what I posted is not how this book deals with things.
    I once read a book, a few chapters, "introduction to topology".
    So I remembered some set language.

    In t Apostol's book I have read only few pages, I tried this problem from Apostol's book because It looked like it could have been done, before reading stuff from the book.
     
    Last edited by a moderator: May 8, 2017
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