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Math Help - GCD and least common multiple Problem

  1. #1
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    GCD and least common multiple Problem

    Please give me help to find out the answer for following question

    With the usual notation show that
    i. For a > 0, and b > 0,[a,b](a,b) = ab
    ii. [a,b, c](ab,bc, ca) = abc if a,b, c are positive integers.
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  2. #2
    Senior Member Shanks's Avatar
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    (i) let (a,b)=d,then there exist positive integers m and n such that a=md,b=nd,and (m,n)=1.
    Thus [a,b]=mnd; ab=mdnd; => [a,b](a,b)=ab. Q.E.D
    (ii)By (i)(proved), we have:
    [a,b,c]=[\frac{ab}{(a,b)},c]=\frac{\frac{abc}{(a,b)}}{(\frac{ab}{(a,b)},c)}=\f  rac{abc}{(a,b)(\frac{ab}{(a,b)},c)}=\frac{abc}{(ab  ,bc,ac)}
    Thus [a,b,c](ab,bc,ac)=abc; Q.E.D
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  3. #3
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    Quote Originally Posted by dhammikai View Post
    Please give me help to find out the answer for following question

    With the usual notation show that
    i. For a > 0, and b > 0,[a,b](a,b) = ab
    ii. [a,b, c](ab,bc, ca) = abc if a,b, c are positive integers.

    These follow directly from unique prime factorization theorem. Do you know that? Can you express (a,b) and [a,b] in terms of prime factors of a,b?
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  4. #4
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    You wrote me:
    "....These follow directly from unique prime factorization theorem. Do you know that? Can you express (a,b) and [a,b] in terms of prime factors of a,b?"

    Please let me know about the unique prime factorization theorem please..
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  5. #5
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    Quote Originally Posted by dhammikai View Post
    You wrote me:
    "....These follow directly from unique prime factorization theorem. Do you know that? Can you express (a,b) and [a,b] in terms of prime factors of a,b?"

    Please let me know about the unique prime factorization theorem please..
    Fundamental theorem of arithmetic - Wikipedia, the free encyclopedia
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  6. #6
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    Quote Originally Posted by Shanks View Post
    (i) let (a,b)=d,then there exist positive integers m and n such that a=md,b=nd,and (m,n)=1.
    Thus [a,b]=mnd; ab=mdnd; => [a,b](a,b)=ab. Q.E.D
    (ii)By (i)(proved), we have:
    [a,b,c]=[\frac{ab}{(a,b)},c]=\frac{\frac{abc}{(a,b)}}{(\frac{ab}{(a,b)},c)}=\f  rac{abc}{(a,b)(\frac{ab}{(a,b)},c)}=\frac{abc}{(ab  ,bc,ac)}
    Thus [a,b,c](ab,bc,ac)=abc; Q.E.D
    @dhammikai - Please note this is a nice proof however Shanks has used some intermediate theorems like

    1. [a,b,c]=[[a,b],c]
    2. (a,b,c)=((a,b),c)
    3. (ka,kb)=k(a,b) etc

    so please ensure you follow them.

    This I feel is much better than the proof I was proposing
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