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Math Help - proving ab|c

  1. #1
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    Post proving ab|c

    I dont know why im having trouble with this problem i sware i did it last semester in number theory now for some reason i cant figure it out in modern algebra.

    Given a|b and b|c and gcd(a,b)=1 then ab|c. I know we get b = aj, c = bk and 1 = am + bn i cant figure out how to put them together to get ab|c
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by ChrisBickle View Post
    I dont know why im having trouble with this problem i sware i did it last semester in number theory now for some reason i cant figure it out in modern algebra.

    Given a|b and b|c and gcd(a,b)=1 then ab|c. I know we get b = aj, c = bk and 1 = am + bn i cant figure out how to put them together to get ab|c
    Since (a,b)=1 there exists x,y\in\mathbb{Z} such that ax+by=1\implies axc+byc=c. So then, since ab\mid acx and ab\mid byc we can conclude that ab\mid acx+byc=c
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  3. #3
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    dont mean to sound stupid but im not seeing how we get ab|acx and ab|byc

    edit: are we using the axc + byc =c then saying byc = ajybk?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by ChrisBickle View Post
    dont mean to sound stupid but im not seeing how we get ab|acx and ab|byc
    ab\mid acx since a\mid a and b\mid c. ab\mid byc since a\mid b and b\mid c.
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  5. #5
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    thanks alot
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  6. #6
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    Well ... if you think about it, if a|b and gcd(a,b)=1 => a=1, so if b|c, then ab|c ....

    Why does a=1?

    a|b => ma = b

    So, gcd(a,b)=gcd(a,ma)=a

    Since gcd(a,b)=1 => a=1
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  7. #7
    Senior Member Shanks's Avatar
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    I think it is a typo. it should be "a|c,b|c,and gcd(a,b)=1...".
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  8. #8
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    ah, haha ... probably ... I was kind of skeptical about the question
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