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Math Help - Proof about integers

  1. #1
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    Proof about integers

    Let a,b,c be integers. Prove the following:

    (a) if b|a, then b|ac

    (b) if b|a and c|b, then c|a

    (c) if c|a and c|b, then c|(ma + nb) for any integer m,n


    Here is what I have come up with:

    (a) I am unsure about how to prove.

    (b) By definition, b|a iff a = qb
    c|b iff b = qc
    Therefore, a = qc => c|a

    (c) By definition, c|a iff a = qc
    c|b iff b = qc

    Can anybody give me some pointers?
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  2. #2
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    Quote Originally Posted by page929 View Post
    Let a,b,c be integers. Prove the following:

    (a) if b|a, then b|ac

    (b) if b|a and c|b, then c|a

    (c) if c|a and c|b, then c|(ma + nb) for any integer m,n


    Here is what I have come up with:

    (a) I am unsure about how to prove.

    (b) By definition, b|a iff a = qb
    c|b iff b = qc
    Therefore, a = qc => c|a

    (c) By definition, c|a iff a = qc
    c|b iff b = qc

    Can anybody give me some pointers?
    What you did for b looks find just remember that the q's are different numbers.

    Here is a) and you will use the same technique for c)

    Since b|a \implies a=q_1b
    Now if we multiply this equation by c \in \mathbb{Z} we get

    ac=cq_1b and since the integers are closed under multiplication we get

    ac=q_2b but this last line implies that

    b|ac

    For c use this and add
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