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  1. #1
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    Math proof help

    I dont quite know how to attack this.

    Here is the problem.

    (a divides b) or (a divides c) implies a divides bc.

    any help at all would be greatly appreciated
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  2. #2
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    Assuming that we are dealing with the integers.
    Now, I ask you: if a divides b then does a divide (bc)?
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  3. #3
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    Quote Originally Posted by Plato View Post
    Assuming that we are dealing with the integers.
    Now, I ask you: if a divides b then does a divide (bc)?
    I know it does, it is just getting started that is hard. It's obvious that no matter what integer you plug in it divides it, I just don't know how to write it out in proof format. Thanks for your help on this.
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    Quote Originally Posted by padsinseven View Post
    I know it does, it is just getting started that is hard. It's obvious that no matter what integer you plug in it divides it, I just don't know how to write it out in proof format. Thanks for your help on this.
    If a|b then b = na, where n is an integer. Does this help you?

    -Dan
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    Quote Originally Posted by padsinseven View Post
    I just don't know how to write it out in proof format.
    Your problem is quite simply: WHAT IS A PROOF?
    If you do not understand that simple question, see your instructor.
    Otherwise, it is pointless to ask for something you do not understand.
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  6. #6
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    Quote Originally Posted by Plato View Post
    Your problem is quite simply: WHAT IS A PROOF?
    If you do not understand that simple question, see your instructor.
    Otherwise, it is pointless to ask for something you do not understand.
    Thanks for the help Plato. I didn't realize that I was able to assume both a divides b and a divides c. Once I understood that the proof was pretty simple. At least I think that I got it.
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by padsinseven View Post
    Thanks for the help Plato. I didn't realize that I was able to assume both a divides b and a divides c. Once I understood that the proof was pretty simple. At least I think that I got it.
    If that is your proof then it isn't correct. You are to assume a|b or a|c, not necessarily both.

    I am going to assume that a divides only b. (We can simply switch the roles of b and c if a divides c and not b, so we are losing no generality here by doing this.)

    So a|b implies that b = na, where n is some positive integer. Thus
    bc = (na)c = (nc)a

    But nc is also a positive integer. Thus a|bc as well.

    -Dan
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  8. #8
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    If a|b the b=an for some n and bc=a(nc), so a|bc.
    On the other hand, if a|c the c=am for some m and bc=a(mb), so a|bc.
    In either case, a|bc.
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