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Math Help - 2 problems involving examples and counter-examples.

  1. #1
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    2 problems involving examples and counter-examples.

    1. if a doesn't divide b and a doesn't divide c , then a doesn't divide b times c.

    give a counterexample to show this is a false statement.



    2. if a divides c and b divides c sometimes, a times b divides c and sometimes it doesn't

    show an example when a times b divides c .


    show and example when a times b doesn't divide c.
    Last edited by mr fantastic; July 20th 2010 at 08:30 PM. Reason: Re-titled.
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  2. #2
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    Quote Originally Posted by lonestarross View Post
    1. if a doesn't divide b and a doesn't divide c , then a doesn't divide b times c.

    give a counterexample to show this is a false statement.


    .
    a can't be divided by i, and a can't be divided by -i, but a can be divided by i . -i
    Note that i= \sqrt{-1}
    I'm not 100% positive with my answer though
    Last edited by nhunhu9; July 20th 2010 at 11:47 AM.
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  3. #3
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    You know of a numerical example of this?
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    Quote Originally Posted by lonestarross View Post
    1. if a doesn't divide b and a doesn't divide c , then a doesn't divide b times c.

    give a counterexample to show this is a false statement.



    2. if a divides c and b divides c sometimes, a times b divides c and sometimes it doesn't

    show an example when a times b divides c .


    show and example when a times b doesn't divide c.
    Usually with this language and nothing else specified, we assume we are working over the integers.

    (1) is very simple.. don't know how to give a hint, here's such a counterexample, learn from it

    a = 2*3
    b = 2*5
    c = 3*5

    (2) this is confusing. Is the first part supposed to mean

    if (a divides c) and (b divides c sometimes)

    or

    if (a divides c and b divides c) sometimes

    ?

    I'll assume the first. But it can be proven with the stricter requirement "a divides c and b divides c always".

    Just use simple examples. For the case ab divides c, let c = ab. For ab does not divide c, let a = b = c > 1.
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  5. #5
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    the first.\\ but when you say a=2*3, b=2*5, c=3*5, 5 does divide 10 and 15...never mind i t is backwards
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    Quote Originally Posted by lonestarross View Post
    the first
    Okay, my response addresses this.
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  7. #7
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    so for 2. ab|c a=2 b=3 c=6 2*3|6

    ab doesn't |c 4*6|12

    still dont get (1) 2*3- 5 2*5- 10 2*3- 15 a will |b 5\10


    on
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  8. #8
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    thank you for your help. got it and understand it now
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