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Divisibility of c by a and b but not ab .

  1. Oct 1, 2003 #1
    Divisibility of "c" by "a "and "b" but not "ab".

    Hello,

    I am having trouble with this question:

    i) Give an example of three positive integers a,b,c such that a|c and b|c but ab does NOT divide c.

    ii) In the situation of part (i), is there a condition that guarentees that if a|c and b|c, then ab|c?

    iii) Is the condition in part (ii) necessary? Either prove that it is necessary, or give an example to show that it is not necessary.


    It took me a long time to find three integers that satisfied conditions in part i) of the question. In fact, I couldn't find any such integers, someone had to tell me.

    The integers that were given to me were 2,4,12. Now that these integers were given to me, I definitely can see how they satisfy conditions in part i).

    However I still am having trouble with (ii) and obviously (iii).

    For part i) I was playing around with prime numbers. And I couldn't find any integers which satisfied part i). Was that my mistake? Was that the reason I couldn't find the integers necessary to answer part i)?

    As well, is that the key to ii) and iii)? That is, to guarantee that, if a|c and b|c, then ab|c, a and b must be prime numbers?

    Any help would be appreciated. Thankyou.
     
  2. jcsd
  3. Oct 1, 2003 #2

    HallsofIvy

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    Did you give this much thought? If a divides c, then c contains all prime factors of a. If b divides c, then c contains all prime factors of b. Doesn't it follow that c contains all prime factors of BOTH a and b and so ab must divide c?
    No, it doesn't follow! Why not? What can you say about those prime factors?

    Try a= 6, b= 4, and c= 12. WHY doesn't ab divide c?

    a= 2, b= 4 and c= 12 also work as you noted. That's because a and b have a FACTOR IN COMMON (in both examples, the common factor is 2). That factor is in c but in ab, we get the factors TOGETHER: i.e. In the first example a= 2*3, b= 2*2, c= 2*2*3 but ab= 2*2*2*3- too many twos. How can you guarentee that that won't happen?
     
  4. Oct 1, 2003 #3
    Hello HallsofIvy,

    I did give this question quite a bit of thought. Suffice it to say that I spent too much time on this last question. I just cannot see how it all fits together. I know NOW that the gcd has something to do with it. But I don't see how.
     
  5. Oct 17, 2003 #4
    Think of this:
    if X=a*b*b
    Y=b*c

    then XY=(a*b*b)*b*c

    then X divides abbc
    Y divides abbc

    look at XY -> how many b's

    how many b's in abbc

    does XY divide abbc

    will XY divide abbbc? why? how many b's

    Aaron
     
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