Divisibility of c by a and b but not ab .

[wubie]

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.

 

[HallsofIvy]

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

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 guarantee that that won't happen?

 

[wubie]

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.

 

[synergy]

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