How to prove that a result does not hold in general

[ver_mathstats]

Homework Statement

Prove that for all a, b, c ∈ ℤ, if a|c and b|c and (a,b)=1 then ab|c. Prove result does not hold in general when (a,b)≠1.

Homework Equations

The Attempt at a Solution

This is not my formal proof it's just the scratch work, for the first part, I have there are integers x and y such that ax + by = 1 which leads us to cax + cby = c. Since a|c and b|c there are integers f and g such that c=fa and c=gb which leads us to gbax + faby = c. Since ab divides the left side it must mean it divides the right side. I am just confused how to prove the second part of the question. I thought instead I could do (a,b)=3 and show that it does not work but I do not think that is right. Any suggestions would be appreciated.

Thank you.

 

[phyzguy]

It only takes a single counter-example to prove the second part.

 

[fresh_42]

Looks good so far. You can work with $(a,b)=3$ as an example. Just find specific numbers $a,b,c$ so that the statement is false.

 

[ver_mathstats]

It only takes a single counter-example to prove the second part.

Thank you, I got it now.

 

[ver_mathstats]

Looks good so far. You can work with $(a,b)=3$ as an example. Just find specific numbers $a,b,c$ so that the statement is false.

Thank you for the reply, I did find values for a, b, and c.