Prove: If gcd(a,b) = 1, a|c, b|c, then prove that (ab)|c.

[bigp0ppa1046]

hopefully someone can help

If gcd(a,b)=1 and a divides c, and b also divides c. Prove that the product a*b divides c.

 

[pka]

It is against my principles to do someone else's homework.
Therefore, I will get you started. Then you can show some work.

Because, gcd(a,b)=1 then lcm(a,b)=ab.
But c is a multiple of both a & b. This means \(\displaystyle ab \le c.\)
So we have \(\displaystyle c = qab + r,\quad 0 \le r < ab.\)
Can you show that \(\displaystyle r=0\)?

 

[daon]

Here's another way.

GCD(a,b)=1 implies that a and b have unique (non-overlapping) prime factorizations. Otherwise, both would share a prime factor and have a GCD of at least that factor.

Say a=P1*P2*P3*...*Pn, and b=R1*R2*R3*...*Rm where Pi and Rj are primes, and Rj does not equal Pi for any i,j.

Since a and b divide c, all the primes Pi from a must appear in the prime factorization of c. Likewise, all the primes from Rj from b must appear in the prime factorization of c. Since no prime from a is the same as any prime from b (and obviously, vise versa), c can be written as k(P1*P2*...Pn)(R1*R2*...*Rm) where k is some integer. Thus c=k(a)(b) which is what we need.