Solve ab=gcd(a,b)*lcm(a,b)

1. Feb 2, 2010

scottstapp

1. The problem statement, all variables and given/known data
If a and b are positive integers, then ab=gcd(a,b)*lcm(a,b).

2. Relevant equations
I am allowed to use the following propositions which have already been proved:

(1) If d is a common divisor of a and b, then ab/d is a common multiple of a and b.
(2) If m is a common multiple of a and b and m divides ab, then ab/m is a common divisor of a and b.

A hint given:
set d=gcd(a,b) and m=lcm(a,b). Use (1) to show that ab/d>=m. Use (2) to show that ab/m<=d.

3. The attempt at a solution
1. Let a and b be positive integers. Suppose d=gcd(a,b) and m=lcm(a,b).
2. By (1) ab/d is a common multiple of a and b so ab/d=aL and ab/d=bK
3. Multiply by m gives mab/d=aLm and mab/d=bKm
4. ab/d>=m

I am missing a step between 2 and 3. Any suggestions?
Thanks,
Scott

2. Feb 2, 2010

VeeEight

Re: ab=gcd(a,b)*lcm(a,b)

You have shown that ab/d is in fact a common multiple. Now you need to show that it is the least common multiple of a and b. To do this, write down the definitions of gcd and lcm and apply them to the equation. You can also try for a contradiction (assume that it is not the least common multiple to contradict d=gcd)

3. Feb 2, 2010

scottstapp

Re: ab=gcd(a,b)*lcm(a,b)

So we know that (ab/d)|a and (ab/d)|b. Therefore because ab is being divided by the greatest common divisor, it must equal its least common multiple. Therefore ab/d>=m. Correct?

To show that ab/m<=d we would say that because ab is being divided by its least common multiple it leaves its greatest common divisor. Therfore ab/m<=d.?

4. Feb 2, 2010

scottstapp

Re: ab=gcd(a,b)*lcm(a,b)

...?