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

[scottstapp]

Homework Statement

If a and b are positive integers, then ab=gcd(a,b)*lcm(a,b).

Homework 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.

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

 

[VeeEight]

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)

 

[scottstapp]

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.?

 

[scottstapp]

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