How Do GCD and LCM Relate in the Product of Two Numbers?

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SUMMARY

The relationship between the greatest common divisor (gcd) and least common multiple (lcm) of two positive integers, a and b, is established by the equation ab = gcd(a, b) * lcm(a, b). The discussion emphasizes the application of two key propositions: (1) if d is a common divisor of a and b, then ab/d is a common multiple, and (2) if m is a common multiple of a and b and m divides ab, then ab/m is a common divisor. The user Scott seeks clarification on demonstrating that ab/d is the least common multiple and how to show that ab/m is less than or equal to d.

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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
 
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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)
 


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.?
 


...?
 

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