Can anyone please review/verify this proof of greatest common divisor?

In summary, the greatest common divisor (GCD) of two positive integers divides their least common multiple (LCM). This can be seen by considering that the LCM is essentially a multiplication of unique prime factors, while the GCD is the common prime factors. Therefore, the GCD must exist in the LCM.
  • #1
Math100
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Homework Statement
Prove that the greatest common divisor of two positive integers divides their least common multiple.
Relevant Equations
None.
Proof: Suppose gcd(a, b)=d.
Then we have d##\mid##a and d##\mid##b for some a, b##\in## ##\mathbb{Z}##.
This means a=md and b=nd for some m, n##\in## ##\mathbb{Z}##.
Now we have lcm(a, b)=##\frac{ab}{gcd(a, b)}##
=##\frac{(md)(nd)}{d}##
=dmn
=dk,
where k=mn is an integer.
Thus, d##\mid##lcm(a, b), and so gcd(a, b)##\mid##lcm(a, b).
Therefore, the greatest common divisor of two positive integers divides their least common multiple.
 
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  • #2
Seems sufficient. Note that the LCM is also basically a multiplication of unique prime factors, for lack of a better phrasing. Take 12 and 28, for example. One is 2*2*3, and the other is 2*2*7. So you end up with 2*2*3*7 as the LCM. The GCD, on the other hand, is prime factors in common, so I this example, 2*2, which are exactly the duplicate factors that get extracted from the LCM, but they must necissarily exist in the LCM.
 
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  • #3
valenumr said:
Seems sufficient. Note that the LCM is also basically a multiplication of unique prime factors, for lack of a better phrasing. Take 12 and 28, for example. One is 2*2*3, and the other is 2*2*7. So you end up with 2*2*3*7 as the LCM. The GCD, on the other hand, is prime factors in common, so I this example, 2*2, which are exactly the duplicate factors that get extracted from the LCM, but they must necissarily exist in the LCM.
Thank you for the help.
 
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