Proof that gcd divides lcm

[Math100]

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.

 

[valenumr]

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.

 

[Math100]

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.