Proof
(related to Proposition: Relationship Between the Greatest Common Divisor and the Least Common Multiple)
 Let \(a > 0\), \(b > 0\) be natural numbers.
 Since \(ab\) is a common multiple of \(a\) and \(b\), it follows from the proposition about the least common multiple that $\operatorname{lcm}(a,b)\mid ab.$
 We set $$d:=\frac{ab}{\operatorname{lcm}(a,b)},\quad\quad( * )$$ and will show that \(d\) is the greatest common divisor of \(a\) and \(b\), i.e. $d=\gcd(a,b).$
 From the definition of multiplication of rational numbers it follows that we can transform the equation \(( * )\) into
\[\frac{a}{d}=\frac{\operatorname{lcm}(a,b)}{b},~\frac{b}{d}=\frac{\operatorname{lcm}(a,b)}{a}.\]
 Since \(a\) and \(b\) divide \(\operatorname{lcm}(a,b)\) , the fractions \(\frac{a}{d}\) and \(\frac{b}{d}\) are, in fact, integers.
 Therefore, \(d\) is a common divisor of \(a\) and \(b\).
 It remains to be shown that $d$ is the greatest common divisor of \(a\) and \(b\).
* Let \(f\) be any common divisor of \(a\) and \(b\).
* Because $a\mid a\frac bf$ and $b\mid b\frac af,$ the number \(\frac{ab}f\) is a common multiple of \(a\) and \(b\).
* It follows from the proposition about the least common multiple that \(\operatorname{lcm}(a,b)\) divides \(\frac{ab}f\).
* Since \(\operatorname{lcm}(a,b)=\frac{ab}d\) (see above), we have that
\[\frac{ab}d\frac{ab}f.\]
* This means that the rational number
\[\frac{\frac{ab}f}{\frac{ab}d}=\frac df\]
is in fact an integer.
* Since \(f\) was an arbitrarily chosen common divisor of both, \(a\) and \(b\), and since it divides the common divisor \(d\), the latter number \(d\) must be the greatest common divisor.
* It follows that $d=\gcd(a,b).$
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References
Bibliography
 Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927