Proof Without Words for $GCD(a,b) \cdot LCM(a,b)=ab$

Question

Is there any proof without words for the identity $GCD (a,b) \cdot LCM(a,b)=ab$ ?

Answer 1

$$ GCD(a,b) = h \rightarrow (a = hr_a) \land (b = hr_b)\land((r_a,h)\times(r_b,h)\times(r_a,r_b)=1) \\ (((((\forall x((x \mid a) \lor (x \mid b) \rightarrow (x \mid g)) \rightarrow ((h \mid g) \land (r_a \mid g) \land (r_b \mid g))) \rightarrow (hr_ar_b \mid g)) \land ((a \mid hr_ar_b) \land b \mid hr_ar_b))) \rightarrow (LCM(a,b)=hr_ar_b)) \rightarrow (GCD(a,b)LCM(a,b)=h\times hr_ar_b =hr_ahr_b=ab) $$ excuse the (informal) symbolism. i have not yet learned to depict the 2-dimensional lattice which would give a wordless explanation for numbers of the type $p^nq^m$, and serve as a basis for imagining the more general picture