State/The GCD times the LCM is the product

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Proposition
Proof

Proposition: (The GCD times the LCM is the product) Suppose that $a,b \in \NN$, and $a,b \neq 0$. Then $$GCD(a,b) \cdot LCM(a,b) = ab.$$

Logical Connections

This statement logically relies on the following definitions and statements: State/Canonical decompositions can be used to find GCD and LCM, State/Minmax addition formula

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State/The GCD times the LCM is the product

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	Suppose that $a,b \in \NN$, and $a,b \neq 0$.
	Recall that canonical decompositions can be used to find the GCD and LCM. Let $(e_p)$ and $(f_p)$ be the exponents in the canonical decompositions of $a$ and $b$, respectively, so that: $$a = \prod_{p \in P} p^{e_p}.$$ $$b = \prod_{p \in P} p^{f_p}.$$ $$ab = \prod_{p \in P} p^{e_p + f_p}.$$ $$GCD(a,b) = \prod_{p \in P} p^{min(e_p,f_p)}.$$ $$LCM(a,b) = \prod_{p \in P} p^{max(e_p, f_p)}.$$ By the "minmax addition formula", one has $e_p + f_p = min(e_p, f_p) + max(e_p, f_p)$.
	It follows immediately that $ab = GCD(a,b) LCM(a,b)$.