Proof of $LCM(a,b)=\prod_{i=1}^{\infty}p_i^{\max(\alpha_i,\beta_i)}$

arithmeticleast-common-multiplenumber theory

For $a,b\in\mathbb{N}$ with prime factorization $a=\prod_{i=1}^{\infty}p_i^{\alpha_i}, b=\prod_{i=1}^{\infty}p_i^{\beta_i}$ with $\alpha_i,\beta_i\in\mathbb{N}_0$ prove: $$LCM(a,b)=\prod_{i=1}^{\infty}p_i^{\max(\alpha_i,\beta_i)}$$

I know, there is another post about this question (Prove that $lcm(a , b) = \prod_{i=1} (P_i)^{\max(\alpha_i,\beta_i)}$) but I'm wondering if there is a more detailed way to prove that the LCM of two natural numbers can be factorized in primes by their highest exponents. I don't know how to start the proof correctly.

Thanks a lot!

Best Answer

I don't think you can have a "more detailed" proof of the linked one:

Obviously, don't consider the primes with exponent $0$ in the $\prod_{i=1}^{\infty}$.

$\prod_{i=1}^{\infty} p_i^{\max(\alpha_i,\beta_i)}$ is a common multiplier.

Suppose that $LCM(a,b) = \prod_{i=1}^{\infty} p_i^{e_i}$ with $e_j < \max(\alpha_j, \beta_j) \exists j$. WLOG $e_j < \alpha_j$, then $LCM(a,b)$ is not a multiplier of $a$, contradiction.

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