# pairwise coprime positive integers where $n \geq 2 .$ Prove $\operatorname{lcm}\left(a_{1}, a_{2}, \ldots, a_{n}\right)=a_{1} a_{2} \cdots a_{n}$

Let $$a_{1}, a_{2}, \ldots, a_{n}$$ be pairwise coprime positive integers where $$n \geq 2 .$$ Prove that

$$\operatorname{lcm}\left(a_{1}, a_{2}, \ldots, a_{n}\right)=a_{1} a_{2} \cdots a_{n}$$

My Attempt. Induction on $$n$$. If $$n=2$$, then $$\operatorname{lcm}(a_1,a_2)=a_1a_2$$. Assume holds for n, show for $$n+1$$:

$$\operatorname{lcm}(a_1,...,a_{n+1})=\operatorname{lcm}(\operatorname{lcm}(a_1,...,a_n),a_{n+1})=\operatorname{lcm}(a_1...a_n,a_{n+1})$$

If $$a_1...a_n$$ and $$a_{n+1}$$ are relatively prime, then so we are done, ıf not how should I do? Can you help? Can you add an answer as different method?

• Since $\,a_{n+1}\,$ is coprime to $\,a_{n},\ldots,a_1\,$ is it coprime to their product by Euclid's Lemma. That concludes your proof. For another method you can use unique prime factorization. Commented Feb 20, 2020 at 1:31

Assume that a prime number $$p_i$$ is a prime divisor of $$a_k$$, and $$v_{p_k}(a_k)=q$$.
Because the numbers are pairwise co-prime, therefore, $$p_i$$ is a prime divisor of $$a_k$$ and only $$a_k$$, and therefore $$v_{p_k}(LCM(a_1,a_2,...,a_m))=q$$
This implies that $$LCM(a_1,a_2,...,a_m)$$ is divisible by $$a_1 \times a_2\times ...\times a_n$$. But by definition of LCM, it is the least non-zero number to be divisible by $$a_1$$ and $$a_2$$ and $$a_3$$ and ... and $$a_n$$. Therefore, we have Q.E.D