$lcm(a_{1},...,a_{n})=lcm(lcm(a_{1},...,a_{n-1}),a_{n})$

Question

I tried to prove this by complete induction on $n$ but I am having problems in the inductive step:

Suppose

$$lcm(a_{1},...,a_{n})=lcm(lcm(a_{1},...,a_{n-1}),a_{n}) \forall k\le n\in \mathbb N$$

Prove that

$$lcm(a_{1},...,a_{n+1})=lcm(lcm(a_{1},...,a_{n}),a_{n+1})$$

Now, we have that

$$lcm(lcm(a_{1},...,a_{n}),a_{n+1})=lcm(lcm((lcm(a_{1},...,a_{n-1}),a_{n}),a_{n+1})$$

but then I don´t know what else to do, do you have any suggestions?

Answer 1

Hint $\,\ a_1,\ldots,a_n\mid m\iff a_1,\ldots, a_{n-1}\mid m\, $ & $\ a_n\mid m,\,$ i.e. if you unwind the definition of lcm then its associativity boils down to the associativity of $ $ & $ $ (logical and).