Prove Corollary (2.24).
Hint: You may find the result of Exercises 1.3, question 15 (ii) useful:
$$\gcd (a, n_1) = \gcd (a, n_2) = \ldots = \gcd (a, n_k) = 1 \implies \gcd (a, n_1 × n_2 \ldots × n_k) = 1$$
Corollary (2.24) is
Let $a_1, a_2, a_3, \ldots , a_n$ be pairwise relatively prime integers then $[a_1, a_2, a_3, \ldots , a_n] = a_1 × a_2 × \ldots × a_n$
We will prove the statement by using induction on $n$.
Let the statement $P(n)$ be
Let $a_1, a_2, a_3, \ldots , a_n$ be pairwise relatively prime integers then $[a_1, a_2, a_3, \ldots , a_n] = a_1 × a_2 × \ldots × a_n$
We need to show the base case $P(2)$, and the inductive step $P(n) \implies P(n+1)$ is true.
Base Case
The base case $P(1)$ is
Let $a_1, a_2$ be pairwise relatively prime integers then $[a_1, a_2] = a_1 × a_2$
Since $a_1$ and $a_2$ share no common factors, the LCM $[a_1, a_2] = a_1 \times a_2$. This is also Proposition (2.20).
So the base case is true.
Inductive Step
We assume the induction hypothesis $P(n)$ is true
Let $a_1, a_2, a_3, \ldots , a_n$ be pairwise relatively prime integers then $[a_1, a_2, a_3, \ldots , a_n] = a_1 × a_2 × \ldots × a_n$.
We assume $a_1, a_2, a_3, \ldots , a_n, a_{n+1}$ are pairwise relatively prime integers and aim to show $[a_1, a_2, a_3, \ldots , a_n, a_n+1] = a_1 × a_2 × \ldots × a_n \times a_{n+1}$
We use Proposition (2.23)
Let $a_1, a_2, a_3, \ldots , a_n$ be non-zero integers, then $[a_1, a_2, a_3, \ldots , a_{n−1}, a_n] = [[a_1, a_2, a_3, \ldots , a_{n−1}] , a_n]$.
This tells us that if $a_1, a_2, a_3, \ldots , a_n, a_{n+1}$ are pairwise relatively prime integers, then
$$ \begin{align} [a_1, a_2, a_3, \ldots , a_n, a_{n+1}] &= [[a_1, a_2, a_3, \ldots , a_n] , a_{n+1}] \\ \\ & = [ a_1 \times a_2 \times \ldots a_n, a_{n+1}] \quad \text{by induction hypothesis} \\ \\ & = a_1 \times a_2 \times \ldots a_n \times a_{n+1} \quad \text{by Proposition 2.20 and given hint}\end{align}$$
The hint tells us that the $\gcd(a_1 \times a_2 \times \ldots a_n, a_{n+1})=1$ and so $(a_1 \times a_2 \times \ldots a_n)$ and $(a_{n+1})$ are both co-prime
Proposition (2.20) tells us that since $(a_1 \times a_2 \times \ldots a_n)$ and $(a_{n+1})$ are both co-prime, their LCM is their product.
We have shown both the base case and inductive step are true, and so by induction, Corollary 2.24 is true.