I'm having trouble wrapping my head around proofs by induction with this problem:
Prove by careful induction on $k$ that, if $p$ is a prime number dividing the product $m_1m_2…m_k$ of (ordinary) integers $m_1,…,m_k$ where $k\geqslant1$, then $p$ divides at least one of the integers $m_i, i =1,…,k.$
Trying again based on feedback
So I start with the base case, that is:
$k = 1$
If $p$ divides $m_1m_2…m_k$, that gives us $1/p$, meaning $p|1$ which supports the hypothesis.
So the base case holds.
Now for the Inductive step, we have $k +1$ which would equate to $2$
Then we have $m_1m_2…m_k = 1*2$ and then $1*2/p$.
Since $p∣ab \implies p∣a$ or $p∣b$, we know that $p|1$ or $p|2$.
So the inductive step holds, and so does the hypothesis.
Best Answer
Base Case: m$_1$...m$_k$ is m$_1$. The hypothesis implies that p thus divides m$_1$. Thus, p divides at least one of the integers of the product m$_1$.
Recursive Step: Suppose that for at least one m$_k$, p divides m$_1$...m$_n$. By Bill Dubuque's hint, p divides m$_1$...m$_n$m$_j$ where j = (n + 1) implies that p divides m$_1$...m$_n$ or m$_j$. If it divides m$_1$...m$_n$, then by assumption for at least one m$_k$ it divides m$_1$...m$_n$. If it divides m$_j$, then for at least one m$_k$ it divides m$_j$. Thus, in either case for at least one m$_k$, p divides m$_1$...m$_n$m$_j$. Therefore, if for at least one m$_k$, p divides m$_1$...m$_n$, then for at least one m$_k$, p divides m$_1$...m$_n$m$_j$.