# Proof by induction with a prime number

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

$k = 1$

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.

• You should think of $p$ as a fixed prime. The induction is on the number of factors in the product $m_1\cdots m_k$. – carmichael561 Aug 25 '16 at 21:30
• $k$ and $p$ do not have anything to do with eachother. – Arthur Aug 25 '16 at 21:35
• Do you already know the theorem that prime $\,p\mid ab\,\Rightarrow\,p\mid a\,$ or $\,p\mid b?\$ That is what yields the inductive step. – Bill Dubuque Aug 25 '16 at 21:36
• Thanks for the pointers! I updated the post with another attempt using the theorem that @BillDubuque mentioned. How does it look now? – kojak Aug 25 '16 at 21:46
• "If $p$ divides $m_1m_2...m_k$, that gives us $1/p$, meaning $p|1$ which supports the hypothesis." How could a prime number $p$ divide $1$? This is false. Maybe to understand better what is even neede dto be proved fix $p=7$. Now you need to show that for whatever integers $m_1, \dots ,m_k$ if $7$ divides their product then $7$ divides one of them. You do induction by the number of $m_i$ not the size of the $m_i$. – quid Aug 25 '16 at 21:53

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$_j$.