Generalized Euclid's Lemma (Proof)

Question

Generalized Euclid's Lemma:

Let $a_1,a_2,a_3,...,a_n \in \mathbb{Z}$.

If $p$ is a prime number such that $p | a_1 \cdot a_2 \cdot a_3 \cdot ... \cdot a_n$,

then $\exists i \in \{1,2,3,...,n\}$ such that $p|a_i$.

According to my book, it says the proof can be done by using induction.

How would I go about writing the proof?

If $p\mid a_1\cdot\ldots\cdot a_n$, then $p\mid a_n$, or $p\mid a_1\cdot\ldots\cdot a_{n-1}$. — Rushabh Mehta, Apr 13 '20 at 18:31

score 1 · Answer 1 · answered Apr 13 '20 at 18:49

The result is obvious for $n=1$ factors: if $p|a_1$, then $p|a_1$.

The case of $n=2$ factors is Euclid’s Lemma: if $p|a_1a_2$, then $p|a_1$ or $p|a_2$.

So assume the result holds for $n=k$ factors (the Induction Hypothesis): if $p$ divides a product with $k$ factors, then it divides at least one of the factors.

You want to prove that then the result holds for $n=k+1$ factors: if $p|a_1\cdots a_{k+1}$, then $p|a_i$ for some $i\in\{1,\ldots,k+1\}$. So, try to use both the $n=2$ and the $n=k$ cases by doing some associating to reach the desired conclusion.

Generalized Euclid's Lemma (Proof)

1 Answers1