# Number Theory Proof

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• Nov 12th 2008, 05:18 PM
Saint22
Number Theory Proof
Could anyone assist me with this problem...

"Use mathematical induction to prove the following generalization.
Suppose $\displaystyle a_1, a_2, ..., a_n$ are integers and p is a prime number. If $\displaystyle p|a_1 a_2 ... a_n$, then $\displaystyle p|a_i$ for some $\displaystyle i = 1, 2, ..., n$." [Hint: The induction step has two cases.]

I believe I can use this theorem without proof:
Suppose a and b are integers and p is a prime number. if p|ab, then p|a or p|b. This theorem comes from the uniqueness of prime factorization section.
• Nov 12th 2008, 07:07 PM
o_O
Let $\displaystyle P_n$ be the statement that if $\displaystyle p \mid a_1 a_2 \cdots a_n$, then $\displaystyle p \mid a_i$ for some $\displaystyle i \in [1, n]$.

Base case: $\displaystyle P_1$ is a trivial statement. $\displaystyle P_2$ is the theorem you have.

Inductive Step:
Assume $\displaystyle P_k$ to be true, that is, if $\displaystyle p \mid {\color{red}a_1 a_2 \cdots a_k}$ then $\displaystyle p \mid a_i$ for some $\displaystyle i \in [1, k]$. It remains to show that $\displaystyle P_{k+1}$ also holds, that is, if $\displaystyle p \mid a_1 a_2 \cdots a_k a_{k+1}$, then $\displaystyle p \mid a_j$ for some $\displaystyle j \in [1, k+1]$.

So if we have that $\displaystyle p \mid ({\color{red}a_1a_2 \cdots a_k}){\color{blue}a_{k+1}}$, by your theorem, we have 2 cases: (1) $\displaystyle p \mid {\color{red}a_1a_2 \cdots a_k}$ or (2) $\displaystyle p \mid {\color{blue}a_{k+1}}$.

(1) By the inductive hypothesis, we already have some $\displaystyle j$ such that $\displaystyle p \mid a_j$, namely $\displaystyle j = i$.

(2) Then let $\displaystyle j = k+1$ and we're done.