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$a_i$ positive integers for $1\le i\le n$

if $p$ prime and $p\mid a_1a_2\cdots a_n$ then $p\mid a_i$ for some $1\le i\le n$:

My thinking is to prove it by contraposition.

$p$ does not divide $a_i$ for all $1\le i\le n$ implies $p$ does not divide $a_1a_2\cdots a_n$:

If $p$ does not divide $a_i$ for all $1\le i\le n$, $p$ being prime it cannot equal the product of any two or more therefore it cannot divide $a_1a_2\cdots a_n$.

Michael Hardy
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miniparser
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    I'd prove it first for $n=2$ and use induction to show it's true for larger values of $n$. ${}\qquad{}$ – Michael Hardy Aug 10 '14 at 21:13
  • You mean use n=2 as the base case. This doesn't strike me as a case where you would need to use induction. – miniparser Aug 11 '14 at 20:00
  • The proposition is vacuously true when $n=1$. All the work is in the case $n=2$. For the $(n+1)$th case, just say if $p\mid (a_1\cdots a_n)a_{n+1}$ then you have two numbers, $a_1\cdots a_n)$ and $a_{n+1}$, and the case $n=2$ imples $p$ divides one of those. Then the induction hypothesis says that if $p$ divides $a_1\cdots a_n$ then $p$ divides one of the factors. This is one of many proofs by induction in which (1) the case $n=1$ is vacuously true and (2) the induction step is trivial and (3) the case $n=2$ is used in the iduction step and (4) the case $n=2$ is the hard part. – Michael Hardy Aug 11 '14 at 21:34

2 Answers2

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The case $n=2$ is where all the work is; the rest is a trivial mathematical induction. You need to prove that if $p$ is prime and $p\mid ab$ then $p\mid a$ or $p\mid b$. That statement is called Euclid's lemma. If you google that term, you should find something on it.

Notice that the assumption that $p$ is prime cannot be dropped: for example, $6\mid 4\cdot3$ but $6\nmid4$ and $6\nmid3$.

Look at this answer that I wrote almost three years ago.

It tells you that if $p\nmid a$ then $a$ has a multiplicative inverse mod $p$, i.e. a number $c$ such that $ac\equiv 1\bmod p$. That means $ac=kp+1$ for some $k$. It follows that $abc= kpb+b$. But if $p\mid ab$ then $ab=\ell p$ for some $\ell$, so we have $\ell p c = kpb+b$. This implies $b = \ell pc-kpb=p(\ell c - kb)$, so $p$ divides $b$.

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Michael Hardy
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  • Induction on Euclid's lemma. That figures. I'll have to get my head around the multiplicative inverse. – miniparser Aug 12 '14 at 17:07
  • @user2008730 : The easiest way to think about this multiplicative inverse and related questions is by modular arithmetic with $p$ as the modulus. I think that's the reason Gauss introduced modular arithmetic in the first place. – Michael Hardy Aug 12 '14 at 21:17
  • 1 mod anything is 1, I believe, except for 1 itself, then it is 0. so if p is 1 c is 0 and if not c is 1/a. but if p is 1 it divides everything,, except 0. think we're dealing with positive ints here which don't include 0. so c is 1/a. so k would have to be 0? – miniparser Aug 18 '14 at 19:03
  • @user2008730 : If you construe $a\bmod b$ as the remainder on division of $a$ by $b$, then $n\bmod 1=0$ if $n\in\mathbb Z$. ${}\qquad{}$ – Michael Hardy Aug 18 '14 at 20:09
  • @user2008730 : I didn't have in mind the possibility that $p=1$, since that's not conventionally included among the prime numbers, and for a good reason: it behaves essentially differently (e.g., imagine factoring the number $3=3\times1\times1\times1\times1\times\cdots$). But in $\mathbb Z\bmod 1$, the multiplicative inverse of $1$ is $1$, and of course $0$ has no such inverse. ${}\qquad{}$ – Michael Hardy Aug 18 '14 at 20:15
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$a_i$ positive integers for $1\le i\le n$

if $p$ prime and $p∣a_1a_2⋯a_n$ then $p∣a_i$ for some $1\le i\le n$:

$n=1$: (vacuously true) $p$ prime and $p|a_1$ -> $p|a_1$

base case:

$n=2$: True by Euclid's Lemma. $p|a_1a_2$ -> $p|a_1$ or $p|a_2$. If $p|a_1$ finished. If not , $p$ and $a_1$ relatively prime so exist $x,y$ so $px + a_1y = 1$. Multiply by $a_2$ to get $pa_2x + a_1a_2y = a_2$. If $p$ divides $pa_2$ and $a_1a_2$ then $p$ divides $a_2$.

$p(n) = p$ prime and $p∣a_1a_2⋯a_n\Rightarrow p∣a_i$ for some $1\le i\le n$

Assume true for n.

$p(n)$ true.

$p(n)\Rightarrow p(n+1)$: $p|(a_1...a_n)a_{n+1}$: If $p$ does not divide $a_1...a_n$, $p$ must divide $a_{n+1}$

miniparser
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