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How to prove a generalized Euclid lemma par induction after proving Euclid lemma?
I want to prove the generalized lemma, to prove by rearranging the product of number and use Euclid lemma as a model. A proof will be nicer if it can use induction principle.

Bill Dubuque
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Davy Giovanni
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  • Maybe you could specify more clearly what it is exactly that you would like to prove? – Dilip Sarwate Aug 31 '12 at 15:00
  • i prove already that if c is prime and divides ab then c divides a or b – Davy Giovanni Aug 31 '12 at 15:02
  • And what is the generalization that you wish to prove? – Dilip Sarwate Aug 31 '12 at 15:03
  • i want to prove that if c is prime and divides any product of number then it divides a least one of the member of the product but i want to use induction, i did without induction but its no really nice – Davy Giovanni Aug 31 '12 at 15:04
  • my idea is using induction Euclid lemma will be a starting point for n=2 my property is true – Davy Giovanni Aug 31 '12 at 15:07
  • now i have to go suppose at the rank n and show the that the property is true at the rank n+1 – Davy Giovanni Aug 31 '12 at 15:07
  • like if c divides a1.....an then its divides ai and prove that is true for a1.....an.an+1 – Davy Giovanni Aug 31 '12 at 15:09
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    normally i edit one giant sentence posts which have 47 words your sentence is such a shocking example so i think i will leave it as is everyone can then see why it is kind of hard to read we hopefully will encourage the use of punctuation that way dont you agree i may as well use at least one punctuation mark, there we go now we have an awesome comma splice to fill out my comment i voted for your post as penance for using your question as a punching bag good luck – rschwieb Aug 31 '12 at 16:24

2 Answers2

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The induction almost writes itself: it is just a matter of writing down your informal reasoning in the language of induction.

We want to prove that for every integer $n\ge 2$, if the prime $p$ divides the product $a_1a_2\cdots a_n$, then $p$ divides at least one of the $a_i$. The result has been already proved for $n=2$. It remains to do the induction step.

Suppose the result is true when $n=k$. We show the result is true when $n=k+1$.

So we are told that $p$ divides $a_1a_2\cdots a_ka_{k+1}$. Thus $p$ divides $(a_1a_2\cdots a_k)a_{k+1}$. This is a product of two numbers. Thus $p$ divides $a_1a_2\cdots a_k$ or $p$ divides $a_{k+1}$.

If $p$ divides $a_{k+1}$ we are finished. If $p$ does not divide $a_{k+1}$, then $p$ divides $a_1a_2\cdots a_k$. Then by the induction hypothesis $p$ divides one of the $a_i$ ($1\le i\le k$), and we are finished.

André Nicolas
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Hint $\ $ The inductive step is: $\rm\:p\:|\:a_{n+1}(a_n\cdots a_1)\:\Rightarrow\:p\:|\:a_{n+1}\ \ or\ \ p\:|\:a_n\cdots a_1\:$ by the prime divisor property, therefore, by induction $\rm\:p\:|\:a_n\cdots a_1\:\Rightarrow\: p\:|\:a_n\ \ or\ \ \cdots\ \ or\ \ p\:|\:a_1.$

More algebraically we could prove contrapositive: $\rm\: mod\ p\!:\ $ $\rm\:x,y\ne 0\:\Rightarrow\:xy\ne 0\:$ inductively extends to: a product of nonzero elements is nonzero. Here is a proof the two-element base case.

Remark $\ $ The proof method works quite generally: if $\rm\:xy\in S\:\Rightarrow\: x\in S\ or\ y\in S\:$ then, by induction, $\rm\:x_1\cdots x_n\in S\:\Rightarrow\: x_1\in S\,\ or\,\cdots\, or\,\ x_n\in S\ \ $ (above $\rm\:S = p\,\Bbb Z).$ If we view the product as a binary tree expression, then the inductive proof essentially lifts the property "$\in S$" up from the root to the leaves.

Bill Dubuque
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