Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $a \gt 1$ and if $a$ has the property that whenever $a|bc$ then $a|b$ or $a|c$, show that $a$ must be a prime.

Not sure how tot prove this, so I started out by checking a few values..

$2|7*4$, where $2 \nmid 7$ but $2|4$ so this $a=2=$prime checks out.

$3|7*6$, where $3 \nmid 7$ but $3|6$ so this $a=3=$prime checks out.

$5|7*15$, where $5 \nmid 7$ but $5|15$ so this $a=5=$prime checks out.

What about not a prime?

$6|4*3$, where $6 \nmid 4$ and $6 \nmid 3$ so $a$ must be a prime

Before, everyone freaks out that "proof by example, isn't a proof"... I know. This is all I could manage. I find writing proofs to be difficult.

share|improve this question
    
You just wrote what is probably the most common definition of prime element without resourcing to ideals and stuff, so: what's your definition of "prime"? –  DonAntonio Dec 7 '12 at 4:07
    
A prime is only divisible by $\pm$ itself and $\pm 1$ –  Dmitri.Mendeleev Dec 7 '12 at 4:09
add comment

2 Answers 2

up vote 4 down vote accepted

If $a$ is composite, then $a=pq$ where $1<p<a$ and $1<q<a$. But although $a$ divides $pq$, it does not divide $p$ or $q$.

share|improve this answer
add comment

Hint $\rm\,\ bc\mid bc\:$ but $\rm\:bc\nmid b,c\:$ if $\rm\:b,c>1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.