Verifying proof for the following gcd question

Asked Apr 28 '19 at 21:42

Active Apr 28 '19 at 23:04

Viewed 58 times

Prove that for all nonzero integers $a$ and $b$, if $\gcd(|a|,|b|)=1$ then for all integers $c$, if $a\mid bc$ then $a\mid c$.

Given that $\gcd(|a|,|b|)=1$ then there exists $m,n\in Z$ such that $am+bn=1$ which implies $acm+bcn=c$.

Given $a|bc$, to show $a\mid c$. Now, $a\mid acm$ and $a\mid bcm$ as $a\mid bc$ implies $a\mid acm +bcn = c$. Therefore $a\mid c$.

Is my proof sufficient? Thank you!

edited Apr 28 '19 at 23:04

N. F. Taussig

asked Apr 28 '19 at 21:42

bengraham12

4

Your proof is fine. – Apr 28 '19 at 21:45
thank you @ChrisCuster – bengraham12 Apr 28 '19 at 21:53
That's the common Bezout-based proof of Euclid's Lemma. See here for a few other forms. – Bill Dubuque Apr 28 '19 at 21:54
You should also be able to do it with the fundamental theorem of arithmetic. – Apr 28 '19 at 22:15
1

@ChrisCuster Yes, unless it's on the way to proving the fundamental theorem of arithmetic. – Ethan Bolker Apr 28 '19 at 23:06
2

@EthanBolker Indeed. See Pierre Samuel's interesting pedagogical article cited here on the many possible paths we can take to prove the Fundamental Theorem of Arithmetic in an elementary course. See also this answer and its links for various "direct" proofs of Euclid's Lemma using only Division (with Remainder), i.e. in arithmetical "assembly language". – Bill Dubuque Apr 28 '19 at 23:59

0 Answers0