1

enter image description here So for example, I have 4x + 3y = 12.

GCD(4, 3) = 1.

1 divides 12..?

But I don't think this helps prove anything.. Advice?

Bill Dubuque
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knowledge_is_power
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    I think you may be misinterpreting a vertical bar. $a/b$ is a fraction. $a | b$ means "$a$ divides $b$". In your case, $1$ divides $12$. The extended Euclidean algorithm will solve your problem - look it up on wikipedia or many other places. – Ethan Bolker Nov 04 '16 at 00:51
  • @EthanBolker my teacher wrote gcd(a,b)/c. He used the fraction bar... Does this change anything? By Euclid's Theorem, ax+by=gcd(a,b) basically., so then gcd(a,b) = c?? – knowledge_is_power Nov 04 '16 at 00:54
  • Are you sure? Maybe it's his handwriting. If he really used a fraction bar he's confused, and confusing. How you deal with that is a problem. Could you ask "why is the fraction $1/12$ useful here? – Ethan Bolker Nov 04 '16 at 00:59
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    @Ethan Some authors do use a slanted bar for divides. In any case it does denotes "divides" and not a fraction. – Bill Dubuque Nov 04 '16 at 01:01
  • I added the teacher's actual pdf. Hope this helps clarify what I'm trying to solve. – knowledge_is_power Nov 04 '16 at 01:01
  • @BillDubuque is right, although I wish authors didn't do that. It means "divides" here. Now you can answer the question with the extended Euclidean algorithm. – Ethan Bolker Nov 04 '16 at 01:08
  • @Ethan Actually we don' need the algorithm, only Bezout's identity (scaled as need be) – Bill Dubuque Nov 04 '16 at 01:11
  • @BillDubuque Fair enough, though most cs students in a discrete mathematics course will probably see and be happier with the algorithm than the pure existence proof minimizing $ax+by$. – Ethan Bolker Nov 04 '16 at 01:17
  • @Ethan But that proof has a natural constructive presentation. – Bill Dubuque Nov 04 '16 at 01:24
  • There is nothing wrong here. An equation $ax+by=c$ with integer coefficients has an integer solution for x,y iff the gcd of a and b divides c

    The interpretation is: when gcd is 1 solution is guaranteed for any c , as 1 divides any c. The theorem says that even if the gcd is not 1 it will have solutions provided RHS is a multiple of that gcd.

     – P Vanchinathan Nov 04 '16 at 02:36

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