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I am looking to prove the converse of the Divisibility of Integer Combination.

I know how to prove the contrapositive of this statement but not the converse ... any help?

Below is my attempt at the solution

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  • $\begingroup$ Someone named this statement? Weird. $\endgroup$
    – rschwieb
    Oct 5 '20 at 13:05
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Hint: In particular 'a' would divide bx+cy for x=0 y=1

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  • $\begingroup$ Thanks ... but for the actual proof, is it valid for me to only showcase 'one' instance of x and y in which the statement is true What I mean to say is that the proof involves a universal quantifier, so is the strategy of using a fixed value for x and y valid? $\endgroup$
    – Guest
    May 28 '14 at 3:53
  • $\begingroup$ Yes, it is enough to show a single case. $\endgroup$ May 28 '14 at 3:56
  • $\begingroup$ Exactly my point, I know from class that with an existential quantifier showing one case is enough, but for a universal case should I not show that for all x and y the statement holds? Extremely sorry for being a total noob in this $\endgroup$
    – Guest
    May 28 '14 at 3:57
  • $\begingroup$ You need to show that a|b. There is no variable 'x' involved. I am not able to understand your doubt! It is given that a|(bx+cy) for all x,y$\in Z$. You need to show a|b, where is the problem? $\endgroup$ May 28 '14 at 4:03
  • $\begingroup$ oh, now I understand. Sorry for making you explain it twice, just had to think it over. Thank you for the help $\endgroup$
    – Guest
    May 28 '14 at 4:04

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