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science0229
 one year ago
Number Theory Help!
science0229
 one year ago
Number Theory Help!

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science0229
 one year ago
Best ResponseYou've already chosen the best response.1For positive integers a,b,c,d, if gcd(a,b)=gcd(c,d)=1, then prove that \[\frac{ a }{ b }+\frac{ c }{ d }\notin Z\] \[b \neq d\]

science0229
 one year ago
Best ResponseYou've already chosen the best response.1I tried to start this problem by saying that since gcd(a,b)=gcd(c,d)=1, there exist x,y,u,w such that \[ax+by=cu+dv=1\]

science0229
 one year ago
Best ResponseYou've already chosen the best response.1Since \[\frac{ a }{ b }+\frac{ c }{ d }=\frac{ ad+bc }{ bd }\], I tried to somehow manipulate the above equation to get this form, but I've failed...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2try by contradiction suppose \(\dfrac{a}{b}+\dfrac{c}{d}=n \implies ad+bc= b*d*n\) \(\implies b\mid ad\) and \(d\mid bc\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2since \(\gcd(a,b)=1\), by euclid lemme, \(b\mid ad \implies b\mid d\) similarly \(d\mid b\) together imply \(b=d\), contradiction, ending the proof.

science0229
 one year ago
Best ResponseYou've already chosen the best response.1So the key to this problem was using contradiction. Does there happen to be an alternate proof?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2sure, but i feel they all are messy... you don't like that divisibility argument is it ?

science0229
 one year ago
Best ResponseYou've already chosen the best response.1No. I do like that elegant solution that you provided using contradiction. It's just that I have a feeling that there might be an alternate solution using the Bezout's Identity...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2btw the key thing is not contradiction, i think the key thing is noticing that \[m\mid n ~~\text{and}~~n\mid m~~\implies n=\pm m\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2sure, lets try using bezout's identity

science0229
 one year ago
Best ResponseYou've already chosen the best response.1Wait. On the other thought, I do have this odd feeling that this will be dirty...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Look at robjohn's reply here http://math.stackexchange.com/questions/1151443/proveforpositiveintegersabcanddwherebdoesnotequaldifgcdab

science0229
 one year ago
Best ResponseYou've already chosen the best response.1Ah. That's exactly what I wanted! It's not that dirty as I thought it would be.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2usually euclid lemma is covered prior to bezout so it is okay to use euclid lemma as it is more fundamental or whatever..

science0229
 one year ago
Best ResponseYou've already chosen the best response.1got it. Thank you!
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