science0229
  • science0229
Number Theory Help!
Mathematics
chestercat
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science0229
  • science0229
For 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
  • science0229
I 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
  • science0229
Since \[\frac{ a }{ b }+\frac{ c }{ d }=\frac{ ad+bc }{ bd }\], I tried to somehow manipulate the above equation to get this form, but I've failed...

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ganeshie8
  • ganeshie8
try 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
  • ganeshie8
since \(\gcd(a,b)=1\), by euclid lemme, \(b\mid ad \implies b\mid d\) similarly \(d\mid b\) together imply \(b=d\), contradiction, ending the proof.
science0229
  • science0229
Bravo!
science0229
  • science0229
Thank you!
science0229
  • science0229
So the key to this problem was using contradiction. Does there happen to be an alternate proof?
ganeshie8
  • ganeshie8
sure, but i feel they all are messy... you don't like that divisibility argument is it ?
science0229
  • science0229
No. 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
  • ganeshie8
btw 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
  • ganeshie8
sure, lets try using bezout's identity
science0229
  • science0229
Wait. On the other thought, I do have this odd feeling that this will be dirty...
ganeshie8
  • ganeshie8
Look at robjohn's reply here http://math.stackexchange.com/questions/1151443/prove-for-positive-integers-a-b-c-and-d-where-b-does-not-equal-d-if-gcda-b
science0229
  • science0229
Ah. That's exactly what I wanted! It's not that dirty as I thought it would be.
ganeshie8
  • ganeshie8
usually euclid lemma is covered prior to bezout so it is okay to use euclid lemma as it is more fundamental or whatever..
science0229
  • science0229
got it. Thank you!

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