Monday, 12 August 2013

elementary number theory - Let amidc and bmidc such that gcd(a,b)=1, Show that abmidc




Let ac and bc such that greatest common divisor (gcd) gcd(a,b)=1, Show that abc.


Answer



HINT   a,b | c  ab | ac,bc  ab | (ac,bc)=(a,b)c=c  via (a,b)=1.



NOTE   This proof works in every domain where GCDs exist, since it doesn't use Bezout's identity. Instead it uses the GCD distributive law  (ac,bc)=(a,b)c,  true in every GCD domain, viz.



LEMMA (a,b) = (ac,bc)/c   if  (ac,bc)  exists.



Proof  d | a,b  dc | ac,bc  dc | (ac,bc)  d | (ac,bc)/c




The above proofs use the universal definitions of GCD, LCM, which often serve to simplify proofs, e.g. see this proof of the GCD * LCM law.


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