Prove that 11, 12 and 13 can never be three terms (not necessarily consecutive) of a geometric progression, irrespective of whether the common ratio is real or complex.

not sure if this
we have
r^x=11
r^y=12
r^z=13  for some integers x,y,z
12/11=r^(y-x)
ln(12/11)=(y-x)*ln(r)
from r^z=13 we have z*ln(r)=ln(13) thus
ln(r)=ln(13)/z
thus
ln(12/11)=(y-x)*ln(13)/z
ln(12/11)/ln(13)=(y-x)/z
but the left hand side is simple the log of 12/11 to the base 13, this is an irrational number thus can not be equal to (y-x)/z for integers x,y,z.  Thus 11,12,13 can not be part of the same geometric series.

Posted by Daniel on 2009-09-19 21:47:49