by george.kourdin Thu May 26, 2011 9:29 am
my general approach to abs value problems is to start with "theory" and than plug in numbers to test solutions. if someone thinks of a more efficient method, please share
start with (2). it doesn't tell us anything about B and we have no way of comparing a to b so eluminate BD- > ACE only
try (1). b < -a. okay test numbers to try and figure out whether this is sufficient. it may be helpful to use a small table to organize your test cases
if a = -1 and b = -10000, then -a = 1 > 1000
if a = -1000 and b = 1, then -a = 1000 > 1
it is unclear whether abs (a) > abs (b) since we can make 2 cases where equation given in (1) holds
use both:
we know based on (2) that a<0 is negative so b <-a means that b is less than some positive number a. again test numbers.
if a = -1000, then -a= 1000 and b = 1, then 1 < 1000 and abs a > abs b
if a = -1, then -a = 1 and b = -1000, then -1000 < 1 and abs a < abs b
there are multiple cases that fit this so we can't answer with certainity whether abs a > abs b.
E