Which positive real number x has the property that x, |x| , and x-|x| form a geometric progression (in that order)?

waffles Mar 15, 2018

#1**0 **# if x=>0:

# if x<0:

x=0 is the only solution.

proof- |x|=q*x, x-|x|=q^{2}*|x|.

|x|=x. therefore x=x*q, x-|x|=x-x=0=q^{2}*x. one solution to this equation is x=0. if x!=0 then we get q^{2}=0 (by dividing the equation 0=x*q^{2} by x) meaning that q=0. but from that we can derive that x=0 (because of the first equation). that is a contradiction, meaning that if x=>0 then x=0.

|x|=-x meaning that q=-1. x-|x|=x-(-x)=2x=q^{2}*|x|=(-1)^{2}*(-x)=-x. this means that 3x=0 but then x=0 and that is a contradiction to x<0.

therefore, the only real number that satisfies the conditions of your question is x=0.

Guest Apr 17, 2018