is there additive and one to one function that is not linear on R?
As Rita mentioned, it really depends on whether you accept the axiom of choice. If you don't accept the axiom of choice, then the answer is who knows?. If you do accept the axiom of choice, then here's an example: Extend {1} to a Hamel basis of R over Q. Then define f(1) 1/2 and, if v is any other element in the Hamel basis, define f(v) v. Extend this linearly over Q to define an additive function f : R - R. To see this function is not linear, suppose it is. Then for any v in the Hamel basis (other than 1), we have v f(v) v*f(1) v*(1/2) but this would imply v0, which cannot be part of any basis (hence a contradiction). Further, this function is one-to-one. To see this, suppose x is a real number such that f(x) 0. Write x q_0 + q_1*v_1 + . + q_n*v_n where q_0, ., q_n are rationals and v_1, ., v_n are elements of the Hamel basis (other than 1). Then the fact that f(x)0 and f is linear over Q implies 0 (1/2 * q_0) * 1 + q_1*v_1 + . + q_n*v_n and then the fact that {1, v_1, ., v_n} is linearly independent over Q implies that q_0 q_1 . q_n 0, so that x0.
The light is warning you that there is a problem with your brakes, it won't cause any problems but should dealt with as soon as possible. Good Luck
if you pump the brakes that will shut off abs until you restart the car. that will not affect anything else and in fact if you know how to drive a car without abs you will probably never notice that there is a problem
If you believe in Hamel basis the answer is yes, I think. However you should read See Theorem 5, p5 of thales.doa.fmph.uniba.sk/sleziak/. and also www.math.rutgers.edu/useminar/ca. and maybe the 1-1 is more restrictive. So, I have to say I'm not really sure.
no - just get a little piece of black tape and stick it over the light. Cheers!
