The discussion revolves around finding all functions f: R->R that satisfy the functional equation f(x+y)=f(x)+f(y)+1 with the condition f(1)=0. Participants conclude that the function f(x) can be expressed as f(x) = x - 1, which satisfies both the functional equation and the initial condition. The conversation emphasizes the importance of continuity in proving that this is the only solution, as well as the use of rational approximations to extend the solution to irrational numbers.
PREREQUISITESMathematics students, educators, and anyone interested in functional equations and their properties, particularly in the context of real analysis and continuity.
Dick said:True. But I was really hoping you'd say limit(q_i-1)=limit(q_i)-1=c-1. That would be the punch line. So f(c)=c-1. You've got all of the parts of this. Can you try and string them together in a logical order??
iceblits said:So:
a sequence of rational numbers q_i approaches any other number. That is, lim(q_i)=c By continuity, lim f(q_i)=f(c). By the formula: f(q_i)=q_i-1. So then lim (q_i-1)=lim(q_i)-1=c-1, so, since f(c)=lim(q_i-1), f(c)=c-1
Dick said:Change the first sentence to "Given any irrational number c there is a sequence of rational numbers q_i that approaches c." From there on, I like it.
iceblits said:yes! so does this complete the proof? thank-you so much for your help by the way..you've been great..hopefully I didn't take up too much of your time and/or frustrate you too much