floor on polynomial Home › Forums › Math Olympiad, I.S.I., C.M.I. Entrance › floor on polynomial Tagged: calculus, floor function, greatest integer function, intermediate value theorem, limit, Polynomial This topic contains 1 reply, has 2 voices, and was last updated by Nitin Prasad 3 months, 3 weeks ago. Viewing 2 posts - 1 through 2 (of 2 total) Author Posts June 19, 2019 at 1:38 am #29332 Alpha BetaParticipant Let P,Q be non constant polynomials with real coefficients such that [P(x)]=[Q(x)] for all real x. ( [ ] denotes floor function). Show that P(x)=Q(x) for all real x. June 23, 2019 at 10:29 pm #29364 Nitin PrasadParticipant Let R(x)=P(x)-Q(x) Now we have $$[Q(x)]+[R(x)]\leq[P(x)]=[Q(x)+R(x)]\leq[Q(x)]+[R(x)]+1$$ Hence we have $$O\leq[R(x)]\leq1$$. Now every non constant polynomial is unbounded. Hence degree of R(x) cannot be greater than 0. Therefore degree of R(x) is zero and hence is a constant. Now if P(x) is a nonconstant polynomial then there exist m such that P(m) is an integer and hence it can be shown that R(x)=O. If degree of P(x) is 0 then the result follows trivially This reply was modified 3 months, 3 weeks ago by Nitin Prasad. This reply was modified 3 months, 3 weeks ago by Nitin Prasad. Author Posts Viewing 2 posts - 1 through 2 (of 2 total) You must be logged in to reply to this topic.