floor on polynomial

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  • #29332

    Alpha Beta
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    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.

    #29364

    Nitin Prasad
    Participant

    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 weeks, 4 days ago by  Nitin Prasad.
    • This reply was modified 3 weeks, 4 days ago by  Nitin Prasad.
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