quasi
Posts:
10,232
Registered:
7/15/05


Re: Does polynomial P need to be an affine mapping
Posted:
May 9, 2013 3:15 PM


Timothy Murphy wrote: >quasi wrote: >>quasi wrote: >>>steinerartur wrote: >>>> >>>>Let P be a polynomial with real coefficients. Suppose >>>>there are nonempty intervals I and J such that P maps >>>>surjectively the rationals of I into the rationals of J. >>>>Does this imply P is an affine mapping? >>> >>>Yes. > >My argument is similar to yours, but possibly simpler. > >As has been established, P(x) has rational coefficients. > >Choose a prime p which does not divide any of the >coefficients.
It suffices for p to not divide the leading coefficient.
>Consider x = n/p^e where n is not divisible by p.
Where in this context, n is allowed to rational.
>Then it is easy to see that P(x) = m/p^{re}, where >p does not divide m, and r is the degree of P.
Yes, where m too is allowed to be rational.
>It follows that if x is rational, and r > 1, P(x) cannot be >of the form c/p (with c not divisible by p).
Nice, and definitely simpler.
quasi

