Theorem: Rational roots of a monic polynomial with integer coefficients are inetegers.
Let f(x) be a monic polynomial given by
f(x) = a0 + a1x + a2x2 + a3x3 + ... + an-1xn-1 + xn.
If ξ is a rational root of the equation f(x)=0, then ξ is an integer.
Proof: If a0=0, then 0 is a root of f, which satisfies this theorem. In that case, f(x) = x·g(x), where g(x), given by
g(x) = a1 + a2x + a3x2 + ... + an-1xn- + xn-1,
has all the same roots as f(x) except possibly 0. If a1=0, then the process can be repeated, stripping off zero-roots, until we find a constant term not equal to zero.
Now we assume a0 ≠ 0. The root, ξ, is rational, so we can let ξ=r/s, where GCD(r,s)=1. Substituting this root in place of x, and multiplying through by sn,
a0sn + a1rsn-1 + a2r2sn-2 + ... + an-1rn-1s + rn = 0
For an arbitrary prime, p, if s were divisible by p, then every term except possibly rn would be divisible by p, so rn would have to be divisible by p as well. Hence p|r, contradicting GCD(r,s)=1, and so the theorem is proved.
Theorem: If n and N are natural numbers, and if N is not the nth power of a natural number, then the nth root of N is irrational.
Proof: The nth root of N is a root of the polynomial,
f(x) = xn − N
Suppose for contradiction that ξ is a rational root of f. The rational root theorem tells us that ξ is a an integer, which means N = ξn , a contradiction.
Polynomial Remainder Theorem
Descartes' Rule of Signs
Formal Power Series
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