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 + ... + a_n-1x^n-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 + ... + a_n-1x^n- + x^n-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 + a₁rs^n-1 + a₂r²s^n-2 + ... + a_n-1r^n-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.

Application of Rational Root Theorem

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. 

Internet References

Related pages in this website

Synthetic Division

Polynomial Division

Polynomial Remainder Theorem

Descartes' Rule of Signs

Formal Power Series

Generating Function

The webmaster and author of the Math Help site is Graeme McRae.
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