**Theorem: ** Rational roots of a monic polynomial with integer
coefficients are inetegers.

Let f(x) be a monic polynomial given by

f(x) = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3}
+ ... + a_{n-1}x^{n-1}
+ x^{n}.

If ξ is a rational root of the equation f(x)=0, then ξ is an integer.

**Proof:** If a_{0}=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) = a_{1} + a_{2}x + a_{3}x^{2} + ... +
a_{n-1}x^{n-}
+ x^{n-1},

has all the same roots as f(x) except possibly 0. If a_{1}=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 s^{n},

a_{0}s^{n} + a_{1}rs^{n-1} + a_{2}r^{2}s^{n-2}
+ ... + a_{n-1}r^{n-1}s + r^{n} = 0

For an arbitrary prime, p, if s were divisible by p, then every term except
possibly r^{n} would be divisible by p, so r^{n} 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) = x^{n} − 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 = ξ^{n}
, 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

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Graeme McRae.