To prove this theorem, each of the properties will be considered individually. To start with the first property, consider a polynomial
P(x) with integer coefficients that has an integer root
xr. This means that
P(xr)=0.
anxrn+an−1xrn−1+⋯+a1xr+a0=0
Now
a0 is subtracted from both sides of the .
anxrn+an−1xrn−1+⋯+a1xr+a0=0
anxrn+an−1xrn−1+⋯+a1xr=-a0
xr(anxrn−1+an−1xrn−2+⋯+a1)=-a0
Since the coefficients of
P(x) are integers and
xr is an integer, the between the parentheses results in an integer. Also,
-a0 is the of an integer and
xr, which means that
xr is a factor of
a0. This completes the proof of the first property.
For the second property, suppose that the polynomial P(x) has a rational root qp, such that the fraction is written in its simplest form. Again, it is obtained that P(qp)=0.
This time the equation will be modified in a different way.
an(qp)n+an−1(qp)n−1+⋯+a1(qp)+a0=0
an(qnpn)+an−1(qn−1pn−1)+⋯+a1(qp)+a0=0
an(pn)+an−1(pn−1q)+⋯+a1(pqn−1)+a0(qn)=0
an(pn)+an−1(pn−1q)+⋯+a1(pqn−1)=-a0(qn)
p(anpn−1+an−1pn−2q+⋯+a1qn−1)=-a0qn
Since the coefficients of
P(x), p, and
q are all integers, the expression between parentheses results in an integer. Also, since
qp is written in its simplest form,
p and
q do not have a common factor, which means that
p and
qn do not have a common factor either. Therefore,
p is a factor of
a0.
The numerator of the root, p, is a factor of a0.
If the modification is done differently, it is possible to come to the other property.
an(pn)+an−1(pn−1q)+⋯+a1(pqn−1)+a0(qn)=0
an−1pn−1q+⋯+a1pqn−1+a0qn=-anpn
q(an−1pn−1+⋯+a1pqn−2+a0qn−1)=-anpn
With the same reasoning as before, it is possible to conclude that
q, the denominator of the root, is a factor of
an, finishing the proof.