I want to find a nice and elegant proof of the Rational Root theorem, but I get stuck. I read some stuff on the Internet, but I have not found a complete proof of the theorem. Here's my try:

Say we have a polynomial:

[tex] F(x) = \sum ^{n}_{r = 0} a_{n}x^{n} = a_{n}x^{n} + a_{n - 1}x^{n-1} + ... + a_{1}x + a_{0} [/tex]

where all of the coefficients are integers and [tex] a_{0} [/tex] does not equal zero.

I want to prove that there is a factor of [tex]a_{n}[/tex], call it q, and a factor of [tex]a_{0}[/tex], call it p, so that:

[tex] F(\frac{p}{q}) = 0 [/tex]

p and q are prime numbers. So I want to prove that p is a factor of [tex]a_{0}[/tex] and q is a factor of [tex]a_{n}[/tex]. Next, I insert p/q into F(x):

[tex] F(\frac{p}{q}) = a_{n}\left(\frac{p}{q}\right)^{n} + a_{n-1}\left(\frac{p}{q}\right)^{n-1} + ... + a_{1}\left(\frac{p}{q}\right) + a_{0} = 0 [/tex]

multiplying by [tex] q^{n} [/tex] we get:

[tex] a_{n}p^{n} + a_{n-1}p^{n-1}q + ... + a_{1}p q^{n-1} + a_{0}q^{n} = 0 [/tex]

and factorizing out p:

[tex] p (a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}) = a_{0} q^{n} [/tex]

As p, q are prime numbers, the Greatest Common Devisor of p and q is 1, i.e. GCD(p, q) = 1. And hence, p devides [tex]a_{0}[/tex] and is then a factor of [tex]a_{0}[/tex].

As the same kind of argument applies to q being a factor of [tex]a_{n}[/tex], I thought that this proved the theorem (p/q are roots of F(x)), but no... The problem is in:

[tex] p (a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}) = a_{0} q^{n} [/tex]

how do I prove that: [tex] a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1} [/tex] is an integer?

Say we have a polynomial:

[tex] F(x) = \sum ^{n}_{r = 0} a_{n}x^{n} = a_{n}x^{n} + a_{n - 1}x^{n-1} + ... + a_{1}x + a_{0} [/tex]

where all of the coefficients are integers and [tex] a_{0} [/tex] does not equal zero.

I want to prove that there is a factor of [tex]a_{n}[/tex], call it q, and a factor of [tex]a_{0}[/tex], call it p, so that:

[tex] F(\frac{p}{q}) = 0 [/tex]

p and q are prime numbers. So I want to prove that p is a factor of [tex]a_{0}[/tex] and q is a factor of [tex]a_{n}[/tex]. Next, I insert p/q into F(x):

[tex] F(\frac{p}{q}) = a_{n}\left(\frac{p}{q}\right)^{n} + a_{n-1}\left(\frac{p}{q}\right)^{n-1} + ... + a_{1}\left(\frac{p}{q}\right) + a_{0} = 0 [/tex]

multiplying by [tex] q^{n} [/tex] we get:

[tex] a_{n}p^{n} + a_{n-1}p^{n-1}q + ... + a_{1}p q^{n-1} + a_{0}q^{n} = 0 [/tex]

and factorizing out p:

[tex] p (a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}) = a_{0} q^{n} [/tex]

As p, q are prime numbers, the Greatest Common Devisor of p and q is 1, i.e. GCD(p, q) = 1. And hence, p devides [tex]a_{0}[/tex] and is then a factor of [tex]a_{0}[/tex].

As the same kind of argument applies to q being a factor of [tex]a_{n}[/tex], I thought that this proved the theorem (p/q are roots of F(x)), but no... The problem is in:

[tex] p (a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}) = a_{0} q^{n} [/tex]

how do I prove that: [tex] a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1} [/tex] is an integer?

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