How do I prove that if $x$ is a root of the polynomial: $$x^m+c_1(x^{m-1})+c_2(x^{m-2})+...+c_m=0$$ then $x$ is either an integer or irrational.
What I thought:
Wouldn't it be the same as proving by supposing that $a/b$ is a root of the polynomial, and that $a$ and $b$ are relatively prime. Plugging into the equation and multiply both sides by $b^m$ then showing that $b$ must be equal to 1.
You're going in the right direction. Suppose $a/b$ is a root of the monic polynomial $f(x)=x^n+a_{n-1}x^{n-1}+\cdots +a_0$, whose coefficients are integers, and $a/b$ is written in lowest terms (i.e. $a,b$ coprime).
Then we have $0=f(a/b)=(a/b)^n+a_{n-1}(a/b)^{n-1}+\cdots +a_0$. By multiplying both sides by $b^n$, we then have $a^n+a_{n-1}ba^{n-1}+\cdots +a_0b^n=0$. By subtracting $a^n$ on both sides, we conclude that $b$ divides $a$, contradicting the assumption that $a/b$ was written in lowest terms if $b\neq \pm 1$.
The same technique can be used to prove the Rational Root Theorem, which states that if we have a polynomial $a_nx^n+\cdots +a_0$ with $a_n\neq 0$ and $a_i$ an integer for each $i$, then if $a/b$ is a rational solution of the polynomial with $a,b$ coprime, then we must have $a| a_0$ and $b| a_n$.
Here's the proof, following in style with the above proof: suppose we have the polynomial $a_nx^n+\cdots +a_0$ with $a_n\neq 0$ and each $a_i$ an integer. Then if $a/b$ is a rational solution with $a,b$ coprime, we have $a_n(a/b)^n+\cdots+a_0=0$, so by multiplying both sides by $b^n$, we have $a_na^n+ba_{n-1}a^{n-1}+\cdots +a_0b^n=0$. Subtracting $a_na^n$ on both sides shows that $a_na^n$ is divisible by $b$. But because $a,b$ are coprime, we find that $a_n$ is divisible by $b$. Likewise, by subtracting $a_0b^n$ on both sides, we see that $a_0b^n$ is divisible by $a$. Again, $a,b$ are coprime, so it must be that $a$ divides $a_0$.
The Rational Root theorem also immediately proves our original result, as with $a_n=1$, we must have $b$ divide $a_n$ and hence is $\pm 1$, and so any rational root is an integer.
