I would like to know whether I have correctly proved the following statement and have correctly extrapolated out a general situation.
We're asked two things:
a) Prove there is no rational number solution to $x^2-3x+1=0$
b) The problem (a) suggests a more general problem. State and outline a proof of this.
a)
Proof: We assume, to the contrary, that there is some rational number solution $x=\frac{a}{b}$, $(a, b\in \mathbb{Z}, \frac{a}{b}\in\mathbb{Q})$ to $x^2-3x+1=0$.
$$\text{Using the quadratic formula we solve for x.} \\D=\sqrt{(-3)^2-4\times1\times1}=\sqrt{9-4}=\sqrt{5}\\x=\frac{3\pm\sqrt{5}}{2}\\\text{Then, according to our assumption, } x=\frac{a}{b} \\\text{Since } \frac{3\pm\sqrt{5}}{2} \notin\mathbb{Q} \text{ and } x\in\mathbb{Q} \text{, it follows that } \frac{a}{b}=\frac{3\pm\sqrt{5}}{2} \text{ is a contradiction. } \blacksquare$$
b) If $D\in \mathbb{R}-\mathbb{Q}$ , then the quadratic equation $ax^2+bx+c=0$ does not have a rational solution. We proceed by a direct proof.
$$\text{Proof: We assume, } D \in \mathbb{R-Q}. \\ D=\sqrt{(b)^2-4ac} \text{ where } a, b, c \in \mathbb{Z}. \\\text{Then, } x=\frac{-b\pm D}{2a}.\\ \text{ Given that } D \text{ is irrational it follows that } x \text{ is irrational.} \\\therefore \text{ if } D \in \mathbb{R} – \mathbb{Q} \text{, then } ax^2+bx+c=0 \text{ does not have a rational solution. } \blacksquare$$
P.S.
This is not homework. Answers are in the back of my book. I am actually trying to improve on my proofs and become more logical.
Best Answer
What you wrote is correct, but I doubt that that's what the person who suggested the problem had in mind.
Let $a,b\in\mathbb Z$, with $b\neq0$, and suppose that $\frac ab$ is a solution of your equation. It is clear that $a\neq0$ and you can assume without loss of generality that $\gcd(a,b)=1$. On the other hand,$$\left(\frac ab\right)^2-3\frac ab+1=0\iff a^2-3ab+b^2=0.$$But it follows from this last equality that $a\mid b$ and that $b\mid a$. But, since $\gcd(a,b)=1$, this can only happen if both $a$ and $b$ are $\pm1$. Therefore, $\frac ab=\pm 1$. But none of these numbers is a root of the equation.
A generalization would be: if $p(x)$ is a polynomial with degree $n$ and integer coefficients such that both the constant term and the coefficient of $x^n$ are $\pm1$ and if $\pm1$ are not roots of $p(x)$, then $p(x) $ has no rational roots.