Prove that there is no rational number solution for an equation.

Question

Prove that there is no rational number solution to the equation $x^2-3x+1=0$. (Note, we do not assume that we know all the solutions of $x^2-3x+1=0$ are given by quadratic formula)

Answer 1

Suppose that $x = \frac{a}{b}$ ($a, b \in \Bbb{Z}$) is a solution to the equation. Without loss of generality, we may assume that $a$ and $b$ are coprime (so that the fraction is in lowest terms). Then, $$ \left( \frac{a}{b} \right)^2 - 3\left( \frac{a}{b} \right) + 1 = 0, $$ and by multiplying by $b^2$, $$ a^2 - 3ab + b^2 = 0. $$ Rearranging a bit, $$ b^2 = a(3b - a), $$ which shows that $a$ divides $b^2$. Since $\gcd(a, b) = 1$, Euclid's Lemma says that $a$ divides $1$. In other words, $a = \pm 1$.

Rearranging the equation a bit differently, $$ a^2 = b(3a - b), $$ so $b$ divides $a^2$, which leads to $b = \pm 1$.

So the assumption that $x \in \Bbb{Q}$ leads inexorably to the conclusion that, in lowest terms, $$ x = \pm \frac{1}{1} = \pm 1. $$ You can check that those are not roots.

Answer 2

As copper.hat mentioned in his comment, assume $x=\frac{a}{b}$ where $a$ and $b$ are relatively prime integers. Thus, $$\begin{eqnarray}\frac{a^2}{b^2}-3\left(\frac{a}{b}\right)+1&=&0\\ a^2-3ab+b^2&=&0\\ a^2-2ab+b^2&=&ab\\ (a-b)^2&=&ab \end{eqnarray}$$ But $a$ and $b$ are relatively prime. So either $a$ and $b$ are both odd or exactly one of them is even. If $a$ and $b$ are both odd, then the left hand side is even and the right hand side is odd. If exactly one of $a$ or $b$ is even, then the left hand side is odd and the right hand side is even. Either case results in a contradiction. Thus $x$ is irrational.

Answer 3

Hint $ $ If $x$ is a root of $\,f(x) = x^2\!-3x+1$ then so too is $\,1/x$ (since, by Vieta, the roots have product $=1).$ If $\,x\in\Bbb Q\,$ is in lowest terms then $x$ or $1/x$ has odd denominator, so reducing mod $2$ it yields a root of $f$ modulo $2$, contra $f$ has no roots mod $2,\,$ since $\,f(0)\equiv 1\equiv f(1).\ \ $ QED