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.

$$\text{Proof: We assume, } D \in \mathbb{R-Q}. \\ D=\sqrt{(b)^2-4ac} \text{ where } a, b, c \in \mathbb{Z}. \\\text{Then, } x=\frac{-b\pm D}{2a}.\\ \text{ Given that } D \text{ is irrational it follows that } x \text{ is irrational.} \\\therefore \text{ if } D \in \mathbb{R} - \mathbb{Q} \text{, then } ax^2+bx+c=0 \text{ does not have a rational solution. } \blacksquare$$


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.

  • $\begingroup$ This looks correct, but I expect it is not what was intended. Often times you can't get such a nice expression for the roots, yet you can still eliminate rational roots. Here, say, if $\frac ab\in \mathbb Q$ were a rational root we'd have $a^2+3ab+b^2=0\implies a\,|\,b$ which, assuming $\gcd (a,b)=1$ implies that $a=\pm 1$. Similarly, deduce that $b=\pm 1$. Work from there. $\endgroup$ – lulu Jul 9 '18 at 12:45
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    $\begingroup$ You might want to check out the rational root theorem too. $\endgroup$ – steven gregory Jul 9 '18 at 12:54
  • $\begingroup$ @lulu what told you to go that route? The rational root theorem as referenced by @stevengregory? I often have a hard time figuring out what is expected.. $\endgroup$ – Cro-Magnon Jul 9 '18 at 13:01
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    $\begingroup$ Aside from already being familiar with the rational root theorem, the point that strikes me is that, as a general rule, it is very difficult to get the roots of polynomials. What's interesting is that you can decide if there are rational roots or not without listing all the roots. You can do this even for high degree polynomials (where there really isn't a strategy for listing the roots in closed form). $\endgroup$ – lulu Jul 9 '18 at 13:14

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.

  • $\begingroup$ Can you explain what told you to go that route? I often have a hard time figuring out what is expected or 'on what string to pull' to make it all work. $\endgroup$ – Cro-Magnon Jul 9 '18 at 13:02
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    $\begingroup$ @Cro-Magnon I don't think that the problem is a good one. It's too vague. I just got the feeling that whover thought about it was thinking about a generalization to equations of arbitrary degree and therefore the resolvent formula shoudn't be used, since it only works for quadratic equations. $\endgroup$ – José Carlos Santos Jul 9 '18 at 13:05
  • $\begingroup$ Ok, that might be true. I will study your answer and the rational root theorem. Thank you! $\endgroup$ – Cro-Magnon Jul 9 '18 at 13:11
  • $\begingroup$ Would you mind looking at my proof over here: math.stackexchange.com/a/2919138/238010 I am unsure about its correctness and completeness. Is there a way to know a proof is correct without help of a more experienced proof writer? $\endgroup$ – Cro-Magnon Sep 17 '18 at 8:15

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