This is a homework question so I'm just looking for some guidance.

Basically we are asked to write a step by step proof in the form of assume/then statements for:

$\forall x \in \mathbb{R}, ax^2 + bx + c = 0 \implies x \not \in \mathbb{Q}$

but with specific coefficients given, for which the statement is true.

I don't really know where to start. I've tried assuming the antecedent and then writing $(x = \frac{-b-\sqrt{b^2-4ac}}{2a}) \lor (x = \frac{-b+\sqrt{b^2-4ac}}{2a})$, and then doing cases, but I don't know where to define $\mathbb{Q}$ within the proof that makes sense.

I've also tried assuming $x \in \mathbb{Q}$ and trying to prove the contrapositive but after making the assumption I don't know how to get to either case.

The main problem I'm having is more of how to format it. As in:

Assume $ax^2 + bx + c = 0$, Then (x = __, or x = __ ), then ..., etc. I have no idea where to go from there to reach the conclusion that $x \not \in \mathbb{Q}$

I apologize if this is a dumb question. Thanks for your time!

  • 1
    $\begingroup$ the user is given specific coefficients that make this work tho' $\endgroup$ – mm-aops Jun 19 '14 at 20:57
  • 2
    $\begingroup$ It depends what coefficients you are given - if $a,b$ are rational and $c$ isn't then for $x$ rational you have $ax^2+bx=-c$: a rational number equal to an irrational one - contradiction, so it can't happen for rational $x$. $\endgroup$ – Mark Bennet Jun 19 '14 at 21:01

(1) The roots of a polynomials are not irrational in general. For instance, take $p(x) = x- 2$, or $p(x)=(x-3)(x-1) = x^2-4x+3$ or given $p(x)=(3x-1)(2x+1)(4x+7)$. You can probably construct a fourth degree or higher polynomial that has rational roots.

For a quadratic equation $ax^2+bx+c=0$ with integer $a,b$ and $c$, we can compute the solutions using the quadratic formula. You will end up with the expression $\sqrt{b^2-4ac}$. It is possible to show that if this can be simplified to a rational (that is integer) number, then your roots are rational. If it cannot be simplified to a rational number, your roots are irrational. (Can you show the last fact using contradiction?) So how can we show that $\sqrt{b^22 - 4ac}$ is irrational? Well, factor out what you can from the radical. Then show that the radical portion is irrational using an argument like the famous proof that $\sqrt{2}$ is irrational. That is, if $p/q=\sqrt{m}$, then $p^2=mq^2$ and so if $n$ is prime and $n|m$, then $n|p^2$ and hence $n|p$...

(2) But there is another, and perhaps better, way. The rational root theorem gives a way to find all rational roots of a polynomial. You may have learned this is highschool. The theorem says, given a polynomial with integer coefficients $$p(x) = a_n x^n + \cdots a_1 x + a_0,$$ look at the factors of $a_n$, call them $f_1, f_2, \dots, f_N$, and the factors of $a_0$, call them $g_1, g_2, \dots, g_M$. Then if $x=p/q$ is a rational root of $p(x)$, it must have the form $$x = \frac{p}{q} = \pm \frac{g_i}{f_k}$$ for some $i$ and $k$.

What you can do is generate a list of all possible rational roots and see if they actually are roots. If none of them are roots, then you know that since some roots do exists, they have to be irrational.

For example, if we have $12x^2+x+2$, all possible rational roots are $$\pm 1/2, \pm 1/3, \pm 1/4, \pm 1/6, \pm 1/12, \pm 2/2, \pm 2/3, \pm 2/4, \pm 2/6, \pm 2/12.$$ We then test to see if any of these are actually roots. If none are roots, then all roots must be irrational.


If the coefficients are rational you can solve any of these problems by one of two ways, the first way is to complete the square, the second way is by using the quadratic formula, which tells you the solutions to $ax^2+bx+c=0$ are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

The solutions are rational if and only if $b^2-4ac$ is a perfect square. This is because the square root of any rational which does not have a perfect square in both the numerator and the denominator in its simplified form is irrational, you can prove this by contradiction, if the root was rational then it could be expressed as $\frac{p}{q}$ with $p$ and $q$coprime, but that would mean the square is of the form $\frac{p^2}{q^2}$ which is a simplified fraction. A contradiction!

  • $\begingroup$ "...is a perfect square of a rational value". It doesn't have to be an integer. $\endgroup$ – DanielV Jun 19 '14 at 21:37
  • $\begingroup$ you're right, I changed it, thanks! $\endgroup$ – Jorge Fernández Hidalgo Jun 19 '14 at 21:55

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