Proposition: Suppose that $a$, $b$, and $c$ are real numbers with $c \not = 0$. Prove, by contradiction, that, if $cx^2 + bx + a$ has no rational root, then $ax^2 + bx + c$ has no rational root.

Hypothesis: $cx^2 + bx + a$ has no rational root where $a$, $b$, and $c$ are real numbers with $c \not = 0$.

Conclusion: $ax^2 + bx + c$ has no rational root

To form a proof by contradiction, we take the negation of the conclusion:

$\neg B$: $ax^2 + bx + c$ has a rational root.

We now have a suitable hypothesis and conclusion for proof by contradiction:

A (Hypothesis): $cx^2 + bx + a$ has no rational root where $a$, $b$, and $c$ are real numbers with $c \not = 0$.

A1: $ax^2 + bx + c$ has a rational root.

Given that this is a proof by contradiction, we can work forward from both the hypothesis and conclusion, as shown above.

My Workings

A2: Let $x = \dfrac{p}{q}$ where $p$ and $q \not = 0$ are integers. This is the definition of a rational number (in this case, $x$): A rational number is any number that can be expressed as the quotient/fraction of two integers.

A3: $a\left(\dfrac{p}{q}\right)^2 + b\left(\dfrac{p}{q}\right) + c = 0$

$\implies \dfrac{ap^2}{q^2} + \dfrac{bp}{q} + c = 0$ where $q \not = 0$.

$\implies ap^2 + bpq + cq^2 = 0$

A4: $ap^2 + bpq + cq^2 = 0$ where $c \not = 0$

$\implies ap^2 + bpq = -cq^2$ where $-cq \not = 0$ since $c \not = 0$ and $q \not = 0$.

A5: $ap^2 + bpq + cq^2 = 0$ where $ap^2 + bpq \not = 0$ and $cq^2 \not = 0$.

But $ap^2 + bpq + cq^2 = 0$? Contradiction. $Q.E.D.$

I would greatly appreciate it if people could please take the time to review my proof and provide feedback on its correctness.

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    $\begingroup$ I would not call A1 a "conclusion". It is another assumption, your $\lnot B$, made for the sake of contradiction. Some words about why $p,q\neq 0$ in A2 would be an improvement. $\endgroup$
    – hardmath
    Feb 11, 2017 at 5:42
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    $\begingroup$ @ThePointer Now I got what you want to say. What you are saying is something like this "$s \neq 0$ and $t \neq 0 \Rightarrow s+t \neq 0$". Did I understand you correctly? $\endgroup$
    – Error 404
    Feb 11, 2017 at 6:05
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    $\begingroup$ @ThePointer Okay.What if I take $s=5$ and $t=-5$. Then what is $s+t=?$ $\endgroup$
    – Error 404
    Feb 11, 2017 at 6:10
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    $\begingroup$ @ThePointer Your A5 inference is wrong. Two rational numbers can be $\ne 0$, yet their sum be $=0$. But you don't need to (and can't) do it that way. Just note that $p \ne 0$ (why?) then divide by $p^2$ to produce a rational root of $c x^2 + b x + a =0$ which is a contradiction.. $\endgroup$
    – dxiv
    Feb 11, 2017 at 6:15
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    $\begingroup$ @ThePointer You are complicating it again (why not stick to my $u,v$ simplest case). But OK, let $a=0, b=1, c=-1$ then they satisfy $a+b+c=0\,$, also $c \ne 0\,$, and all you can derive is that $a+b \ne 0\,$. Where is the contradiction in that? You did not assume nor prove that $a+b=0\,$ at any previous step. $\endgroup$
    – dxiv
    Feb 11, 2017 at 6:36

3 Answers 3


The roots of $ax^2+bx+c = 0 $ are $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} $ and the roots of $cx^2+bx+a = 0 $ are $\dfrac{-b\pm\sqrt{b^2-4ac}}{2c} $.

If the first are rational and the second are not, then their sum and ratio are irrational.

Since the roots of the first equation are rational, their sum and product are rational. These are $\dfrac{b}{a}$ and $\dfrac{c}{a}$.

Since the ratio of the two equations' roots is irrational, $\dfrac{a}{c}$ is irrational.

Since their sum is irrational, $\dfrac{b}{c}$ is irrational.

But these both contradict the previously proven rationality of these two ratios.

Therefore the roots of the second equation are also rational.


To continue what you started

$ap^2 + bpq + cq^2 = 0$ divide both sides by $p^2$.

$a + b\frac qp + c(\frac {q^2}{p^2}) = 0$

So $\frac pc$ is a rational solution to $cx^2 + bx + a = 0$. A contradiction.

Worth noting: if $w$ is a solution to $ax^2 + bx + c = 0$ ($a \ne 0; c\ne 0$) then $\frac 1w$ is a solution to $cx^2 +bx +a=0$. And if $w \ne 0$ then $w$ is rational if and only $\frac 1w $ is rational.


Since you have proved that $ap^2+bpq \neq 0$ and $cq^2 \neq 0$, you are very close to the answer.

Notice that $p(ap+bq)=-cq^2$. But $p(ap+bq) \neq 0 \Rightarrow p \neq 0$.

Divide both sides by $p^2$. Then you get $a+b\frac qp=-c\frac {q^2}{p^2}.$

This shows that $cx^2+bx+a=0$ has a rational root $\frac qp$. WHICH is contradiction.

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    $\begingroup$ Actually $p \ne 0$ follows directly from A2 since $x=0$ is not a root of $a x^2+b x +c =0$ when $c \ne 0$. $\endgroup$
    – dxiv
    Feb 11, 2017 at 6:39
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    $\begingroup$ @dxiv :D awesome!! +1 $\endgroup$
    – Error 404
    Feb 11, 2017 at 6:44

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