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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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    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. – 2017-02-11
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    I read your whole proof. It's really well written. But I did not understand $A_5$. – 2017-02-11
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    @VikrantDesai thanks for the response. From **A4**, we have $ap^2 + bpq = -cq^2$ where $-cq \not = 0$ since $c \not = 0$ and $q \not = 0$. Therefore, $ap^2 + bpq \not = 0$ since $ap^2 + bpq = -cq^2$ where $-cq^2 \not = 0$. Is that more understandable? – 2017-02-11
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    @hardmath thanks for the response. I will make adjustments now. If you could spare the time, any comments on correctness would be greatly appreciated. – 2017-02-11
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    @ThePointer Sorry but I really did not get the point. To get contradiction is to get that the polynomial $cx^2+bx+a$ also has a rational root. So one way would be showing that $\frac qp$ is a rational root of $cx^2+bx+a$. – 2017-02-11
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    @VikrantDesai Notice that, if it was true that $ax^2 + bx + c$ has a rational root, then the rational number $x = \dfrac{p}{q}$, which is a rational number by definition, must be a valid root of the polynomial. However, after substituting it into the equation, we found that $ap^2 + bpq + cq^2 \not = 0$; in other words, there is no rational number $x = \dfrac{p}{q}$ that will make $ap^2 + bpq + cq^2 = 0$ true. – 2017-02-11
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    @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? – 2017-02-11
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    @ThePointer Okay.What if I take $s=5$ and $t=-5$. Then what is $s+t=?$ – 2017-02-11
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    @VikrantDesai In this case, it is $s + t = c$ where $c = 0$; therefore, regardless of the value of $s$ or $t$, $s + t \not = 0$. The corresponding section form the proof is $ap^2 + bpq = -cq^2$ where $-cq \not = 0$ since $c \not = 0$ and $q \not = 0$. – 2017-02-11
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    @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.. – 2017-02-11
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    @dxiv thanks for the response. But the sum of $ap^2 + bpq$ in $ap^2 + bpq = -cq^2$ where $-cq^2 \not = 0$ cannot be $0$? – 2017-02-11
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    @ThePointer All you have at **A4** is that a sum of two terms is $0\,$, where you know one of them to be non-$0$. Then at **A5** you try to somehow derive a contradiction from just that. It doesn't hold, write it step by step using $u=ap^2+bpq$ and $v=cq^2$ to keep it simple. Just because you know that $u+v=0$ and $v \ne 0$ there is no contradiction to be derived from just that. – 2017-02-11
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    @dxiv But in this case, we have $a + b + c = 0$ where $c \not = 0$ ($c = cq^2$) $\implies a + b = -c$ where $c = cq^2 \not = 0$ $\implies a + b = c \not = 0$? – 2017-02-11
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    @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. – 2017-02-11
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    @dxiv Ahh, I understand now. Thank you for taking the time to thoroughly explain this. :) – 2017-02-11
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    How is A5 different from A3 and why is A5 a contradiction? – 2017-02-11
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    @fleablood it isn't. I seemed to have made a subtle (to me) error and had convinced myself that it was correct. I've fixed the proof now. – 2017-02-11
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    Don't see the fix. – 2017-02-11
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    @fleablood I didn't mean by editing the question. Is it not better to leave the original error coupled with the appropriately chosen answer? – 2017-02-11

3 Answers 3

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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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    Actually $p \ne 0$ follows directly from **A2** since $x=0$ is not a root of $a x^2+b x +c =0$ when $c \ne 0$. – 2017-02-11
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    @dxiv :D awesome!! +1 – 2017-02-11
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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.

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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.