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.
Best Answer
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.