Prove or disprove that if one root of a quadratic equation is rational, then the other root must be rational as well. I'm taking an introduction to discrete math course and I'm having some trouble with this homework problem. I think we're supposed to assume that the coefficients are integers based on other examples we've been given and if they are I think I understand how to prove this "informally" as if I have the equation:
$$ ax^2 + bx + c = 0 $$
Then I can say 
$$ ax^2 + bx + c = (px - r)(qx - s) $$
so it seems like $b = - qr - ps$ and so:
$$
r = \frac{-b -ps}{q}\\
\text{and} \\
s = \frac{-b -qr}{p}
$$
So if either r or s are rational then both must be rational. So my question is, is my proof even valid and secondly, how could I write this more "formally"? We're supposed to write our proof in this style (this is unrelated to this problem, I'm just showing this example of style):
$$
\textbf{Theorem:}\\
\forall m,n \in \mathbb{Z}, (m ~\equiv~ 0(mod ~2)) ~\wedge~ (n ~\equiv~ 0(mod ~2)) ~\rightarrow~ (m + n) ~\equiv~ 0(mod ~2)\\
\begin{array}{| c || l | l |} \hline
 \textbf{Step} & \textbf{Derivation} & \textbf{Justification} \\ \hline
 1 & m,n ~\in~ \mathbb{Z} ~\equiv~ 0(mod ~2) & \text{Select generic particulars} \\ \hline
 2 & \exists r ~\in~ \mathbb{Z} ~\mid~ m = 2 \cdot r & \text{Definition of} \equiv ~0(mod ~2) \\ \hline
 3 & \exists s ~\in~ \mathbb{Z} ~\mid~ n = 2 \cdot s & \text{Definition of} \equiv ~0(mod ~2) \\ \hline
 4 & m + n ~=~ 2 \cdot r ~+~ 2 \cdot s & \text{Substitution} \\ \hline
 5 & m + n ~=~ 2 \cdot (r + s) & \text{Distributive property} \\ \hline
 6 & 2 \cdot (r + s) ~\in~ \mathbb{Z} & \text{Closure of } \mathbb{Z} \text{ on} \cdot and + \\ \hline
 7 & \therefore m + n ~\equiv~ 0(mod ~2)$ & \text{Definition of} \equiv ~0(mod ~2) \\ \hline
\end{array}
$$
I really have no idea how to even state the theorem in formal terms. All the other problems are stated more formally and I guess I don't get how to write "for the two possible answers of a quadratic equation" as part of a formal universal statement. The furthest I've come so far is:
$$ \forall a,b,c \in \mathbb{Z}, x \in \mathbb{R}, ax^2 + bx + c = 0 \land \text{one of the roots } \in \mathbb{Q} \rightarrow \text{the other root} \in \mathbb{Q} $$
But I need to replace the English text with the actual correct symbols. I'm also not completely sure how to write the justifications for what I did above, it's been 10 years since I've taken any sort of math class and I don't really remember the names for a lot of stuff.
 A: If you want it simple, my suggestion is to proceed in the reverse direction. That is, start with the two roots r,s of your polynomial and derive the coefficients from there.
Alternatively, a formalization of your claim could be:
$$\forall a, b, c \in \mathbb{Z} \left (  \exists x \in \mathbb{Q} \ \ ax^2 + bx + c = 0 \Rightarrow \left ( \forall y \in \mathbb{R} \ \ ay^2 + by + c = 0 \Rightarrow y \in \mathbb{Q} \right ) \right )$$
Your starting point seems reasonable, but very incomplete.
A: If I understand your problem the product of the two roots is the integer $c$. If $c \neq 0$ and you know one root is rational can you figure out the other root? Suppose that $c = 0$. What does that tell you about one of the roots? Take note of André Nicolas' comment.
A: Fact:- Let p + √q be an irrational. Then its conjugate (p – √q) is also an irrational.
Lemma:- If p + √q is a root of ($ax^2 + bx + c = 0$), the its conjugate (p – √q) is also a root.---[Proof can be provided on request.]
An alternate form of your statement is:-
Prove that the statement "If one root is rational, the other can be irrational." is NOT true.
From the lemma, the conjugate of that irrational root must also be a root. Thus both roots are now irrational. This contradicts the assumption.
