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Let ${ x }_{ 1 },{ x }_{ 2 }\in\mathbb{R}$ be the roots of the quadratic equation $ax^2+bx+c=0$ with rational coefficients such that ($a,b,c\in\mathbb{Q}, a\neq 0$). Prove using the Vieta formulas that ${x}_{1}\in\mathbb{Q}\Leftrightarrow{x}_{2}\in\mathbb{Q}$.

My Proof:

$$\frac { -b+\sqrt { b^{ 2 }-4ac } }{ 2a } +\frac { -b-\sqrt { b^{ 2 }-4ac } }{ 2a } =\frac { -2b }{ 2a } =-\frac { b }{ a } $$

$$\Rightarrow { x }_{ 1 }+{ x }_{ 2 }=-\frac { b }{ a } $$

1) ${ x }_{ 1 }\in\mathbb{Q}\Rightarrow { x }_{ 2 }\in\mathbb{Q}$

Assume: ${ x }_{ 2 }\notin\mathbb{Q}$ (by contradiction)

Let: ${ x }_{ 1 }= \frac { m }{ n }$ and ${ x }_{ 2 }=p$ such that $m,n\in\mathbb{Z}$ such that $n\neq0$; $p\notin\mathbb{Q}$

$$\frac { m }{ n } +p=-\frac { b }{ a }$$

$$p=-\frac { b }{ a } -\frac { m }{ n } \Rightarrow p=-1(\frac { bn }{ an } +\frac { ma }{ an } )\Rightarrow p=-\frac { bn+ma }{ an } $$

This is contradictory to $p$ being irrational, so $p$ must be rational; therefore, ${x}_{2}\in\mathbb{Q}$

2) ${ x }_{ 2 }\in\mathbb{Q}\Rightarrow { x }_{ 1 }\in\mathbb{Q}$

Assume: ${ x }_{ 1 }\notin\mathbb{Q}$ (by contradiction)

Let: ${ x }_{ 2 }= \frac { k }{ l }$ such that $k,l\in\mathbb{Z}$ and $l\neq0$; ${x}_{1}=q$ such that $q\notin\mathbb{Q}$

$$q+\frac { k }{ l } =-\frac { b }{ a } $$

$$q=-\frac { b }{ a } -\frac { k }{ l } \Rightarrow q=\frac { bl+ka }{ al } $$

This is contradictory to $q$ being irrational; therefore, $q$ must be rational, so, ${x}_{1}\in\mathbb{Q}$

Is my proof any good? In not, where are its flaws and how can I improve it?

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    $\begingroup$ First note that you were asked to use the Vieta relations. So immediately we have that $x_1+x_2=-\frac{b}{a}$. No computation of the roots. The second direction is essentially the same as the first, so it would be enough to say "similarly, if $x_2$ is rational then $x_1$ is rational." $\endgroup$ – André Nicolas Oct 21 '15 at 21:37
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Your proof is good. It proves what you needed to prove, and it does so without making any logical mistakes.


That said, the proof is overcomplicated.

First of all, you don't need to use a proof by contradiction.

You can simply say, in step $1$, that if $x_1=\frac mn$, then, because $x_1 + x_2 = -\frac ba$, you know that $x_2 = -\frac ba -\frac mn = -\frac{bn + ma}{an}$, and since $\frac{bn + ma}{an}\in \mathbb Q$, you conclude that $x_2\in \mathbb Q$.

Secondly, you don't need the two steps. Basically, your first step did not prove the statement

If the first root is rational, then the second root is rational.

It actually proved the statement

If one of the roots is rational, then the other one is too.


Here is a rule of thumb:

Say I have a statement I want to prove, say something of the shape $A\implies B$.

If I want to prove it by contradiction, I usually assume that $A$ is true, and that $B$ is not true, and then arrive to a contradiction.

However, it is then a good idea to look at your proof again and ask yourself:

Is the proof by contradiction really necesary?

For example, sometimes, your proof looks something like this:

  • Let's assume $A$ is true.
  • Let's assume that $\neg B$ is true.
  • Then, because (somethingsomething), and therefore, $B$ is true.
  • But since $\neg B$ is also true, that is a contradiction.
  • Therefore, $B$ is true.
  • So, $A\implies B$ is true.

Now, if the somethingsomething does not contain any reference to $\neg B$, or can be proven even if $\neg B$ is not true, then basically, the proof can be reduced to: - Let's assume $A$ is true - Then, somethingsomething, and therefore, $B$ is true. - So, $A\implies B$ is true

which is much more elegant.

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  • $\begingroup$ I'm confused by what you are trying to say. At first, you said my proof is good, proves what was asked, and has no logical flaws. Then, it seems that you are saying it didn't prove what needed to be proved. Or do you just mean that it was a very inefficient, inelegant way of proving it? $\endgroup$ – Cherry_Developer Oct 21 '15 at 21:51
  • $\begingroup$ @Cherry_Developer Where, exactly, did I say that it did not prove what needed to be proved? $\endgroup$ – 5xum Oct 21 '15 at 22:01
  • $\begingroup$ @Cherry_Developer I said, and I quote, That said, the proof is overcomplicated.. That does not mean the proof is wrong. It just means that you spent 2 pages to prove something that can be proven in 5 rows, and the 5 row proof is much easier to follow, and also easier to understand. $\endgroup$ – 5xum Oct 21 '15 at 22:03
  • $\begingroup$ @Cherry_Developer Yeah. But that means that it also proved the statement you think it proved. Because the statement you claim was proven is weaker than the statement you actually proved. $\endgroup$ – 5xum Oct 21 '15 at 22:04
  • $\begingroup$ Thank you for clarifying. I'm going to sleep on this and then I will return to turn that monstrosity into a more efficient, and elegant proof. Hopefully, with your guidance. $\endgroup$ – Cherry_Developer Oct 21 '15 at 22:05
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The first part of your proof is that $x_1+x_2=-b/a$ and, from the assignment text, I think you don't need it, because $x_1+x_2=-b/a$ is one of Viète’s formulas.

On the other hand, the argument is correct.

For the main part you're doing too much work.

First note that, since $a,b,c\in\mathbb{Q}$ by assumption, also $-b/a\in\mathbb{Q}$.

Since $x_2=(-b/a)-x_1$ and $x_1=(-b/a)-x_2$, by Viète’s formula,

  • if $x_1\in\mathbb{Q}$, then $x_2=(-b/a)-x_1\in\mathbb{Q}$
  • if $x_2\in\mathbb{Q}$, then $x_1=(-b/a)-x_2\in\mathbb{Q}$

This just uses the arithmetic properties of $\mathbb{Q}$.

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  • $\begingroup$ So just assume the fact that $rational + x = rational$ always means that $x$ will be rational? $\endgroup$ – Cherry_Developer Oct 23 '15 at 10:57
  • $\begingroup$ @Cherry_Developer No, I use the fact that rational+rational=rational. $\endgroup$ – egreg Oct 23 '15 at 10:58
  • $\begingroup$ Isn't that the same thing that I meant? I meant that when you add a number to a rational number, and the result is rational, the number you added must have been rational. $\endgroup$ – Cherry_Developer Oct 23 '15 at 11:04
  • $\begingroup$ @Cherry_Developer I was in a hurry, when typing in my previous comment. Indeed, the two formulations are equivalent, but the direct one is easier to use. $\endgroup$ – egreg Oct 23 '15 at 11:37

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